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Mathematics LibreTexts

13.4: Right Triangle Trigonometry

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  • Page ID 40841

Learning Objectives

  • Use right triangles to evaluate trigonometric functions.
  • Find function values for 30°(\(\dfrac{\pi}{6}\)),45°(\(\dfrac{\pi}{4}\)),and 60°(\(\dfrac{\pi}{3}\)).
  • Use equal cofunctions of complementary angles.
  • Use the definitions of trigonometric functions of any angle.
  • Use right-triangle trigonometry to solve applied problems.

Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task and, in fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles and a branch of mathematics known as trigonometry. In this section, we will define a new group of functions known as trigonometric functions, and find out how they can be used to measure heights, such as those of the tallest mountains.

We have previously defined the sine and cosine of an angle in terms of the coordinates of a point on the unit circle intersected by the terminal side of the angle:

\[ \begin{align*} \cos t &= x \\ \sin t &=y \end{align*} \]

In this section, we will see another way to define trigonometric functions using properties of right triangles .

Using Right Triangles to Evaluate Trigonometric Functions

In earlier sections, we used a unit circle to define the trigonometric functions . In this section, we will extend those definitions so that we can apply them to right triangles. The value of the sine or cosine function of \(t\) is its value at \(t\) radians. First, we need to create our right triangle. Figure \(\PageIndex{1}\) shows a point on a unit circle of radius 1. If we drop a vertical line segment from the point \((x,y)\) to the x -axis, we have a right triangle whose vertical side has length \(y\) and whose horizontal side has length \(x\). We can use this right triangle to redefine sine, cosine, and the other trigonometric functions as ratios of the sides of a right triangle.

Graph of quarter circle with radius of 1. Inscribed triangle with an angle of t. Point of (x,y) is at intersection of terminal side of angle and edge of circle.

\[ \cos t= \frac{x}{1}=x \]

Likewise, we know

\[ \sin t= \frac{y}{1}=y \]

These ratios still apply to the sides of a right triangle when no unit circle is involved and when the triangle is not in standard position and is not being graphed using \((x,y)\) coordinates. To be able to use these ratios freely, we will give the sides more general names: Instead of \(x\),we will call the side between the given angle and the right angle the adjacent side to angle \(t\). (Adjacent means “next to.”) Instead of \(y\),we will call the side most distant from the given angle the opposite side from angle \(t\). And instead of \(1\),we will call the side of a right triangle opposite the right angle the hypotenuse . These sides are labeled in Figure \(\PageIndex{2}\).

A right triangle with hypotenuse, opposite, and adjacent sides labeled.

Understanding Right Triangle Relationships

Given a right triangle with an acute angle of \(t\),

\[\begin{align} \sin (t) &= \dfrac{\text{opposite}}{\text{hypotenuse}} \label{sindef}\\ \cos (t) &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \label{cosdef}\\ \tan (t) &= \dfrac{\text{opposite}}{\text{adjacent}} \label{tandef}\end{align}\]

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “ S ine is o pposite over h ypotenuse, C osine is a djacent over h ypotenuse, T angent is o pposite over a djacent.”

how to: Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle

  • Find the sine as the ratio of the opposite side to the hypotenuse.
  • Find the cosine as the ratio of the adjacent side to the hypotenuse.
  • Find the tangent is the ratio of the opposite side to the adjacent side.

Example \(\PageIndex{1}\): Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure \(\PageIndex{3}\), find the value of \(\cos α\).

A right triangle with side lengths of 8, 15, and 17. Angle alpha also labeled which is opposite to the side labeled 8.

The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17, so via Equation \ref{cosdef}:

\[\begin{align*} \cos (α) &= \dfrac{\text{adjacent}}{\text{hypotenuse}} \\[4pt] &= \dfrac{15}{17} \end{align*}\]

Exercise \(\PageIndex{1}\)

Given the triangle shown in Figure \(\PageIndex{4}\), find the value of \(\sin t\).

A right triangle with sides of 7, 24, and 25. Also labeled is angle t which is opposite the side labeled 7.

\(\frac{7}{25}\)

Relating Angles and Their Functions

When working with right triangles, the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure \(\PageIndex{5}\). The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

Right triangle with angles alpha and beta. Sides are labeled hypotenuse, adjacent to alpha/opposite to beta, and adjacent to beta/opposite alpha.

We will be asked to find all six trigonometric functions for a given angle in a triangle. Our strategy is to find the sine, cosine, and tangent of the angles first. Then, we can find the other trigonometric functions easily because we know that the reciprocal of sine is cosecant, the reciprocal of cosine is secant, and the reciprocal of tangent is cotangent.

how to: Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles

  • If needed, draw the right triangle and label the angle provided.
  • Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
  • sine as the ratio of the opposite side to the hypotenuse
  • cosine as the ratio of the adjacent side to the hypotenuse
  • tangent as the ratio of the opposite side to the adjacent side
  • secant as the ratio of the hypotenuse to the adjacent side
  • cosecant as the ratio of the hypotenuse to the opposite side
  • cotangent as the ratio of the adjacent side to the opposite side

Example \(\PageIndex{2}\): Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure \(\PageIndex{6}\), evaluate \( \sin α, \cos α, \tan α, \sec α, \csc α,\) and \( \cot α\).

Right triangle with sides of 3, 4, and 5. Angle alpha is also labeled which is opposite the side labeled 4.

\[ \begin{align*} \sin α &= \dfrac{\text{opposite } α}{\text{hypotenuse}} = \dfrac{4}{5} \\ \cos α &= \dfrac{\text{adjacent to }α}{\text{hypotenuse}}=\dfrac{3}{5} \\ \tan α &= \dfrac{\text{opposite }α}{\text{adjacent to }α}=\dfrac{4}{3} \\ \sec α &= \dfrac{\text{hypotenuse}}{\text{adjacent to }α}= \dfrac{5}{3} \\ \csc α &= \dfrac{\text{hypotenuse}}{\text{opposite }α}=\dfrac{5}{4} \\ \cot α &= \dfrac{\text{adjacent to }α}{\text{opposite }α}=\dfrac{3}{4} \end{align*}\]

Exercise \(\PageIndex{2}\)

Using the triangle shown in Figure \(\PageIndex{7}\), evaluate \( \sin t, \cos t,\tan t, \sec t, \csc t,\) and \(\cot t\).

Right triangle with sides 33, 56, and 65. Angle t is also labeled which is opposite to the side labeled 33.

\[\begin{align*} \sin t &= \frac{33}{65}, \cos t= \frac{56}{65},\tan t= \frac{33}{56}, \\ \\ \sec t &= \frac{65}{56},\csc t= \frac{65}{33},\cot t= \frac{56}{33} \end{align*}\]

Finding Trigonometric Functions of Special Angles Using Side Lengths

We have already discussed the trigonometric functions as they relate to the special angles on the unit circle. Now, we can use those relationships to evaluate triangles that contain those special angles. We do this because when we evaluate the special angles in trigonometric functions, they have relatively friendly values, values that contain either no or just one square root in the ratio. Therefore, these are the angles often used in math and science problems. We will use multiples of \(30°, 60°,\) and \(45°\), however, remember that when dealing with right triangles, we are limited to angles between \(0° \text{ and } 90°\).

Suppose we have a \(30°,60°,90°\) triangle, which can also be described as a \(\frac{π}{6}, \frac{π}{3},\frac{π}{2}\) triangle. The sides have lengths in the relation \(s,\sqrt{3}s,2s.\) The sides of a \(45°,45°,90° \)triangle, which can also be described as a \(\frac{π}{4},\frac{π}{4},\frac{π}{2}\) triangle, have lengths in the relation \(s,s,\sqrt{2}s.\) These relations are shown in Figure \(\PageIndex{8}\).

Two side-by-side graphs of circles with inscribed angles. First circle has angle of pi/3 inscribed, radius of 2s, base of length s and height of length . Second circle has angle of pi/4 inscribed with radius , base of length s and height of length s.

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

Given trigonometric functions of a special angle, evaluate using side lengths.

  • Use the side lengths shown in Figure \(\PageIndex{8}\) for the special angle you wish to evaluate.
  • Use the ratio of side lengths appropriate to the function you wish to evaluate.

Example \(\PageIndex{3}\): Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of \(\frac{π}{3}\), using side lengths.

\[\begin{align*} \sin (\dfrac{π}{3}) &= \dfrac{\text{opp}}{\text{hyp}}=\dfrac{\sqrt{3}s}{2s}=\dfrac{\sqrt{3}}{2} \\ \cos (\dfrac{π}{3}) &= \dfrac{\text{adj}}{\text{hyp}}=\dfrac{s}{2s}=\dfrac{1}{2} \\ \tan (\dfrac{π}{3}) &= \dfrac{\text{opp}}{\text{adj}} =\dfrac{\sqrt{3}s}{s}=\sqrt{3} \\ \sec (\dfrac{π}{3}) &= \dfrac{\text{hyp}}{\text{adj}} = \dfrac{2s}{s}=2 \\ \csc (\dfrac{π}{3}) &= \dfrac{\text{hyp}}{\text{opp}} =\dfrac{2s}{\sqrt{3}s}=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3} \\ \cot (\dfrac{π}{3}) &= \dfrac{\text{adj}}{\text{opp}}=\dfrac{s}{\sqrt{3}s}=\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{3}}{3} \end{align*}\]

Exercise \(\PageIndex{3}\)

Find the exact value of the trigonometric functions of \(\frac{π}{4}\) using side lengths.

\( \sin (\frac{π}{4})=\frac{\sqrt{2}}{2}, \cos (\frac{π}{4})=\frac{\sqrt{2}}{2}, \tan (\frac{π}{4})=1,\)

\( \sec (\frac{π}{4})=\sqrt{2}, \csc (\frac{π}{4})=\sqrt{2}, \cot (\frac{π}{4}) =1 \)

Using Equal Cofunction of Complements

If we look more closely at the relationship between the sine and cosine of the special angles relative to the unit circle, we will notice a pattern. In a right triangle with angles of \(\frac{π}{6}\) and \(\frac{π}{3}\), we see that the sine of \(\frac{π}{3}\), namely \(\frac{\sqrt{3}}{2}\), is also the cosine of \(\frac{π}{6}\), while the sine of \(\frac{π}{6}\), namely \(\frac{1}{2},\) is also the cosine of \(\frac{π}{3}\) (Figure \(\PageIndex{9}\)).

\[\begin{align*} \sin \frac{π}{3} &= \cos \frac{π}{6}=\frac{\sqrt{3}s}{2s}=\frac{\sqrt{3}}{2} \\ \sin \frac{π}{6} &= \cos \frac{π}{3}=\frac{s}{2s}=\frac{1}{2} \end{align*}\]

A graph of circle with angle pi/3 inscribed with a radius of 2s, a base with length s and a height of.

This result should not be surprising because, as we see from Figure \(\PageIndex{9}\), the side opposite the angle of \(\frac{π}{3}\) is also the side adjacent to \(\frac{π}{6}\), so \(\sin (\frac{π}{3})\) and \(\cos (\frac{π}{6})\) are exactly the same ratio of the same two sides, \(\sqrt{3} s\) and \(2s.\) Similarly, \( \cos (\frac{π}{3})\) and \( \sin (\frac{π}{6})\) are also the same ratio using the same two sides, \(s\) and \(2s\).

The interrelationship between the sines and cosines of \(\frac{π}{6}\) and \(\frac{π}{3}\) also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π, π,and the right angle is \(\frac{π}{2}\), the remaining two angles must also add up to \(\frac{π}{2}\). That means that a right triangle can be formed with any two angles that add to \(\frac{π}{2}\)—in other words, any two complementary angles. So we may state a cofunction identity : If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure \(\PageIndex{10}\).

Right triangle with angles alpha and beta. Equivalence between sin alpha and cos beta. Equivalence between sin beta and cos alpha.

Using this identity, we can state without calculating, for instance, that the sine of \(\frac{π}{12}\) equals the cosine of \(\frac{5π}{12}\), and that the sine of \(\frac{5π}{12}\) equals the cosine of \(\frac{π}{12}\). We can also state that if, for a certain angle \(t, \cos t= \frac{5}{13},\) then \( \sin (\frac{π}{2}−t)=\frac{5}{13}\) as well.

COFUNCTION IDENTITIES

The cofunction identities in radians are listed in Table \(\PageIndex{1}\).

how to: Given the sine and cosine of an angle, find the sine or cosine of its complement.

  • To find the sine of the complementary angle, find the cosine of the original angle.
  • To find the cosine of the complementary angle, find the sine of the original angle.

Example \(\PageIndex{4}\): Using Cofunction Identities

If \( \sin t = \frac{5}{12},\) find \(( \cos \frac{π}{2}−t)\).

According to the cofunction identities for sine and cosine,

\[ \sin t= \cos (\dfrac{π}{2}−t). \nonumber\]

\[ \cos (\dfrac{π}{2}−t)= \dfrac{5}{12}. \nonumber\]

Exercise \(\PageIndex{4}\)

If \(\csc (\frac{π}{6})=2,\) find \( \sec (\frac{π}{3}).\)

Using Trigonometric Functions

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

how to: Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides

  • For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
  • Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
  • Using the value of the trigonometric function and the known side length, solve for the missing side length.

Example \(\PageIndex{5}\): Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure \(\PageIndex{11}\).

A right triangle with sides a, c, and 7. Angle of 30 degrees is also labeled which is opposite the side labeled 7.

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

\[ \tan (30°)= \dfrac{7}{a} \nonumber\]

We rearrange to solve for \(a\).

\[\begin{align} a &=\dfrac{7}{ \tan (30°)} \\ & =12.1 \end{align} \nonumber\]

We can use the sine to find the hypotenuse.

\[ \sin (30°)= \dfrac{7}{c} \nonumber\]

Again, we rearrange to solve for \(c\).

\[\begin{align*} c &= \dfrac{7}{\sin (30°)} =14 \end{align*}\]

Exercise \(\PageIndex{5}\):

A right triangle has one angle of \(\frac{π}{3}\) and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

\(\mathrm{adjacent=10; opposite=10 \sqrt{3}; }\) missing angle is \(\frac{π}{6}\)

Using Right Triangle Trigonometry to Solve Applied Problems

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height. Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure \(\PageIndex{12}\).

Diagram of a radio tower with line segments extending from the top and base of the tower to a point on the ground some distance away. The two lines and the tower form a right triangle. The angle near the top of the tower is the angle of depression. The angle on the ground at a distance from the tower is the angle of elevation.

how to: Given a tall object, measure its height indirectly

  • Make a sketch of the problem situation to keep track of known and unknown information.
  • Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
  • At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
  • Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
  • Solve the equation for the unknown height.

Example \(\PageIndex{6}\): Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57° 57° between a line of sight to the top of the tree and the ground, as shown in Figure \(\PageIndex{13}\). Find the height of the tree.

A tree with angle of 57 degrees from vantage point. Vantage point is 30 feet from tree.

We know that the angle of elevation is \(57°\) and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of \(57°\), letting \(h\) be the unknown height.

\[\begin{array}{cl} \tan θ = \dfrac{\text{opposite}}{\text{adjacent}} & \text{} \\ \tan (57°) = \dfrac{h}{30} & \text{Solve for }h. \\ h=30 \tan (57°) & \text{Multiply.} \\ h≈46.2 & \text{Use a calculator.} \end{array} \]

The tree is approximately 46 feet tall.

Exercise \(\PageIndex{6}\):

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of \(\frac{5π}{12}\) with the ground? Round to the nearest foot.

About 52 ft

Access these online resources for additional instruction and practice with right triangle trigonometry.

  • Finding Trig Functions on Calculator
  • Finding Trig Functions Using a Right Triangle
  • Relate Trig Functions to Sides of a Right Triangle
  • Determine Six Trig Functions from a Triangle
  • Determine Length of Right Triangle Side

Visit this website for additional practice questions from Learningpod.

Key Equations

Cofunction Identities

Key Concepts

  • We can define trigonometric functions as ratios of the side lengths of a right triangle. See Example .
  • The same side lengths can be used to evaluate the trigonometric functions of either acute angle in a right triangle. See Example .
  • We can evaluate the trigonometric functions of special angles, knowing the side lengths of the triangles in which they occur. See Example .
  • Any two complementary angles could be the two acute angles of a right triangle.
  • If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa. See Example .
  • We can use trigonometric functions of an angle to find unknown side lengths.
  • Select the trigonometric function representing the ratio of the unknown side to the known side. See Example .
  • Right-triangle trigonometry permits the measurement of inaccessible heights and distances.
  • The unknown height or distance can be found by creating a right triangle in which the unknown height or distance is one of the sides, and another side and angle are known. See Example .
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Problem Solving With Trigonometry Lesson 13 4

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Trigonometry

Unit 1: right triangles & trigonometry, unit 2: trigonometric functions, unit 3: non-right triangles & trigonometry, unit 4: trigonometric equations and identities, review articles.

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Trigonometry : Solving Word Problems with Trigonometry

Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : solving word problems with trigonometry.

lesson 13 4 problem solving with trigonometry

You can draw the following right triangle using the information given by the question:

1

Since you want to find the height of the platform, you will need to use tangent.

lesson 13 4 problem solving with trigonometry

You can draw the following right triangle from the information given by the question.

2

In order to find the height of the flagpole, you will need to use tangent.

lesson 13 4 problem solving with trigonometry

You can draw the following right triangle from the information given in the question:

3

In order to find out how far up the ladder goes, you will need to use sine.

lesson 13 4 problem solving with trigonometry

Example Question #4 : Solving Word Problems With Trigonometry

In right triangle ABC, where angle A measures 90 degrees, side AB measures 15 and side AC measures 36, what is the length of side BC?

lesson 13 4 problem solving with trigonometry

This triangle cannot exist.

lesson 13 4 problem solving with trigonometry

Example Question #5 : Solving Word Problems With Trigonometry

A support wire is anchored 10 meters up from the base of a flagpole, and the wire makes a 25 o angle with the ground. How long is the wire, w? Round your answer to two decimal places.

23.81 meters

lesson 13 4 problem solving with trigonometry

28.31 meters

21.83 meters

To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25 o , the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w. 

Screen shot 2020 07 13 at 12.54.08 pm

Now, we just need to solve for w using the information given in the diagram. We need to ask ourselves which parts of a triangle 10 and w are relative to our known angle of 25 o . 10 is opposite this angle, and w is the hypotenuse. Now, ask yourself which trig function(s) relate opposite and hypotenuse. There are two correct options: sine and cosecant. Using sine is probably the most common, but both options are detailed below.

We know that sine of a given angle is equal to the opposite divided by the hypotenuse, and cosecant of an angle is equal to the hypotenuse divided by the opposite (just the reciprocal of the sine function). Therefore:

lesson 13 4 problem solving with trigonometry

To solve this problem instead using the cosecant function, we would get:

lesson 13 4 problem solving with trigonometry

The reason that we got 23.7 here and 23.81 above is due to differences in rounding in the middle of the problem. 

lesson 13 4 problem solving with trigonometry

Example Question #6 : Solving Word Problems With Trigonometry

When the sun is 22 o above the horizon, how long is the shadow cast by a building that is 60 meters high?

To solve this problem, first set up a diagram that shows all of the info given in the problem. 

Screen shot 2020 07 13 at 1.38.59 pm

Next, we need to interpret which side length corresponds to the shadow of the building, which is what the problem is asking us to find. Is it the hypotenuse, or the base of the triangle? Think about when you look at a shadow. When you see a shadow, you are seeing it on something else, like the ground, the sidewalk, or another object. We see the shadow on the ground, which corresponds to the base of our triangle, so that is what we'll be solving for. We'll call this base b.

lesson 13 4 problem solving with trigonometry

Therefore the shadow cast by the building is 150 meters long.

If you got one of the incorrect answers, you may have used sine or cosine instead of tangent, or you may have used the tangent function but inverted the fraction (adjacent over opposite instead of opposite over adjacent.)

Example Question #7 : Solving Word Problems With Trigonometry

From the top of a lighthouse that sits 105 meters above the sea, the angle of depression of a boat is 19 o . How far from the boat is the top of the lighthouse?

423.18 meters

318.18 meters

36.15 meters

110.53 meters

To solve this problem, we need to create a diagram, but in order to create that diagram, we need to understand the vocabulary that is being used in this question. The following diagram clarifies the difference between an angle of depression (an angle that looks downward; relevant to our problem) and the angle of elevation (an angle that looks upward; relevant to other problems, but not this specific one.) Imagine that the top of the blue altitude line is the top of the lighthouse, the green line labelled GroundHorizon is sea level, and point B is where the boat is.

Screen shot 2020 07 13 at 3.07.05 pm

Merging together the given info and this diagram, we know that the angle of depression is 19 o  and and the altitude (blue line) is 105 meters. While the blue line is drawn on the left hand side in the diagram, we can assume is it is the same as the right hand side. Next, we need to think of the trig function that relates the given angle, the given side, and the side we want to solve for. The altitude or blue line is opposite the known angle, and we want to find the distance between the boat (point B) and the top of the lighthouse. That means that we want to determine the length of the hypotenuse, or red line labelled SlantRange. The sine function relates opposite and hypotenuse, so we'll use that here. We get:

lesson 13 4 problem solving with trigonometry

Example Question #8 : Solving Word Problems With Trigonometry

Angelina just got a new car, and she wants to ride it to the top of a mountain and visit a lookout point. If she drives 4000 meters along a road that is inclined 22 o to the horizontal, how high above her starting point is she when she arrives at the lookout?

9.37 meters

1480 meters

3708.74 meters

10677.87 meters

1616.1 meters

As with other trig problems, begin with a sketch of a diagram of the given and sought after information.

Screen shot 2020 07 13 at 5.37.06 pm

Angelina and her car start at the bottom left of the diagram. The road she is driving on is the hypotenuse of our triangle, and the angle of the road relative to flat ground is 22 o . Because we want to find the change in height (also called elevation), we want to determine the difference between her ending and starting heights, which is labelled x in the diagram. Next, consider which trig function relates together an angle and the sides opposite and hypotenuse relative to it; the correct one is sine. Then, set up:

lesson 13 4 problem solving with trigonometry

Therefore the change in height between Angelina's starting and ending points is 1480 meters. 

Example Question #9 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 50 feet apart. The shorter building is 40 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 48 o . How high is the taller building?

To solve this problem, let's start by drawing a diagram of the two buildings, the distance in between them, and the angle between the tops of the two buildings. Then, label in the given lengths and angle. 

Screen shot 2020 07 13 at 5.56.45 pm

Example Question #10 : Solving Word Problems With Trigonometry

Two buildings with flat roofs are 80 feet apart. The shorter building is 55 feet tall. From the roof of the shorter building, the angle of elevation to the edge of the taller building is 32 o . How high is the taller building?

Screen shot 2020 07 13 at 5.58.09 pm

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  1. PDF 13.4 Problem Solving with Trigonometry

    Explain. Suppose you used a trigonometric ratio in terms of ∠B, h, and a different side length. How would this change your findings? What does this tell you about the choice of sides and included angle? Explain 1 Using the Area Formula Area Formula for a Triangle in Terms of its Side Lengths

  2. PDF LESSON Problem Solving with Trigonometry 13-4 Practice and Problem

    Problem Solving with Trigonometry Practice and Problem Solving: Modified If you know the measures of two sides of a right triangle or one side and one acute angle, you can figure out the other measures. Use trigonometric ratios and the Pythagorean Theorem to find the ... LESSON 13-4 m ∠J = 90° − 32° = ...

  3. PDF LESSON Problem Solving with Trigonometry 13-4 Practice and Problem

    Problem Solving with Trigonometry Practice and Problem Solving: A/B Use a calculator and inverse trigonometric ratios to find the unknown side lengths and angle measures. Round lengths to the nearest hundredth and angle measures to the nearest degree. ... LESSON 13-4 . Author: Dhirendra Singh Created Date: 3/18/2018 3:18:10 PM ...

  4. PDF LESSON Problem Solving with Trigonometry 13-4 Practice and Problem

    Problem Solving with Trigonometry Practice and Problem Solving: A/B Use a calculator and inverse trigonometric ratios to find the unknown ... LESSON 13-4 Practice and Problem Solving: A/B 1. 3 yd; 37°; 53° 2. 13.99 ft; 11.33 ft; 39° 3. 8 km; 62°; 28° 4. 5. 3; 5; 5.83

  5. Module 13 : Trigonometry with Right Triangles Flashcards

    Learn Test Match Q-Chat Ace-The-Test2 Top creator on Quizlet Lesson 13.1 : Tangent Lesson 13.2 : Sine and Cosine Ratios Lesson 13.3 : Special Right Triangles Lesson 13.4 : Problem Solving with Trigonometry Terms in this set (16) tangent opposite/adjacent Given an acute angle ∠A, if tan A = x, then tan⁻¹ x = m∠A trigonometric ratio

  6. PDF 13_4 Problem Solving with Trigonometry 2018-19

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  7. PDF LESSON Problem Solving with Trigonometry 13-4 Reteach

    Problem Solving with Trigonometry Reteach A triangle is solved when the measures of all three angles and all three sides are known. We can solve a right triangle using trigonometry ratios and the Pythagorean Theorem. Using the given information in the figure, we can solve the triangle by finding DF and the measures of all three angles. 4 cos ...

  8. 13.4: Right Triangle Trigonometry

    Figure 13.4.9: The sine of π 3 equals the cosine of π 6 and vice versa. This result should not be surprising because, as we see from Figure 13.4.9, the side opposite the angle of π 3 is also the side adjacent to π 6, so sin( π 3) and cos( π 6) are exactly the same ratio of the same two sides, √3s and 2s.

  9. PDF CHAPTER Solutions Key 13 Trigonometric Functions

    13-1 RIGHT-ANGLE TRIGONOMETRY, ... PRACTICE AND PROBLEM SOLVING 13. sin ... 4 = 1 24. Beginner: about 13 m; intermediate: about 27 m; advanced: about 52 m; possible answer: for each category, I estimated the typical slope angle by finding the mean of the least and greatest angle

  10. Geometry Lesson 13.4 Problem Solving with Trigonometry

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  11. Lesson 13.4 Problem Solving With Trigonometry

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  12. Right triangles & trigonometry

    Trigonometry 4 units · 36 skills. Unit 1 Right triangles & trigonometry. Unit 2 Trigonometric functions. Unit 3 Non-right triangles & trigonometry. Unit 4 Trigonometric equations and identities. Course challenge. Test your knowledge of the skills in this course. Start Course challenge. Math.

  13. PDF Chapter 13: Trigonometric Functions

    698C Chapter 13 Trigonometric Functions Mathematical Connections and Background Right Triangle Trigonometry To solve many of the problems in this lesson, stu-dents will need to remember the Pythagorean Theorem, which states that the sum of the squares of the legs of a right triangle equals the square of the hypotenuse. The

  14. PDF LESSON Problem Solving with Trigonometry 13-4 Practice and Problem

    Problem Solving with Trigonometry Practice and Problem Solving: A/B Use a calculator and inverse trigonometric ratios to find the unknown side lengths and angle measures. Round lengths to the nearest hundredth and angle measures to the nearest degree. 1. 2. 3. AC = ___________ DE = ___________ m∠B = ___________ EF = ___________

  15. Problem Solving With Trigonometry Lesson 13 4

    1. Lesson 13: Proof of the Pythagorean Theorem 2. LESSON Problem Solving with Trignometry 13-4 Practice and ... 3. Chapter 13: Trigonometric Functions 4. Right Triangle Trig Missing Sides and Angles Loading… 5. Prerequisites Trigonometry Practice 6. Grade 11 Mathematics Practice Test 7. Key Vocabulary Lessons 7-1, 7-2, and 7-3 Lessons 7-4 and 7 ...

  16. Trigonometry

    Trigonometry 4 units · 36 skills. Unit 1 Right triangles & trigonometry. Unit 2 Trigonometric functions. Unit 3 Non-right triangles & trigonometry. Unit 4 Trigonometric equations and identities. Course challenge. Test your knowledge of the skills in this course. Start Course challenge. Math.

  17. Lesson 13 4

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  18. Mathway

    Free math problem solver answers your trigonometry homework questions with step-by-step explanations.

  19. PDF LESSON Problem Solving with Trigonometry 13-4

    trigonometry to find the area. For Problems 4 7, follow the steps to derive an area formula, and then apply the formula to find the areas. 4. If you know AB and the measure of A, you can find the height of the triangle. Write a trigonometric equation to relate A, h, and c. _____ 5. Solve for h. h _____ Substitute your value for h into

  20. Solving Word Problems with Trigonometry

    Possible Answers: Correct answer: 23.81 meters. To make sense of the problem, start by drawing a diagram. Label the angle of elevation as 25, the height between the ground and where the wire hits the flagpole as 10 meters, and our unknown, the length of the wire, as w. Now, we just need to solve for w using the information given in the diagram.

  21. PDF CHAPTER 13 and Trigonometry

    theorems, and examples for each lesson. Stack sheets of paper with edges 1 4 inch apart. Fold up bottom edges. All tabs should be the same size. Crease and staple along fold. Turn and label the tabs with the lesson names. 13-1 Simplifying Square Roots 13-2 45º -45º -90º Triangles 13-3 30º -60º -90º Triangles 13-4 Tangent Ratios 13-5 Sine ...

  22. problem solving with trigonometry lesson 13 4

    In this lesson, we will apply prior knowledge of bearings and trigonometry to solve problems. Before we start this lesson, let's see what you can remember from this topic. These slides will take you through some tasks for the lesson... STEPS in solving problems involving RIGHT TRIANGLES. STEP 3 Substitute the given values to the variables in the formula and solve for the unknown.

  23. LESSON 13 4 PROBLEM SOLVING WITH TRIGONOMETRY.docx

    X ( 4, 5), Y ( 3, 5), Z ( 3, 4)_ Practice and Problem Solving: C (LESSON 13.4) For Problems 1-6, use trigonometry and the Pythagorean theorem to solve the right triangles on the coordinate plane. Show your work. 1. First use the slope formula to verify that ∆ ABC is a right triangle. _____ 2.