Advanced Trigonometry Review Study Guide

Updated on Oct 2, 2011

Word Problems

At this point, we are now able to solve just about any problem that involves a triangle. The main trick is to draw the triangle and include all the information that is given. The word problems in this lesson will require knowledge and formulas from throughout this book.

Example 1

A car is driven down a straight desert highway. When the trip begins, a mesa off in the distance is 20° to the right. After the car has been driven for 45 miles, the mesa appears to be 50° to the right. At this point, how far away is the mesa?

All of our angles are relative to the direction of the highway, so we can draw it running from left to right as in Figure 20.1.

Figure 20.1

Thus, our task is to find the length x of the triangle in Figure 20.2.

Figure 20.2

This can be solved by the Law of Sines as soon as we calculate the measure of the angle opposite the 45-mile side. This angle is 180° – 20° – 130° = 30°. Thus, we have

Thus, the mesa is about 30.8 miles away from the car after the 45 miles have been driven.

Example 2

A spy watches the enemy launch a rocket straight upward. The spy is hidden 2 miles from the launch site. When the rocket first comes into sight, it is 37° above the horizontal. Two seconds later, it is 43° above the horizontal. How far did the rocket travel in those two seconds?

Figure 20.3 illustrates this problem.

Figure 20.3

We have two right triangles, and thus, we can use basic trigonometric functions. The height h of the rocket when it was 37° above the horizontal can be calculated as

h = 2 · tan(37°) ≈ 1.5 miles

The height of the rocket when it was 43° above the horizontal is thus h + x, which is calculated by

x + h = 2 · tan(43°) ≈ 1.87 miles


x = (x + h) – h ≈ 1.87 – 1.5 = 0.37 miles

In those two seconds, the rocket traveled 0.37 miles, or about 1,954 feet.

Example 3

Suppose that everyone in town knows that a 30-foot-tall steeple has been built on top of a particularly large church. When you go to look at it, you notice that the tip of the steeple is 38° above the horizontal while the base of the steeple is 32° above the horizontal. How tall is the church together with the steeple?

Figure 20.4 illustrates the given information. We need to find the length x and then add the 30 feet for the steeple.

Figure 20.4

We will need to use the distance D to the church, which forms the base of the two right triangles in the figure. The first one has height x and angle 32°, thus

The larger right triangle has a height of x + 30 and an angle of 38°, thus

If we put these two equations together, we get

With a calculator, we estimate that , thus,

1.25x = x + 30
0.25x = 30

Thus, the church is 120 + 30 = 150 feet tall with the steeple.

Example 4

What is the area of a pentagon with all sides 4 inches in length and all angles the same (a regular pentagon)?

With two diagonal lines, the pentagon can be split into three triangles, as shown in Figure 20.5. Each of these triangles has a sum of 180° worth of angles. Thus, the total sum of the angles of the pentagon is 3 · 180° = 540°. As all five of the angles are the same, each one measures .

Figure 20.5

We could use trigonometry to find the lengths and then the areas of the three triangles in Figure 20.5. However, it is much easier to use the center of the pentagon to split it into triangles, as shown in Figure 20.6.

Figure 20.6

This divides the pentagon into 10 right triangles, each with base 2 inches, height h, and angle 54°. Thus,

h = 2 · tan(54°) ≈ 2.75 inches

The area of each of these 10 triangles is square inches. Thus, the area of the pentagon is about 10(2.75) = 27.5 square inches.

Example 5

A marshy piece of property is triangular in shape. One angle measures 34°, and the two sides adjacent are 800 feet and 1,000 feet in length. What is the length of the third side? How many acres is the piece of property? (Note: An acre measures 43,560 square feet.)

The third side can be found directly by the Law of Cosines. If the third length is C, then
C2 = A2 + B2 – 2ABcos(θ)
C2 = 8002 + 1,0002 – 2(800)(1,000)cos(34°)
C = √1,640,000 – 1,600,000 · cos(34°) ≈ 560 feet

The area can then be found by Heron's formula:

area = √1,180(1,180 – 1,000)(1,180 – 800)(1,180 – 560)
area = √1,180(1,180)(180)(380)(620) ≈ 223,699 square feet, or about 5.13 acres

Practice problems for this study guide can be found at:

Advanced Trigonometry Review Practice Questions

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