Applications of Trigonometric Functions

Solving a right triangle means finding the missing lengths of its sides and the measurements of its angles. We will label right triangles so that side a is opposite angle A, side b is opposite angle B, and side c is the hypotenuse opposite right angle C. CA B a b c Solving Right Triangles

Solve the right triangle shown. Solution We begin by finding the the measure of angle B. We do not need a trigonometric function to do so. Because C = 90º and the sum of a triangle’s angles is 180, we see that A + B = 90º. Thus, B = 90º – A = 90º – 34.5º = 55.5º. Now we need to find a. Because we have a known angle, and unknown opposite side, and a known adjacent side, we use the tangent function. tan34.5º = a/10.5 Now we solve for a. A = 10.5tan34.5  =7.22 CA B a b = 10.5 c 34.5º Text Example

Solve the right triangle shown. Solution Finally, we need to find c. Because we have a known angle, a known adjacent side, and an unknown hypotenuse, we use the cosine function. cos34.5  = 10.5/c c=10.5/cos34.5  = In summary, B = 55.5º, a = 7.22, and c = CA B 7.22 b = 10.5 c 34.5º 55.5º Text Example cont.

Example

Example 1: A bridge is to be constructed across a small river from post A to post B. A surveyor walks 100 feet due south of post A. She sights on both posts from this location and finds that the angle between the posts is 73 . Find the distance across the river from post A to post B. It follows that x = 327. The distance across the river from post A to post B is 327 feet. Use a calculator to find tan 73 o = Post B Post A 100 ft. x 73 ○ Example 1: Application 3.27 = tan 73  = = adj opp

Use the figure to find: a. the bearing from O to B. b. the bearing from O to A. Solution a.To find the bearing from O to B, we need the acute angle between the ray OB and the north-south line through O. The measurement of this angle is given to be 40º. The figure shows that the angle is measured from the north side of the north-south line and lies west of the north- south line. Thus, the bearing from O to B is N 40º W. W N E S A B C D O 40º 75º 25º 20º Text Example

Use the figure to find: a. the bearing from O to B. b. the bearing from O to A. Solution W N E S A B C D O 40º 75º 25º 20º To find the bearing from O to A, we need the acute angle between the ray OA and the north-south line through O. This angle is specified by the voice balloon in the figure. The figure shows that this angle measures 90º – 20º, or 70º. This angle is measured from the north side of the north-south line. This angle is also east of the north-south line. This angle is also east of the north-south line. Thus the bearing from O to A is N 70º E. b. Text Example cont.

Example A boat leaves the entrance of a harbor and travels 40 miles on a bearing of S64E. How many miles south and east did the boat travel? S64E 40 miles a b

angle of elevation When an observer is looking downward, the angle formed by a horizontal line and the line of sight is called the: Angle of Elevation and Angle of Depression When an observer is looking upward, angle of elevation. the angle formed by a horizontal line and the line of sight is called the: observer object line of sight horizontal observer object line of sight horizontal angle of depression angle of depression. Angle of Elevation and Angle of Depression

Example 2: A ship at sea is sighted by an observer at the edge of a cliff 42 m high. The angle of depression to the ship is 16 . What is the distance from the ship to the base of the cliff? The ship is 146 m from the base of the cliff. line of sight angle of depression horizontal observer ship cliff 42 m 16 ○ d Example 2: Application d = =

Example 3: A house painter plans to use a 16 foot ladder to reach a spot 14 feet up on the side of a house. A warning sticker on the ladder says it cannot be used safely at more than a 60  angle of inclination. Does the painter’s plan satisfy the safety requirements for the use of the ladder? Next use the inverse sine function to find .  = sin  1 (0.875) = The painter’s plan is unsafe! ladder house The angle formed by the ladder and the ground is about 61 . θ Example 3: Application sin  = = 0.875

Simple Harmonic Motion An object that moves on a coordinate axis is in simple harmonic motion if its distance from the origin, d, at time t is given by either d = a cos  t or d = a sin  t. The motion has amplitude |a|, the maximum displacement of the object from its rest position. The period of the motion is 2  / , where  > 0. The period gives the time it takes for the motion to go through one complete cycle.

Frequency of an Object in Simple Harmonic Motion An object in simple harmonic motion given by d=acos  t or d = asin  t has frequency f given by f =  /2 ,  >0. Equivalently, f = 1/period.

Example A mass moves in simple harmonic motion described by the following equation, with t measured in seconds and d in centimeters. Find the maximum displacement, the frequency, and the time required for one cycle.

Example cont. Since a=8, the maximum displacement is 8 cm.

Example cont. The frequency is 1/6 cm per second.

Example cont. The time required for one cycle is 6 seconds.