# Trigonometry Problems

Trigonometry is the subject that deals with the measurement of triangles. Using the concepts of trigonometry many real world problems are solved. The initial applications of trigonometry were restricted to astronomy, navigation and surveying. The view of trigonometric relationships as functions, expanded the applications to include many a physical phenomena related to rotations and vibrations.

In this page, we will solve few word problems to show how the real world problems can be solved using trigonometric ratios.

## Trigonometry  Sample Problems

Angular speed
A car is moving at a speed of 75 mph.  If the diameter of the wheels is 2.5 ft ,
1. Find the number of revolutions per minute the wheels are making.
2. Find the angular speed of the wheels in radians per minute.

1. The wheels make one revolution when the car moves a distance equal to their circumference.
So the number of revolutions the wheels are making in a minute is the distance moved in one minute divided by the circumference.

# revolutions = Distance moved in one minute / circumference of the wheels.
Distance moved in one minute = speed per minute = $\frac{75}{60}$ = 1.25 miles = 1.25 x 5280 = 6600 ft.
Circumference of the wheels = πd = 2.5 x π = 7.854 ft
Number of revolutions the wheels are making in a minute = 6600/7.854 ≈ 1050 revolutions

2.  The central angle made by the wheels in one minute = # revolutions per minute x 2π radians
Angular speed per minute = 1050 x 2 π = 2100 π radians per minute.

Right triangle models:
Roger wanted to measure the width of a stream. First he stood directly opposite to a pole fixed right on the opposite bank as shown in the diagram. Then he walked 100 ft along the bank and viewed the pole again. He measured the angle this line of observation was making
with the line along which he had walked, as 52 º. Use trigonometric ratios to find the width of the stream assuming the width
does not change over the distance walked.

In the diagram, the width of the river is given by the leg AC of the right right triangle ABC. The length of the other led AB,the distance walked is
known and equal = 100ft.  With reference to the angle measure given these two are the opposite and adjacent sides.
Hence we can use sine function to solve the problem.

Sin B = Opposite leg / Adjacent leg = AC / AB
Sin 52 º = AC / 100
AC = 100 x sin 52  ≈ 78.81 ft.
The width of the stream = 78.81 ft.

Harmonic Motion

The displacement from equilibrium of an oscillating weight suspended by a spring is given
y(t) = $10sin(\frac{\pi }{2}t)$
where y(t) is displacement in feet for time 't' seconds.

1. Find the displacements when t =0, 1/2 and 1 seconds.
2. What is the period and frequency of this harmonic motion?

1. To find the displacements, all we need to do is to plug the values of time given in the equation.
y(t) = $10sin(\frac{\pi }{2}t)$
y(0) = $10sin(\frac{\pi }{2}(0))$ = 10 x sin(0) = 0

y(1/2) = $10sin(\frac{\pi }{2}(\frac{1}{2}))=10sin(\frac{\pi }{4})$ = 10 (0.70711) = 7.071 ft.
y(1) = $10sin(\frac{\pi }{2}.1)=10sin(\frac{\pi }{2})$ = 10 x 1 = 10 ft.
Hence the displacements when t = 0, 1/2 and 1 second are 0 ft, 7.07 ft and 10 ft.

2. For an equation of the form y = a.sin(bt),  the period is given by
p = $\frac{2\pi }{b}$ .
In the given equation y(t) = $10sin(\frac{\pi }{2}t)$,  b = $\frac{\pi }{2}$
hence the period = $\frac{2\pi }{\frac{\pi }{2}}$ = 4 seconds
Frequency = 1/p = 1/4 cycles per second.

Bearing Problem:
A ship leaves a port at noon and has a bearing of S 32 W. If the ship sails at 22 knots.

1. Find the nautical miles the ship will have traveled in the directions South and West at 5.00 P.M
2. At 5 P.M, the ship changed course and was moving westward. Find the bearing and distance of the ship from the port at 7.00 P.M.

1. The above sketch depicts the ships course. O represents the port and A is the position of the ship at 5.00 PM
Distance traveled by the ship in 5 hrs = Length OA = 5 x 22 = 110 Nautical miles.
OB is the distance traveled by the ship southward and AB is the distance traveled westward.
OB and AB are the adjacent and opposite legs of the right triangle OAB with reference to the angle known, which is the bearing of the ship's     course to the port. The length of hypotenuse OA is known already calculated.
Hence the two distances can be found using the sine and the cosine functions.

Sin 32 = opposite leg / hypotenuse = AB / OA
AB = OA sin 32 = 110 x sin 32 ≈ 58.29 nautical miles

Cos 32 = adjacent leg / hypotenuse = OB / OA
OB = OA cos 32 = 110 x cos 32 ≈ 93.29 nautical miles.
The ship had traveled 58.29 miles west and 93.29 miles south at 5.00 PM

2.  Two hours have lapsed between 5.00 PM and 7.00 PM and its position is indicated by C.
Distance traveled by the ship in two hours AC = 2 x 22 = 44
In right right triangle OBC, length BC = BA + AC = 58.29 + 44 = 102.29 nautical miles
The bearing angle with the North south line at the port = Angle CAB
Since the lengths of the opposite and adjacent legs are known, the angle can be found using inverse tan function.
tan θ = opposite leg / adjacent leg = BC / OB = 102.29 / 93.29 = 1.0965
θ = tan-1 (1.0965) = 47.63 º

The distance of the ship from the port is OC.  Using Pythagorean theorem,
OC2 = OB2 + BC2
=  (93.29)2 + (102.29)2
= 19166.2682
OC = √ 19166.2682 ≈ 138.44 nautical miles

At 7.00 PM the ship was at a distance of 138.44 nautical miles from the Port, with a bearing to the Port as S 47.63 W.