In ancient times Trigonometry is the branch of mathematics that dealt purely with the measurement of triangles. With the advancement in Physical sciences and Calculus, the trigonometric relations were started to be viewed as functions. A good skill in solving trigonometric problems is a necessary pre-requisite for understanding the more subtle concepts of Calculus. This page provides help on solving some of the typical Trigonometry Homework problems.

Help with Trigonometry Problems

Right triangle trigonometry:
The first level problems in Trigonometry are to be solved using right triangles. The trigonometric relationships are defined using right triangles. The question could be either to find the values of trigonometric ratios from a given diagram, or you need to make a diagram for a given ratio and find the rest. The key to solving this problem lies in identifying the apt trigonometric ratio to be used for solving the problem.

Example:

Find the lengths of the sides b and c in the
given right triangle rounded to the nearest cm
if the measure of angle A = 80º


For the given measure of angle A, the length of the adjacent leg CA is known. So If we want to find the value of ' a ' first, what trig ratio should we use?  Note here we need to find the length of the opposite leg related to angle A.
This means we need to use the tan function which involves the opposite and adjacent legs related to an angle.

Tan A = opposite leg / adjacent leg
Tan 80 = $\frac{a}{6}$   ⇒    a = 6 x tan 80 ≈ 34 cm

To find we can either use the cosine function ( as b is the hypotenuse) or use the Pythagorean theorem since we already know the lenghts of two sides of the right triangle.
Using cosine function

Cos A = adjacent leg / hypotenuse
cos 80 = $\frac{6}{b}$     ⇒   b = $\frac{6}{cos 80}$ ≈ 35 cm.

Trigonometric Identities:
The four types of simple identities used in trigonometry are
  1. Reciprocal identities
  2. Quotient identities
  3. Cofunction identities
  4. Pythagorean identities.

Verify the following trigonometric identity:
$\frac{tan^{2}\theta }{1-cos^{2}\theta }= sec^{2}\theta $
Left side of the identity = $\frac{tan^{2}\theta }{1-cos^{2}\theta }=\frac{tan^{2}\theta }{sin^{2}\theta +cos^{2}\theta -cos^{2}\theta }$          Using sin2 θ + cos2 θ = 1
                                  = $\frac{tan^{2}\theta }{sin^{2}\theta }$
                                  = $\frac{\frac{sin^{2}\theta }{cos^{2}\theta }}{sin^{2}\theta }$                                           Using quotient identity tan θ = $\frac{sin\theta }{cos\theta }$

                                  = $\frac{1}{cos^{2}\theta }$ =  Sec2 θ                             Using the reciprocal identity.
                                  = Right side.
The identity is hence verified.

Measuring Heights and Distances
A very long advertising banner of length 50 meters is dropped from the top of a tall building and fixed on one of its wall.  Joe wanted to estimate
the height of the building.  He stood at some distance from the base of the building and measured the angles of elevation of the bottom and top of the banner as 68º and 75º. What is the height of the Building calculated using these measures?

In the diagram above, AB represents the building and BD
the length of the banner. The height of the building is given
by the expression X + 50 meters.

We need to solve for X eliminating Y.
Measure of < ACD = 68º
Measure of < ACB = 75º

The two tan ratios that are to be considered in the right triangles ACD and ACB are
tan 68 = $\frac{AD}{AC}$ =$ \frac{X}{Y} $                     ---------------------(1)
tan 75 = $\frac{AB}{AC}=\frac{X+50}{Y}$             ----------------------(2)

y = $\frac{x}{tan 68}$                               Solved for Y using equation (1)
tan 75 = $\frac{X+50}{\frac{X}{tan68}}$                       Eliminated Y using substitution
tan 75 = $\frac{(X+50)tan68}{X}$
X.tan 75 = X tan 68 + 50 tan 68
X tan75 - X tan68 = 50.tan68
X(tan75 - tan68) = 50 tan68
X = $\frac{50tan68}{tan75-tan68}$                    Solved expression for X
   ≈ 98.45 meters.                  Calculated value of X.

Hence the height of the building = X + 50 = 98.45 + 50 = 148.45 meters.

Law of sines - Law of cosines
The law of sines is used to solve a triangle when two sides and one angle opposite to one of the sides or two angles and one side opposite to the one of the angles are known.
The Law of cosines is used to solve a triangle either when three sides of the triangle are known or two sides and the angle included between them are given.
Let us solve a word problem based on this.
Maria was riding on the bicycle along a main road. She took a turn to a side road which was inclined at an angle 36º to the main road to reach
a store which was situated 500 meters down the road. After making some purchases she took another side turned around 65º and took another side road and reached the main road again.

  1. How far Maria rode from the Store to get back to the main road. (Round to the nearest meter)
  2. When she reached back the main road, how far was she away from the point she took the deviation?

A is the point where Maria took the turn to the store from the main Road.  B represents the position of the store and C is the point
where Maria joined the main road again.

1.  We need to fin the distance BC. Measure of angle A and length of side AB are known.
     If we could compute the measure of the angle opposite to the side AB we can use sine rule to find BC.
     Measure of angle C in triangle ABC = 65 - 36 = 29º     Exterior angle theorem

    $\frac{BC}{SinA}= \frac{AB}{sinC}$                    Applied Sine rule
                  BC = Sin A .$ \frac{AB}{sinC}$
                       = Sin 36 . $\frac{500}{sin29}$
                       =  606 meters.

2. We need to find AC, the distance she has passed on the main Road.
    Measure of angle B = 180 - 65 = 115º

   $\frac{AC}{Sin B}=\frac{AB}{SinC}$                   Applied Sine rule.
                  AC = Sin B . $\frac{AB}{SinC}$
                       = Sin115 , $\frac{500}{sin29}$
                       = 935 meters.
        
Bearing Problems:
bearing problems generally pose difficulty to students as many notations are used to represent it. In general bearing is the angle made by the line of direction with the North South line. The notation N 40º E means the the angle is measured from North towards the east, whilst E 55º N would mean the angle 55 is measured clockwise from E to North. In air navigation, bearings are measure clock wise from North.

Example:
Two fire towers A and B are located 40 kilometers apart, where tower A is due west of tower B. A fire is spotted by both the towers. The bearings of the fire from A and B are E 16º N and W 36º N. Find the shortest distance to the fire, from the line joining the two Fire towers.
The sketch for the problem looks as follows


The distance d is to be found. This can be achieved in two steps as follows
In triangle ABC measure of angle C = 180 - (16 +36) = 128º.
$\frac{BC}{Sin16}= \frac{40}{sin 128}$                     Applied Sine rule
BC = sin 16 , $ \frac{40}{sin 128}$ 
     ≈ 14 KM
 In the right triangle close to the tower B,
sin 36 = $\frac{d}{14}$
d = 14 x sin36 ≈ 8.23 Km.

Hence the fire is at a distance of 8.23 KM from the line joining the two towers.