Trigonometry refers to calculations using triangles. In trigonometry we study how the angles and sides of a triangle are related to each other. There are many applications of trigonometry such has measuring heights of monuments or structures; distances between ship positions at the sea; video games such as Mario; in construction for measuring sizes and dimensions of structures; in flight as well as maritime engineering; etc.

Trigonometry Formulas:

$sin(x)$ = $\frac{opposite\ side}{hypotenuse}$

$cos(x)$ = $\frac{adjacent\ side}{hypotenuse}$

$tan(x)$ = $\frac{opposite\ side}{adjacent\ side}$

$cosec(x)$ = $\frac{1}{sin(x)}$

$sec(x)$ = $\frac{1}{cos(x)}$ 

$cot(x)$ = $\frac{1}{tan(x)}$ 
Trigonometric Identities:

$sin^2 (x)\ +\ cos^2 (x)$ = $1$

$tan^2 (x)\ +\ 1$ = $sec^2 (x)$

$1 + cot^2 (x)$ = $csc^2 (x)$
Law of Sines:

$\frac{a}{sin (A)}$ = $\frac{b}{sin (B)}$ = $\frac{c}{sin (C)}$
Law of Cosines:

$a^2$ = $b^2\ +\ c^2\ -2bc\ Cos(A)$

$b^2$ = $a^2\ +\ c^2\ -2ac\ Cos(B)$

$c^2$ = $a^2\ +\ b^2\ -2ab\ Cos(C)$
Area of Triangle:

Area = $\sqrt{(s(s-a)(s-b)(s-c)}$

Where, $s$ = $\frac{1}{2}$ $(a + b + c)$
(a,b and c are the sides of a triangle)
Example 1: 

From the top of a $150 m$ tower, two cars are seen at an angle of depression of $45$ degrees and $60$ degrees on either side of the tower. Find the distance between the two cars.

Trigonometry Word Problem

Solution:
 
This is what the picture would look like. Now we know that the angles at $A$ and $B$ would also be $45$ and $60$ degrees respectively. So if the distance between car $A$ and tower is $x$ and that between car $B$ and tower is $y$, then the following figure holds:
 
Problem on Trigonometry

From the above figure we see that:

$Tan(A)$ = $\frac{opposite\ side}{hypotenuse}$

So, 

$tan(45)$ = $\frac{150}{x}$

$1$ = $\frac{150}{x}$

$x$ = $150 m$

Similarly,

$Tan(B)$ = $\frac{opposite\ side}{hypotenuse}$

$tan(60)$ = $\frac{150}{y}$

$1.732$ = $\frac{150}{y}$

$y$ = $\frac{150}{1.732}$ = $86.61 m$

So the total distance between the two cars is  = $x + y$ = $150 + 86.61$ = $236.61 m$
Example 2: 

A tree grows at an angle of $83$ degrees with respect to the horizontal ground. The angle of elevation from a peg $100 m$ from the base of the tree is $33$ degrees. Find the height of the tree.

Solution:

First let us sketch a picture of the situation described in the question. It would look like this:

Trigonometry applications
 
We see that in the above triangle we know two angles and the included side. We can find the third angle using the angle sum property of triangle. 

$m < X + m < Y + m < Z$ = $180$

$33 + 83 + Z$ = $180$

$116 + Z$ = $180$

$Z$ = $180 - 116$

$Z$ = $64$

Now using the law of sines we can write an equation for the height of the tree $x$ as:

$\frac{x}{Sin(X)}$ = $\frac{z}{Sin(Z)}$

Now substituting the values of $x,\ z,\ X$ and $Z$ we have:

$\frac{x}{Sin(33)}$ = $\frac{100}{Sin(64)}$

Substituting the sine values for 33 and 64 degree angles we have:

$\frac{x}{0.5446}$ = $\frac{100}{0.8988}$

Cross multiplying here we have:

$0.8988x$ = $54.46$

Dividing both the sides by $0.8988$ gives us the value of $x$ as:

$x$ = $\frac{54.46}{0.8988}$ = $60.59 m$

So height of the tree is $60.59  m$.
Example 3:

There is an approximately circular lake. A surveyor wants to measure the approximate diameter of this lake. He starts from one end of the lake and walks a distance of $265$ yards towards a tree that lies a little away from the edge of the lake. Then from there he turns $115^{\circ}$ and walks towards the other end of the lake, reaching it in $290$ yards. Find the approximate diameter of the lake.

Solution:

First let us sketch a picture of the situation described in the question.

Trigonometry Word Problems
 
We see that the end points of the lake $A$ and $B$ along with the position of the tree $T$ form a triangle $BAT$. In this triangle two sides and the included angle is known to us. So we can use the law of cosines to find the length of the third side $x$.

The law of cosines is like this:

$a^2$ = $b^2\ +\ c^2\ -\ 2bcCos(A)$

So for our problem it would be like this:

$x^2$ = $a^2\ +\ b^2\ -\ 2abCos(T)$

Substituting the values from our triangle we have:

$x^2$ = $290^2\ +\ 265^2\ -\ 2(290)(265)Cos(65)$

Simplifying that we have:

$x^2$ = $84100 + 70225 - 64956$

Evaluating that we have:

$x^2$ = $89369$

Taking square root on both the sides we have:

$x$ = $\sqrt{89369}$

$x$ = $298.95$  yards

Thus the approximate diameter of the lake would be: $299$ yards or $300$ yards.
Example 4: 

A triangular pen for livestock is to be constructed in the back yard. The fencing has to be used as one side of the pen. The length of the fencing is $12$ meters. One of the other sides of the pen would be the $15 m$ wall of the barn which is at an angle of $50$ degrees to the fence. What area would this pen cover?

Solution:

First let us make a picture of the pen that we are trying to construct.

Word problem on trigonometry
 
The triangle $ABC$ represents our pen. The side $AB$ of this triangle is the fence which is $12 m$ long. The side $BC$ of this triangle is our barn wall which is $15 m$ long. Now we know that we have the length of two sides of the triangle and the measure of the included angle. So we can use the formula for area of a triangle which is:

$A$ = $\frac{1}{2}$ $abSin(C)$

For our problem it would be:

$A$ = $\frac{1}{2}$ $(AB)(BC)Sin(B)$

Substituting the values we have:

$A$ = $\frac{1}{2}$ $(12)(15)Sin(50)$

Substituting the value for $sin(50)$ we have:

$A$ = $\frac{1}{2}$ $(12)(15)(0.766)$

Multiplying out the numbers we have:

$A$ = $6(15)(0.766)$

$A$ = $68.94 m^2$

Thus the area of the pen would be $68.94 m^2$.
Example 5:

The average time between a high tide and a low tide at a port is $6$ hours. The average depth of water at the port is $50$ meters. However at high tide the water level rises to $70$ meters. Find the trigonometric function that represents the depth of the water over time if it's peak high tide at $12:00 AM$. Also sketch the graph of the function.

Solution:

The general form of a sine curve function is like this:

$y(t)$ = $A sin(Bt + C) + D$

Here, $A$ is the amplitude of the curve. Amplitude refers to half the difference between the highest and lowest point on the curve. $B$ is the horizontal stretch or shrink that can be found from the period of the function. $C$ is the horizontal translation from the parent function. It is used to find the phase shift. $D$ is the vertical shift of the function. 

The parent function that we need to start working form would be:

$y(t)$ = $sin(t)$

Now since the average depth is $50$ meters, so we know that the vertical shift is $D$ = $50$ meters. Thus, using this value of $D$ our function now becomes:

$y(t)$ = $sin(t) + 50$

Next, we are given that the difference between average depth and high tide depth is $70 - 50$ = $20$ meters. So that becomes our amplitude $A$. Thus using this our equation now becomes:

$y(t)$ = $20\ sin(t) + 50$

We are given that the time between high and low tide is $6$ hours. So the time between two high tides would be $12$ hours. So the period of the function would be $12$ hours. We know that:

$period$ = $\frac{2 \pi}{B}$

Thus using the period of $6$ hours we have:

$6$ = $\frac{2 \pi}{B}$

Solving that for $B$ we have:

$B$ = $\frac{2 \pi}{6}$ = $\frac{p}{3}$

So now our function becomes:
$y(t)$ = $20sin$ $(\frac{\pi}{3}$ $t + C) + 50$

Now we need to find the value for that $C$. We are given that at $12:00AM$ when $t$ = $0$, the tide is peak high at $70$ meters. Thus,

$y(t$ = $0)$ = $20 sin$ $(\frac{\pi}{3}$ $(0)\ +\ C)\ +\ 50$

Solving that for $y$ = $70$ we have:

$70$ = $20Sin(C)\ +\ 50$

$70 - 50$ = $20Sin(C)$

$20$ = $20Sin(C)$

$Sin(C)$ = $1$

$C$ = $\frac{p}{2}$

Thus our function is:

$y(t)$ = $20Sin$ $(\frac{\pi}{3}$ $t\ +$ $\frac{p}{2})$ $+\ 50$

Its graph would look like this:

Trigonometric Word Problems