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

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:

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


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

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

$x$ = $150 m$


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

$tan(60)$ = $\frac{150}{y}$

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

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

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.


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.


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?


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.


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