Trigonometry is a branch of math which deals exclusively with triangles and the relation between their sides and angles. This discipline can be traced back to ancient Babylon and Egypt, where it aided the construction of the pyramids. Modern trigonometry has evolved to include measurement of three-dimensional spaces, which is studied under spherical trigonometry.

## Learn Trigonometry

### Solved Example

**Question:**A right triangle with a hypotenuse of 8 includes a 60$^o$ angle. Find the measure of the other two angles and the lengths of the other two sides.

**Solution:**

Since it is a right angle triangle, one angle must be of 90$^O$.

So, third angle = 180 - (90 + 60) = 30$^o$.

(Sum of the angles of a triangle is 180$^o$)

Now, lets find the unknown sides of the triangle.

Sin 60$^o$ = $\frac{AB}{AC}$ = $\frac{a}{8}$

$\frac{\sqrt{3}}{2}$ = $\frac{a}{8}$

[Sin 60$^o$ = $\frac{\sqrt{3}}{2}$]

a = 8 * $\frac{\sqrt{3}}{2}$ = 4 * $\sqrt{3}$ $\approx$ 6.92

Again

Cos 60$^o$ = $\frac{BC}{AC}$ = $\frac{b}{8}$

$\frac{1}{2}$ = $\frac{b}{8}$

[Cos 60$^o$ = $\frac{1}{2}$]

b = 8 * $\frac{1}{2}$ = 4

**Given:**Right triangle with a hypotenuse of 8 includes a 60$^o$ angle.So, third angle = 180 - (90 + 60) = 30$^o$.

(Sum of the angles of a triangle is 180$^o$)

Now, lets find the unknown sides of the triangle.

Sin 60$^o$ = $\frac{AB}{AC}$ = $\frac{a}{8}$

$\frac{\sqrt{3}}{2}$ = $\frac{a}{8}$

[Sin 60$^o$ = $\frac{\sqrt{3}}{2}$]

a = 8 * $\frac{\sqrt{3}}{2}$ = 4 * $\sqrt{3}$ $\approx$ 6.92

Again

Cos 60$^o$ = $\frac{BC}{AC}$ = $\frac{b}{8}$

$\frac{1}{2}$ = $\frac{b}{8}$

[Cos 60$^o$ = $\frac{1}{2}$]

b = 8 * $\frac{1}{2}$ = 4

## Basic Trigonometry Help

### Solved Examples

**Question 1:**Find Cos A and Tan A by using the given information to construct a reference triangle.

Sin A = $\frac{3}{5}$ and Tan A < 0

**Solution:**

**Given:**Sin A = $\frac{3}{5}$ and Tan A < 0

Since Sin A is positive, the terminal side is either in first quadrant or in second quadrant.

In second condition, Tan A is negative means that the terminal side is in second quadrant.

By Pythagorean Theorem

$5^2 = 3^2 + x^2$

25 = 9 + $x^2$

x = - 4 (x is in 2nd quadrant)

Now

Cos A = $\frac{BC}{AC}$ = $\frac{-4}{5}$

Tan A = $\frac{P}{B}$ = $\frac{3}{-4}$ = $\frac{-3}{4}$

**Question 2:**The angle of depression of a boat from the top of the lighthouse 150 feet above the surface of the water is 10$^o$. Find the distance x from the base of the lighthouse to the boat.

**Solution:**

Angle of elevation from the boat = Angle of depression from the lighthouse.

Let us find algebraically the value of x using the tangent function.

Tan A = Tan 10$^o$ = $\frac{AB}{BC}$ = $\frac{150}{x}$

x = $\frac{150}{Tan\ 10^o}$ $\approx$ 850.82.

(Using Tan 10$^o$ = 0.1763)

The value of x is 850.82 (approx).

Thus, the boat is about 850 feet from the base of the lighthouse.

Let us find algebraically the value of x using the tangent function.

Tan A = Tan 10$^o$ = $\frac{AB}{BC}$ = $\frac{150}{x}$

x = $\frac{150}{Tan\ 10^o}$ $\approx$ 850.82.

(Using Tan 10$^o$ = 0.1763)

The value of x is 850.82 (approx).

Thus, the boat is about 850 feet from the base of the lighthouse.