Home > Triangles > Trigonometry and Right-angled triangles

## Trigonometry and Right-angled triangles

#### Conceptual Understanding

If we find the ratio between a pair of sides on a triangle the value will be a constant even on larger triangles provided the triangles are similar to each other.

### The right-angled trigonometry triangle

The hypotenuse is the longest side and is opposite the right-angle

The opposite is the side that is opposite the given angle. (Or the angle we wish to find)

The adjacent is the side next to the given angle.  (It is between the given angle and the right angle)

### The ratios SohCahToa

$\displaystyle sin \theta = \frac{opp}{hyp}$

$\displaystyle cos \theta = \frac{adj}{hyp}$

$\displaystyle tan \theta = \frac{opp}{adj}$

When solving a trigonometry right-angled triangle problem we need two pieces of information and then use the relevant ratio to find the third.

There are three possible scenarios.

1. We are finding an angle
2. We are finding a side that is a numerator in the ratio
3. We are finding a side that is a denominator in the ratio

### Examples

Type 1 (The unknown is the numerator)

We have the hypotenuse and we want the opposite side.  The only ratio with hypotenuse and opposite in it is the sine ratio.

$\displaystyle sin\theta = \frac{opp}{hyp}$

Let’s fill out the equation

$\displaystyle sin 35^o=\frac{x}{17}$

$\displaystyle \implies 0.374=\frac{x}{17}$

$\implies x=9.75cm$

Type 2 (The unknown is the denominator)

We have the opposite and we want the adjacent side.  The only ratio with adjacent and opposite in it is the tangent ratio.

$\displaystyle tan \theta = \frac{opp}{adj}$

Let’s fill out the equation

$\displaystyle tan 27^o=\frac{28}{x}$

$\implies x\times tan 27^o=28$

$\displaystyle \implies x=\frac{28}{tan 27^o}$

$\therefore x =54.95mm$

Type 3 (The angle is unknown)

We have the hypotenuse and adjacent side.  The only ratio with adjacent and hypotenuse in it is the cosine ratio.

$\displaystyle cos \theta = \frac{adj}{hyp}$

Let’s fill out the equation

$\displaystyle cos x=\frac{15}{32}$

$\implies \displaystyle x=cos^{-1}\left(\frac{15}{32}\right)$

$\implies x =62.0^o$