Skip to Main Content

Module 9: Geometry

Solving Right Triangles Using Trigonometry

There are three trigonometric functions to know, each being equivalent to a certain ratio. For the same triangle, this ratio may change based on the reference point. Below are the trigonometric functions:

Sine Ratio

s i n θ = o p p o s i t e h y p o t e n u s e {"version":"1.1","math":"sin\theta=\dfrac{opposite}{hypotenuse}"}

Cosine Ratio

c o s θ = a d j a c e n t h y p o t e n u s e {"version":"1.1","math":"cos\theta=\dfrac{adjacent}{hypotenuse}"}

Tangent Ratio

t a n θ = o p p o s i t e a d j a c e n t {"version":"1.1","math":"tan\theta=\dfrac{opposite}{adjacent}"}

To remember each ratio, remember the abbreviation:

SOH CAH TOA

We can solve for an unknown by rearranging and isolating. Refer to Module 7: Algebra for more help. How do we rearrange and solve for angle θ {"version":"1.1","math":"\theta"} if we know the side lengths? Much like how rearranging requires opposite operations, moving trigonometric functions across an equal sign requires the inverse of these trigonometric functions:

s i n → s i n − 1 {"version":"1.1","math":"sin\rightarrow sin^-1"}

c o s → c o s − 1 {"version":"1.1","math":"cos\rightarrow cos^-1"}

t a n → t a n − 1 {"version":"1.1","math":"tan\rightarrow tan^-1"}

All inverses can be computed via calculator. Ensure that the calculator is set to the right mode (degrees or radians) depending on what is to be worked in. Examples here will feature degrees.

As an example of rearranging, let’s say we had a right triangle as shown below. To find θ {"version":"1.1","math":"\theta"}, we’ll need to use the given information and the correct trig ratio.

From the reference point, the only information that is available is the length of the opposite side and hypotenuse. Selecting the correct ratio can obtain θ {"version":"1.1","math":"\theta"}.

s i n θ = o p p o s i t e h y p o t e n u s e {"version":"1.1","math":"sin\theta=\dfrac{opposite}{hypotenuse}"}

θ = s i n − 1 o p p o s i t e h y p o t e n u s e {"version":"1.1","math":"\theta=sin^{-1} \dfrac{opposite}{hypotenuse}"}

θ = s i n − 1 1 2 {"version":"1.1","math":"\theta=sin^{-1} \dfrac{1}{2}"}

θ = 30 ∘ {"version":"1.1","math":"\theta=30^{\circ}"}

The triangle above is one of two special right triangles that will be very handy to know. These triangles will hold true and can be scaled up/down to account for triangles with proportional sides that have the same set of angles. The angle sets themselves are also relatively simple numbers to remember:

Let’s imagine there was a new triangle with angles 45-45-90, but had a leg length of 2 instead of 1. What does this mean for the other side lengths? As we know it has the same angle set as one of the special cases, solving the triangle would involve determining the scaling factor of said leg, and then applying it to all the other side lengths of the special case to scale the whole thing. 2 is double of 1, which means that the length of the other leg would be 2, and the length of the hypotenuse would be 2 2 {"version":"1.1","math":"2\sqrt2"}.

To recap, special cases apply when a special angle set exist in another triangles, causing the sides of said triangle to be proportional to the certain special case.

Connect with us