# Solving Right Triangle in Trigonometry

Solving right triangle in trigonometry is really easy. We just need to remember some mnemonics and a little background in Pythagorean Theorem.

This topic is very powerful in many ways. Right triangle has a very huge application in both basic and advanced mathematics. The foundation in solving this easily is to use SOH-CAH-TOA formula.

Given the right triangle, Right Triangle in Trigonometry

Parts: $\Theta$ – this is the angle itself

Adjacent Side (A) – this is the side adjacent to the angle, so if $\theta$ is in $\angle C$, the adjacent side will be the side $\overline{BC}$

Opposite Side (O) – this is the side always opposite to the given angle

Hypotenuse (H) – this is constant part, the hypotenuse is always the longest side of the triangle opposite the right angle.

When solving this problem, we need to make sure the location of the angle. If it is on the 1st, 2nd, 3rd or 4th quadrant because the sign of cosine and sine varies depending on the location of the angle. Take a look of the photo below. Signs of Sine and cosine in each quadrant

SOH-CAH-TOA mnemonics: $sin\theta=\displaystyle\frac{O}{H}$ $cos\theta=\displaystyle\frac{A}{H}$ $tan\theta=\displaystyle\frac{O}{A}$

Worked Problem 1:

Given that $sin\theta =\displaystyle\frac{3}{5}$ and $\theta$ is in quadrant 2. Find the value of the following:

a. $cos\theta$

b. $tan\theta$

c. $sin2\theta$

Solution:

Draw a quick draft of the figure so that you can identify where is the adjacent and opposite sides. Since we’re dealing with right triangle, we can always use Pythagorean Theorem to solve for the unknown side. $H^2=A^2+O^2$ $5^2=A^2+3^2$ $A^2=25-9=16$ $A=\sqrt{16}=\pm4$

Solving for $cos\theta$: We use CAH, consider that cosine in quadrant 2 is negative $cos\theta=\displaystyle\frac{A}{H}$ $cos\theta=\displaystyle\frac{-4}{5}$

Solving for $tan\theta$: In quadrant 2, sine is positive and cosine is negative $tan\theta=\displaystyle\frac{O}{A}$ $tan\theta=\displaystyle\frac{3}{-4}=-\displaystyle\frac{3}{4}$

Solving for $sin2\theta$:

Now, we need to break up the a little $sin2\theta$

Recall that $sin2\theta=2sin\theta cos\theta$, from double angle identity. $sin2\theta=2sin\theta cos\theta$ $sin2\theta=2\cdot \displaystyle\frac{3}{5}\cdot -\displaystyle\frac{4}{5}$ $sin2\theta=-\displaystyle\frac{24}{25}$

Worked Problem 2:

Find $sin(x-y)$   if $sin(x)=\displaystyle\frac{3}{5}$   and $sin(y)=\displaystyle\frac{5}{13}$. If $x,y$    are on the 1st quadrant.

Solution:

I know what you’re thinking, $sin(x-y)\ne sin(x)-sin(y)$. We cannot use the distributive property because we’re dealing with angles. Not algebraic quantities.

Since we have two angles x and y, we also need to create 2 triangles.

In figure 1, we need to complete the side of the triangle. $H^2=A^2+O^2$ $5^2=3^2+O^2$ $A^2=5^2-3^2=16$ $A^2=\pm 4$

Same goes in figure 2, $13^2=A^2+5^2$ $A^2=13^2-5^2$ $A^2=\pm 12$

Now, recall the sum and difference formula of $sin(x\pm y)=sin(x)cos(y)\pm cos(x)sin(y)$ (sincoscosin mnemonic)

Thus, $sin(x-y)=sin(x)cos(y)-cos(x)sin(y)$

Now, we don’t have $cos(x)$   and $cos(y)$,  so let’s look for it.

Solving for $cos(x)$:  Refer to figure 1, remember that location of the angle, since it is on quadrant 1, all values of A and O are positive. $cos(x) =\displaystyle\frac{A}{H}$ $cos(x) =\displaystyle\frac{4}{5}$

Solving for $cos(y)$: Refer to figure 2, again the angle is in quadrant 1 so the values of opposite and adjacent are all positive. $cos(y) =\displaystyle\frac{A}{H}$ $cos(y) =\displaystyle\frac{12}{13}$

Going back to $sin(x-y)$ $sin(x-y)=sin(x)cos(y)-cos(x)sin(y)$ $sin(x-y)=\displaystyle\frac{3}{5}\cdot\displaystyle\frac{12}{13}-\displaystyle\frac{4}{5}\cdot\displaystyle\frac{5}{13}$ $sin(x-y)=\displaystyle\frac{16}{65}$

To outline the steps and guide you to solve right triangle in trigonometry, here are some few tips and steps that you might want to consider so that you would where to start.

Step 1: DRAW the figure. Unless you have the power of wild imagination of the images, you can skip this, but if you’re in the learning process, always draw!

Step 2: Complete the parts of right triangle using Pythagorean Theorem. Easy right?

Step 3: Location of angle. The sign of your answer is dependent on the location of the angle. Mathematics is so strict; always take into account the sign after solving the problem. If your sign is wrong, you are as good as your 3-year old brother. Your answer is wrong.

Step 4: Give the answer to the question if what is being asked. Remember that, because you might be pre occupied with the series of steps and you have forgotten that the problem is just asking for the hypotenuse. Do not overthink!

Step 5: Be cool. The key to making your life easier in trigonometry is to memorize the basic trigonometric identities and learn how to use them properly. Trigonometric Identities is like knowing how to add, subtract, multiply, and divide in arithmetic.

So, you think you can do these simple steps?

### Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.