# How do i solve this trig question??

Mar 18, 2018

You use the definition of sine of an angle, in this case W, to solve this problem in almost a single step.

#### Explanation:

The definition of sine of an angle is:

$\sin \left(\theta\right) = \left(\text{Opposite")/("Hypotenuse}\right)$ where,

$\theta$ is the angle you're looking to solve,

Opposite is the measure of the side directly opposite of the angle

Hypotenuse is the side across form the 90 degree angle

Remember that this will only work for right triangles. A more complex version of this identity is used for non-right triangles.

Now, for your problem specifically, we take

$\sin \left(W\right)$ to equal your opposite side, 7, over divided by your hypotenuse, 8:

$\sin \left(W\right) = \frac{7}{8}$

IMPORTANT: Your calculator must be set to "Degrees" mode for this next part!

Take ${\sin}^{-} 1$ of both sides to solve for W and you get:

$W = {\sin}^{-} 1 \left(\frac{7}{8}\right)$ Plug that into your calculator and you get:

$W = 61.04497563$

Then applying the appropriate rounding for the problem you get:

$W = 61.0$

Mar 18, 2018

$\angle W = {61.0}^{\circ}$

#### Explanation:

$\text{Since VWX is a right triangle use "color(blue)"trig. ratio}$

•color(white)(x)sin hatW="opposite"/"hypotenuse"=(VX)/(VW)=7/8

$\Rightarrow \angle W = {\sin}^{-} 1 \left(\frac{7}{8}\right) = {61.0}^{\circ}$

Mar 18, 2018

$m \angle W = {61}^{\circ}$

#### Explanation:

Since this is a right triangle, we are able to use the $\tan , \sin ,$and $\cos$ functions.

We know the lengths of the opposite side from $\angle W$ and the hypotenuse, so we will use the $\sin$ function.

$\sin \angle W = \text{opposite"/"hypotenuse} \rightarrow \sin \angle W = \frac{7}{8}$

Using your calculator, find the value of $\frac{7}{8}$, which should be $0.875$. The equation should now look like this:

$\sin \angle W = 0.875$

Still using your calculator, hit $2 n d$ and then hit $\sin$, this will show you the measure of $\angle W$. (You may not have a $2 n d$ button in which case use ${\text{sin}}^{-} 1$.)

$\angle W = 61.045 \rightarrow {61}^{\circ}$

Here are a couple things to remember when you see questions like this:

1. Determine which side lengths you know so that you can decide which function to use ($\tan , \sin , \cos$)
2. When you have the form like the one that we had in this question ($\sin \angle W = 0.875$) remember to hit $2 n d$ and then the function. Or, depending on your calculator, you should use ${\sin}^{-} 1 , {\cos}^{-} 1 , {\tan}^{-} 1$ instead.