Question

Need help QUICK! Given triangle ABC, which equation could be used to find the measure of angle B?

Answer 1

Answer:

second option

Step-by-step explanation:

We are going to use the acronym:

"Soh Cah Toa".

Why? It tells us the right-triangle definitions of sine, cosine, and tangent.

sine is opposite over hypotenuse.

cosine is adjacent over hypotenuse.

tangent is opposite over adjacent.

So looking at our triangle with respect to B tells us that 3 is the opposite measurement and 6 is the adjacent.  No matter what angle we are looking for in this triangle, the hypotenuse is constantly going to by $3\sqrt{5}$.

So let's look at cos(B).

$\cos(B)=\frac{6}{3\sqrt{5}}$

We need to rationalize the denominator by multiplying top and bottom by sqrt(5):

$\cos(B)=\frac{6\sqrt{5}}{3(5)}=\frac{2\sqrt{5}}{5}$

So now looking at sin(B).

$\sin(B)=\frac{3}{3\sqrt{5}}$

We have to rationalize again by multiplying top and bottom by sqrt(5):

$\sin(B)=\frac{3\sqrt{5}}{3(5)}=\frac{\sqrt{5}}{5}$.

So looking at our triangle with respect to A tells us that 3 is the adjacent measurement and 6 is the opposite.  No matter what angle we are looking for in this triangle, the hypotenuse is constantly going to by $3\sqrt{5}$.

We don't have to use any trigonometric ratios with A.