What is the angle of b?

Question

So first off, I know how to find the missing length of the leg of the triangle using the pythagorean theorem. $6^2 + b^2 = c^2$ $36 + b^2 = 100$ $100 - 36 = 64$ $\sqrt{64} = 8$.

So angle angle $A$ is going to be $\frac{\text{adjacent}}{\text{hypotenuse}}$

$\cos ( \text{angle} ) = \frac{\text{adjacent}}{ \text{hypotenuse}}$

$\cos ( \text{angle} ) = \frac{8}{10}$

$\cos^{-1} ( 0.8 ) = 36.86989765^\circ$.

Here is where I run into my problem. So assuming that $B$ is going to be $\frac{\text{opposite}}{\text{hypotenuse}}$

$\sin (\text{angle}) = \frac{6}{10}$

$\sin^{-1}(0.6) = 36.86989765^\circ$.

However, when I use the law of cosines I get the same answer for angle $A$, but a different angle for $B$, which comes out as $53.13$. It would be greatly appreciated if someone could help me figure out where my mistakes are. Thanks!

Answer 1

B is ~$53.13^\circ$

When you took opposite by adjacent with regard to $B$, it should be : $$\sin(B)=\frac{8}{10} \implies B=\sin^{-1}\left(\frac{8}{10}\right)\approx 53.13^\circ$$

You took opposite/adjacent to be 6/10, when the side opposite to angle B is actually 8 units long. That's why you calculated it wrong the first time.