# How do you find the angles of a right triangle given Triangle ABC with AB=5√26 BC=5, CA=25 and angle BCA is 90 degrees?

Nov 18, 2016

Use $A = {\sin}^{-} 1 \left(\frac{b}{c}\right)$ and $B = {\sin}^{-} 1 \left(\frac{a}{c}\right)$.

#### Explanation:

The trigonometric function $\sin \theta$ is equal to the ratio $\left(\text{opposite side")/("hypotenuse}\right)$. This is often shortened to $\sin \theta = \left(\text{opp")/("hyp}\right)$.

To solve for the value of $\angle B A C$ (also just called $\angle A$), we note that the side opposite of this angle is $B C = 5$ and the hypotenuse is $A B = 5 \sqrt{26}$. Our equation then becomes

$\sin A = \left(\text{opp")/("hyp}\right) = \frac{5}{5 \sqrt{26}} = \frac{1}{\sqrt{26}}$

So now we have
$\sin A = \frac{1}{\sqrt{26}}$
and we want to solve for $A$. This is done by applying the inverse function of $\sin$ (that is, ${\sin}^{-} 1$) to both sides:

${\sin}^{-} 1 \left(\sin A\right) = {\sin}^{-} 1 \left(\frac{1}{\sqrt{26}}\right)$

Applying ${\sin}^{-} 1 \left(\right)$ to $\sin \left(\right)$ undoes the $\sin$ function, so the LHS simply reduces to $A$, and we get

A=sin^-1(1/sqrt26)approx11.31° (by calculator)

I'll leave $\angle B$ as practice for you; the answer is
Bapprox78.69°.
(Hint: there's a really easy shortcut.)

Remember: trig functions take in an angle and return a number between 0 and 1. Inverse trig functions thus take in a number between 0 and 1 and return an angle.

Side note: it is often sufficient to jump right from
$\sin A = \frac{b}{c}$
to
$A = {\sin}^{-} 1 \left(\frac{b}{c}\right)$.