Question

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?

Answer 1

Use `A=sin^-1(b/c)` and `B=sin^-1(a/c)`.

Explanation:

The trigonometric function `sintheta` is equal to the ratio `("opposite side")/("hypotenuse")`. This is often shortened to `sintheta=("opp")/("hyp")`.

To solve for the value of `angle BAC` (also just called `angle A`), we note that the side opposite of this angle is `BC=5` and the hypotenuse is `AB=5sqrt26`. Our equation then becomes

`sinA=("opp")/("hyp")=5/(5sqrt26)=1/sqrt26`

So now we have
`sinA=1/sqrt26`
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(sinA)=sin^-1(1/sqrt26)`

Applying `sin^-1()` to `sin()` 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 `angleB` 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
`sinA=b/c`
to
`A=sin^-1(b/c)`.