Question

Find the measure of A to the nearest degree using sine.

PLEASE ASAPPPP!!

Answer 1

well, since we're using Sine, we'll need the opposite side to ∡A, and that'd be side BC, so let's first find BC

`\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ a^2+o^2=c^2\implies o=√(c^2 - a^2) \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{61}\\ a=\stackrel{adjacent}{47}\\ o=\stackrel{opposite}{BC} \end{cases} \\\\\\ BC=√( 61^2 - 47^2)\implies BC=√( 3721 - 2209 ) \implies BC=√( 1512 )`

now let's get ∡A

`\sin( A )=\cfrac{\stackrel{opposite}{√(1512)}}{\underset{hypotenuse}{61}} \implies \sin^(-1)(~~\sin( A )~~) =\sin^(-1)\left( \cfrac{√(1512)}{61} \right) \\\\\\ A =\sin^(-1)\left( \cfrac{√(1512)}{61} \right)\implies A \approx 39.60^o`

Make sure your calculator is in Degree mode.