Question

URGENT EXPLANATION AND SOLVING

Answer 1

$\cfrac{a}{\sin(a)} = \cfrac{b}{\sin(b)}$

$\cfrac{31}{\sin(42)} = \cfrac{37}{\sin(x)}$

$31\sin(x) = 37 \sin(42)$

$\sin(x) = \cfrac{37 \sin(42) }{31}$

$x = \sin^{-1}(\cfrac{37 \sin(42) }{31} )$

$x = 53 \textdegree$

Answer: 53°

Answer 2

We've moved on to inverting the sine.

First let's criticize the question. Question writers: x is a terrible name for an angle; help the kids out and use capital letters like A and B and C for angles and small letters for the sides.

Second, when we have two sides and and an angle not included, there are often two solutions, two angles, supplementary to each other which solve the problem. Supplementary means they add up to 180 degrees, so one acute, one obtuse. Asking to type "the number" doesn't even recognized the possibility.

Let's call this triangle XYZ with vertices X,Y,Z and respective opposite sides x,y,z. We're solving for X and given x=37, z=31, Z=42 deg.

The Law of Sines says

$\dfrac{x}{\sin X} = \dfrac{z}{\sin Z}$

$\sin X = \dfrac{x \sin Z}{z}$

$\sin X =\dfrac{37 \sin 42}{31} \approx 0.7986$

That's a perfectly regular sine, so there will be two triangle angles whose sine is that. The acute one is the principal value,

$X_1 = \arcsin 0.7986 \approx 53.0 ^\circ$

That's probably the answer they're looking for but of course

$X_2 = 180^\circ - 53.0^\circ =127.0^\circ$

has the same sine and has an equally valid triangle with those given sides and angle. Bad question.

Will the green box accept 127.0?  Inquiring minds want to know.