Question

What is the length of AB¯¯¯¯¯, to the nearest tenth of a centimeter?

Answer 1

Answer:

$AB\approx13.7cm$ to the nearest  tenth.

Step-by-step explanation:

We know two angles and a given side, we can use the sine rule to find the required length.

$\frac{AB}{\sin(50\degree)}=\frac{12}{\sin(42\degree)}$

We solve for the AB by multiplying both sides by $\sin(50\degree)$.

This implies that;

$AB=\frac{12}{\sin(42\degree)}\times \sin(50\degree)$

$AB=13.738$

$AB\approx13.7cm$ to the nearest  tenth.

Answer 2

Answer:

13.43

Step-by-step explanation:

Law of Sines

x/sin50 = 12/sin42

solve for x and don't forget to put calculator into degree mode