Use the law of sines to solve the following non right angles. help me please with these 2

Question

asked May 20, 2020 102k views

1 Answer

Marianella · Answer 1 · 2020-05-26T02:08:58+0000

1. The interior angles of any triangle add to 180 degrees in measure. So

$m\angle A=(180-85-47)^\circ=48^\circ$

Then by the law of sines,

$(\sin48^\circ)/(19)=\frac{\sin85^\circ}b=\frac{\sin47^\circ}c\implies b\approx25.5,c\approx18.7$

2. By the law of sines,

$\frac{\sin\angle A}a=(\sin\angle B)/(16)=(\sin104^\circ)/(25)$

so that

$\sin\angle B=(16\sin104^\circ)/(25)\implies m\angle B\approx38.4^\circ$

Then

$m\angle A\approx(180-104-38.4)^\circ=37.6^\circ$

so that

$\frac{\sin\angle A}a=(\sin104^\circ)/(25)\implies a=(25\sin\angle A)/(\sin104^\circ)\approx15.7$