Question

HELP PLZZ will give brainliest <3

Answer 1

Answer:

0

1

Step-by-step explanation:

First question:

You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$

$\dfrac{10}{\sin 30^\circ} = \dfrac{40}{\sin B}$

$\dfrac{1}{0.5} = \dfrac{4}{\sin B}$

$\sin B = 2$

The sine function can never equal 2, so there is no triangle in this case.

Answer: no triangle

Second question:

You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.

$\dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

$\dfrac{10}{\sin 63^\circ} = \dfrac{}{\sin C}$

$\sin C = \dfrac{8.9\sin 63^\circ}{10}$

$C = \sin^{-1} \dfrac{8.9\sin 63^\circ}{10}$

$C \approx 52.5^\circ$

One triangle exists for sure. Now we see if there is a second one.

Now we look at the supplement of angle C.

m<C = 52.5°

supplement of angle C: m<C' = 180° - 52.5° = 127.5°

We add the measures of angles B and the supplement of angle C:

m<B + m<C' = 63° + 127.5° = 190.5°

Since the sum of the measures of these two angles is already more than 180°, the supplement of angle C cannot be an angle of the triangle.

Answer: one triangle