Question

I got 2 questions.
1. In triangle ABC, find B, to the nearest degree, given a=7, b=10, and C is a right angle.

2. Solve the right triangle ABC with right angle C if B=30 and C=10

Answer 1

1) answer would be 12.2
2) answer would be 28.2888

hope this helps

Answer 2

Answer:

1. In triangle ABC,

AC = a = 10, BC =b= 7 and ∠ C = 90°

By cosine law,

$AB^2 = 7^2 + 10^2 - 2\times 7\times 10 cos 90^{\circ}$

$AB^2 = 49 + 100 - 0$

$AB^2 = 149$

$AB= \sqrt{149}$

Now, by the law of sine,

$\frac{sin B}{10} = \frac{sin90^{\circ}}{\sqrt{149} }$

$sin B = \frac{10}{\sqrt{149} }$

$\angle B = 55.0079798014\approx 55.008^{\circ}$

2. In triangle ABC,

∠B = 30°,  AB=c=10 and ∠C = 90°

∠A = 180°-(30+90)°=60°

$\frac{AC}{10}=sin30^{\circ}$

⇒ $AC = \frac{10}{2} = 5$

By Pythagoras,

$CB^2 = AB^2 - AC^2=10^2 - 5^2 = 100 - 25 = 75$

⇒ $CB = \sqrt{75} =8.66025403784\approx 8.66$