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

1. In triangle ABC,

AC = a = 10, BC =b= 7 and ∠ C = 90°

By cosine law,

`AB^2 = 7^2 + 10^2 - 2* 7* 10 cos 90^(\circ)`

`AB^2 = 49 + 100 - 0`

`AB^2 = 149`

`AB= √(149)`

Now, by the law of sine,

`(sin B)/(10) = (sin90^(\circ))/(√(149) )`

`sin B = (10)/(√(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°

`(AC)/(10)=sin30^(\circ)`

⇒
`AC = (10)/(2) = 5`

By Pythagoras,

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

⇒
`CB = √(75) =8.66025403784\approx 8.66`