How do you solve the triangle given ∠B = 151°, ∠C = 7°, a = 31?

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Answer 1 · 2016-12-23T18:13:30

b= 40.12, c=10.09 and $angle A$ = $22^o$

The triangle would appear to like in the figure given below. Angle A would be 180- (151+7) = $22^o$
enter image source here
Using sine formula,

$sin22 /31 = sin 151 /b = sin 7 /c$

b= $31* sin 151 /sin22$ = 40.12

c= $31* sin7 / sin 22$ =10.09

Answer 2 · 2016-12-23T18:20:02

Subtract $angle B and angle C$ from $180^@$ , to find $angle A$ , then use The Law of Sines to find the lengths of sides "b" and "c".

$angle A = 180^@ - angle B - angle C$

$angle A = 180^@ - 151^@ - 7^@$

$angle A = 22^@$

$a/sin(A) = b/sin(B) = c/sin(C)$

$b = asin(B)/sin(A)$

$b = 31sin(151^@)/sin(22^@)$

$b~~ 40$

$c = asin(C)/sin(A)$

$c = 31sin(7^@)/sin(22^@)$

$c~~ 10$