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

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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 $∠ A$ $angle A$ = $22^{o}$ $22^o$

Explanation:

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

$\frac{sin 22}{31} = \frac{sin 151}{b} = \frac{sin 7}{c}$ $sin22 /31 = sin 151 /b = sin 7 /c$

b= $31 \cdot \frac{sin 151}{sin 22}$ $31* sin 151 /sin22$ = 40.12

c= $31 \cdot \frac{sin 7}{sin 22}$ $31* sin7 / sin 22$ =10.09

Answer 2 · 2016-12-23T18:20:02

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

Explanation:

$∠ A = 180^{\circ} - ∠ B - ∠ C$ $angle A = 180^@ - angle B - angle C$

$∠ A = 180^{\circ} - 151^{\circ} - 7^{\circ}$ $angle A = 180^@ - 151^@ - 7^@$

$∠ A = 22^{\circ}$ $angle A = 22^@$

Use The Law of Sines

$\frac{a}{sin (A)} = \frac{b}{sin (B)} = \frac{c}{sin (C)}$ $a/sin(A) = b/sin(B) = c/sin(C)$

$b = a \frac{sin (B)}{sin (A)}$ $b = asin(B)/sin(A)$

$b = 31 \frac{sin (151^{\circ})}{sin (22^{\circ})}$ $b = 31sin(151^@)/sin(22^@)$

$b \approx 40$ $b~~ 40$

$c = a \frac{sin (C)}{sin (A)}$ $c = asin(C)/sin(A)$

$c = 31 \frac{sin (7^{\circ})}{sin (22^{\circ})}$ $c = 31sin(7^@)/sin(22^@)$

$c \approx 10$ $c~~ 10$

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

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