Question

Solve triangle ABC which has
angleA=250251angleB=600511anda=382Findb

Answer 1

Answer on Question #58942 – Math – Trigonometry

Question

1. Solve triangle $ABC$ which has

angle $A = 250251$ angle $B = 600511$ and $a = 382$. Find $b$.

Solution

So as angle $A = 250251{}^\circ$ and angle $B = 600511{}^\circ$, $\sin A = \sin 250251{}^\circ = \sin (360{}^\circ \cdot 695 + 51{}^\circ) = \sin 51{}^\circ$ and $\sin B = \sin 600511{}^\circ = \sin (360{}^\circ \cdot 1668 + 31{}^\circ) = \sin 31{}^\circ$. Then by the law of sines we have

$$\frac{a}{\sin A} = \frac{b}{\sin B},$$

hence

$$b = \frac{a \cdot \sin B}{\sin A} = \frac{382 \cdot \sin 31{}^\circ}{\sin 51{}^\circ} \approx 253.2$$

Answer: $b \approx 253.2$.

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