Can someone help me solve this problem?

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Can someone help me solve this problem?

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Answer 1 · 2018-03-26T06:06:42

Two Triangles are possible.

$Case 1 : ˆ B = {60.6}^{\circ}, ˆ C = {70.4}^{\circ}, c = 162.3 m$ $color(brown)("Case 1 : " hat B = 60.6^@, hat C = 70.4^@, c = 162.3 " m"$
$Case 2 : ˆ B = {119.4}^{\circ}, ˆ C = {11.6}^{\circ}, c = 34.6 m$ $color(orange)("Case 2 : " hat B = 119.4^@, hat C = 11.6^@, c = 34.6 " m"$

Explanation:

According to law of Sines,

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

Given : $a = 130 m, b = 150 m, ˆ A = 49^{\circ}$ $a = 130 m, b = 150m , hat A = 49^@$

$sin B = \frac{150 \cdot sin 49}{130} = 0.8708$ $sin B = (150 * sin 49) / 130 = 0.8708$

$ˆ B = {sin}^{- 1} 0.8708 = {60.6}^{\circ} or {119.4}^{\circ}$ $hat B = sin ^-1 0.8708 = 60.6^@ " or " 119.4^@$

Hence, Two Triangles are possible.

Case 1 : $ˆ B = {60.6}^{\circ}$ $hat B = 60.6^@$

$ˆ C = 180 - ˆ A - ˆ B = 180 - 49 - 60.6 = {70.4}^{\circ}$ $hat C = 180 - hat A - hat B = 180 - 49 - 60.6 = 70.4^@$

$c = \frac{a sin C}{sin A} = \frac{130 \cdot sin 70.4}{sin 49} = 162.3 m$ $c = (a sin C) / sin A = (130 * sin 70.4) / sin 49 = 162.3 " m"$

Case 2 : $ˆ B = {119.4}^{\circ}$ $hat B = 119.4^@$

$ˆ C = 180 - ˆ A - ˆ B = 180 - 49 - 119.4 = {11.6}^{\circ}$ $hat C = 180 - hat A - hat B = 180 - 49 - 119.4 = 11.6^@$

$c = \frac{a sin C}{sin A} = \frac{130 \cdot sin 11.6}{sin 49} = 34.6 m$ $c = (a sin C) / sin A = (130 * sin 11.6) / sin 49 = 34.6 " m"$

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