# Can someone help me solve this problem?

Mar 26, 2018

Two Triangles are possible.

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

#### Explanation:

According to law of Sines,

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

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

$\sin B = \frac{150 \cdot \sin 49}{130} = 0.8708$

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

Hence, Two Triangles are possible.

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

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

$c = \frac{a \sin C}{\sin} A = \frac{130 \cdot \sin 70.4}{\sin} 49 = 162.3 \text{ m}$

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

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

$c = \frac{a \sin C}{\sin} A = \frac{130 \cdot \sin 11.6}{\sin} 49 = 34.6 \text{ m}$