# How do you use Law of sines, given A=102, b=13, c=10?

Mar 26, 2018

color(maroon)(a = 17.97, hat B = 45.03^@, hat C = 32.97^@

color(brown)("Area of the triangle " A_t = (1/2) b c sin A = color(brown)(63.58 " sq units"

#### Explanation:

$\hat{A} = {102}^{\circ} , b = 13 , c = 10$

We cannot directly use Law of Sines.

$\text{First let's use Law of Cosines to find the third side } \textcolor{b r o w n}{a}$

${a}^{2} = {b}^{2} + {c}^{2} - \left(2 b c \cos A\right)$

$a = \sqrt{{13}^{2} + {10}^{2} - \left(2 \cdot 13 \cdot 10 \cdot \cos 102\right)} = 17.97 \text{ units}$

Only one triangle is possible, having $\text{ } \hat{A} = {102}^{\circ}$ an obtuse angle.

Since we know all the three sides and one angle, we can apply law of sines to find the other two angles.

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

$\hat{B} = {\sin}^{-} 1 \left(\frac{b . \sin A}{a}\right) = {\sin}^{-} 1 \left(\frac{13 \cdot \sin 102}{17.97}\right) = {45.03}^{\circ}$

Similarly, $\hat{C} = {\sin}_{1} \left(\frac{10 \cdot \sin 102}{17.97}\right) = {32.97}^{\circ}$

$\text{Area of the triangle " A_t = (1/2) b c sin A = (1/2) * 13 * 10 * sin 102 = color(maroon)(63.58 " sq units}$