# How do you solve the triangle given C=103^circ, a=3/8, b=3/4?

Mar 25, 2018

color(purple)(c = 0,9109, hat A = 23.65^@, hat B = 53.35^@

color(indigo)("Area of Triangle " A_t = (1/2) a b sin C = 0.137 "sq units"

#### Explanation:

$\hat{C} = {103}^{\circ} , a = \frac{3}{8} , b = \frac{3}{4} , \text{ To solve the triangle}$

Applying Law of cosines,

$c = \sqrt{{a}^{2} + {b}^{2} - 2 a b \cos C}$

$c = \sqrt{{\left(\frac{3}{8}\right)}^{2} + {\left(\frac{3}{4}\right)}^{2} - \left(2 \cdot \left(\frac{3}{8}\right) \cdot \left(\frac{3}{4}\right) \cdot \cos 103\right)}$

$c = 0.9109$

Applying Law of sines,

$\sin \frac{A}{\frac{3}{8}} = \sin \frac{B}{\frac{3}{4}} = \sin \frac{103}{0.9109}$

$\sin A = \frac{\left(\frac{3}{8}\right) \cdot \sin 103}{0.9109} = 0.4011$

$\hat{A} = {\sin}^{-} 1 0.4011 = {23.65}^{\circ}$

Similarly, $\sin B = \frac{\left(\frac{3}{4}\right) \cdot \sin 103}{0.9109} = 0.8023$

$\hat{A} = {\sin}^{-} 1 0.8023 = {53.35}^{\circ}$

color(indigo)("Area of Triangle " A_t = (1/2) a b sin C = (1/2) * (3/8) * (3/4) * sin 103 = 0.137 "sq units"