How do you solve the triangle given A=56, B=22, a=12.2?

Mar 25, 2018

$\hat{C} = 102 , b = 5.42 , c = 14.39$

color(brown)("Perimeter of the triangle " = P = 32.01 " units"

color(purple)("Area of the triangle " A_t ~~ 32.34 " sq units"

Explanation:

Given : $\hat{A} = 56 , \hat{B} = 22 , a = 12.2$

$\hat{C} = 180 - \left(\hat{A} + \hat{B}\right) = 180 - 56 - 22 = 102$

Applying Law of Sines,

$\frac{12.2}{\sin} 56 = \frac{b}{\sin} 22 = \frac{c}{\sin} 102$

$b = \frac{12 \cdot \sin 22}{\sin} \left(56\right) \approx 5.42$

$c = \frac{12.2 \cdot \sin 102}{\sin} 56 \approx 14.39$

color(brown)("Perimeter of the triangle " = P = 12.2 + 5.42 + 14.39 = 32.01 " units"

color(purple)("Area of the triangle " A_t = (1/2) a b sin C = (1/2) 12.2 * 5.42 * sin 102 ~~ 32.34 " sq units"