Question

Solve the following triangles. Identify the case # in each given
triangle.

2.) Given parts:

a= 20 b= 10 A= 30°

Answer 1

$a = 20\\ b = 10\\ A = 30°$

To solve for angle B, we use the sine rule,

$\begin{aligned} \dfrac{b}{sinB} &= \dfrac{a}{sinA}\\ \\ \dfrac{10}{sinB} &= \dfrac{20}{sin30}\\ \\ sinB &= \dfrac{10×sin30}{20}\\ \\ sinB &= 0.25\\ B &= 14.5° \end{aligned}$

To solve for angle C,

Sum of internal angles in a triangle = 180°

$\therefore$ A + B + C = 180°

C = 180° - B - A

C = 180° - 14.5° - 30°

$\therefore$ C = 135.5°

To solve for side c, we use the sine rule;

$\begin{aligned} \dfrac{a}{sinA} &= \dfrac{c}{sinC}\\ \\ \dfrac{20}{sin30} &= \dfrac{c}{sin135.5}\\ \\ c &= \dfrac{20× sin135.5}{sin30}\\ \\ c &= 28 \end{aligned}$

$\therefore$ a = 20, b = 10, c = 28

A = 30°, B = 14.5°, C= 135.5°