Question

use the law of sines to solve for all possible triangles that satisfy the given conditions. (if an answer does not exist, enter dne. round your answers to one decimal place.) a = 26, b = 14, ∠a = 116° ∠b = ° ∠c = ° c = resources read it watch it submit answer

Explanation:

Step1: Use Law of Sines to find ∠B

By the Law of Sines, $\frac{\sin B}{b}=\frac{\sin A}{a}$. So, $\sin B=\frac{b\sin A}{a}$. Substituting $a = 26$, $b = 14$, and $\angle A=116^{\circ}$, we have $\sin B=\frac{14\sin116^{\circ}}{26}\approx\frac{14\times0.8988}{26}\approx0.484$. Then $B=\sin^{- 1}(0.484)\approx29.0^{\circ}$.

Step2: Find ∠C

Since the sum of angles in a triangle is $180^{\circ}$, $\angle C = 180^{\circ}-\angle A-\angle B$. Substituting $\angle A = 116^{\circ}$ and $\angle B\approx29.0^{\circ}$, we get $\angle C=180^{\circ}-116^{\circ}-29.0^{\circ}=35.0^{\circ}$.

Step3: Use Law of Sines to find side c

Again, by the Law of Sines $\frac{c}{\sin C}=\frac{a}{\sin A}$. So, $c=\frac{a\sin C}{\sin A}$. Substituting $a = 26$, $\angle A = 116^{\circ}$, and $\angle C = 35.0^{\circ}$, we have $c=\frac{26\sin35.0^{\circ}}{\sin116^{\circ}}\approx\frac{26\times0.5736}{0.8988}\approx16.6$.

Answer:

$\angle B = 29.0^{\circ}$
$\angle C = 35.0^{\circ}$
$c = 16.6$