Question

question 7
using the law of sines to solve for all possible triangles if $\angle b = 50^\circ$, $a = 102$, $b = 49$.
assume $\angle a$ is opposite side $a$, $\angle b$ is opposite side $b$, and $\angle c$ is opposite side $c$.

• if no triangles exist, enter dne for all answers. if multiple triangles exist, input all angles and sides as a list.
• round your solutions to the nearest tenth place.
• a solution write-up is required for this problem.

answers:

• $\angle a = $
• $\angle c = $
• $c = $

Explanation:

Step1: Apply Law of Sines for ∠A

The Law of Sines states $\frac{a}{\sin A}=\frac{b}{\sin B}$. We know $∠B = 50^\circ$, $a = 102$, $b = 49$. Plugging in: $\frac{102}{\sin A}=\frac{49}{\sin 50^\circ}$. Solve for $\sin A$: $\sin A=\frac{102\sin 50^\circ}{49}$. Calculate $\sin 50^\circ\approx0.7660$, so $\sin A\approx\frac{102\times0.7660}{49}\approx\frac{78.132}{49}\approx1.5945$. Since the sine of an angle cannot exceed 1, no triangle exists.

Answer:

∠A: DNE, ∠C: DNE, c: DNE