Question

question 5 of 15, step 1 of 1
how many triangles exist that fit the following criteria?

$b = 40^{circ}, a = 6, b = 5$

Explanation:

Step1: Apply the Law of Sines

By the Law of Sines, $\frac{\sin A}{a}=\frac{\sin B}{b}$. We know $B = 40^{\circ}$, $a = 6$, and $b = 5$. So $\sin A=\frac{a\sin B}{b}=\frac{6\sin40^{\circ}}{5}$.

Step2: Calculate $\sin A$ value

$\sin40^{\circ}\approx0.6428$, then $\sin A=\frac{6\times0.6428}{5}=\frac{3.8568}{5}=0.7714$.

Step3: Analyze the number of solutions for angle A

Since $0 < A<180^{\circ}$, and $\sin A = 0.7714$, there are two possible values for $A$: one acute angle $A_1=\sin^{- 1}(0.7714)\approx50.4^{\circ}$ and one obtuse - angle $A_2 = 180^{\circ}-50.4^{\circ}=129.6^{\circ}$. When $A_1\approx50.4^{\circ}$, $C_1=180^{\circ}-(40^{\circ}+50.4^{\circ}) = 89.6^{\circ}$. When $A_2 = 129.6^{\circ}$, $C_2=180^{\circ}-(40^{\circ}+129.6^{\circ})=10.4^{\circ}$.

Answer:

2