Question

question 7 of 10
how many solutions does a triangle with side lengths ( a = 4 ), ( a = 112^{circ} ), and ( b = 9 ) have?
use a numeral for your answer.
answer here

Explanation:

Step1: Recall the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. We can use it to find $\sin B$ first.
Given $a = 4$, $A=112^{\circ}$, $b = 9$. From the Law of Sines, $\sin B=\frac{b\sin A}{a}$.

Step2: Calculate $\sin B$

Substitute the values: $\sin B=\frac{9\times\sin(112^{\circ})}{4}$.
We know that $\sin(112^{\circ})=\sin(180^{\circ} - 68^{\circ})=\sin(68^{\circ})\approx0.9272$.
So $\sin B=\frac{9\times0.9272}{4}=\frac{8.3448}{4} = 2.0862$.

Step3: Analyze the range of sine function

The range of the sine function is $[- 1,1]$. Since $\sin B = 2.0862>1$, there is no angle $B$ that satisfies this equation. So, there are no solutions for this triangle.

Answer:

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