Question

question 8 of 10
in the following triangle, find the values of the angles ( b ) and ( b ), which are the best approximations to the solutions of this ambiguous case.

a. ( b = 70.5^{circ} ) or ( b = 109.5^{circ} )
b. ( b = 65^{circ} ) or ( b = 115^{circ} )
c. ( b = 67^{circ} ) or ( b = 102.2^{circ} )
d. ( b = 56.2^{circ} ) or ( b = 123.8^{circ} )

Explanation:

Step1: Apply Law of Sines

We know in a triangle, the Law of Sines states $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Here, we have a triangle with angle $A = 45^\circ$, side $a = 22$, and side $b = 16.5$ (opposite angle $B$). So, using $\frac{\sin B}{b}=\frac{\sin A}{a}$, we substitute the values: $\frac{\sin B}{16.5}=\frac{\sin 45^\circ}{22}$.

Step2: Solve for $\sin B$

First, calculate $\sin 45^\circ=\frac{\sqrt{2}}{2}\approx0.7071$. Then, $\sin B=\frac{16.5\times\sin 45^\circ}{22}=\frac{16.5\times0.7071}{22}\approx\frac{11.667}{22}\approx0.5303$.

Step3: Find the acute angle $B$

Now, find the angle whose sine is approximately $0.5303$. Using a calculator, $B\approx\sin^{-1}(0.5303)\approx32.0^\circ$? Wait, no, wait, maybe I mixed up the sides. Wait, wait, the side opposite angle $A$ (45°) is 22? Wait, no, looking at the diagram, the side with length 22 is opposite to angle $B$? Wait, no, let's re - examine. The angle at the left is 45°, the side adjacent to it (the base) has two segments: $B'$ and $B$, and the other sides: the left - most side is 22, and the two equal sides are 16.5. Wait, maybe the triangle is an ambiguous case (SSA). Let's correct the Law of Sines application. Let's denote: angle $A = 45^\circ$, side $a = 16.5$ (the side opposite angle $A$? No, no, the side with length 22 is opposite angle $B$? Wait, no, the angle at the left is 45°, the side from the left vertex to the top vertex is 22, and the two sides from the top vertex to the base are 16.5. So, in the triangle with angle 45°, side 22 (opposite angle $B$) and side 16.5 (opposite angle 45°? No, that can't be. Wait, maybe the correct notation is: angle $A = 45^\circ$, side $a = 16.5$, side $c = 22$, and we need to find angle $C$ (which is $B$). Wait, I think I made a mistake in the side - angle correspondence. Let's start over.

Let's consider the triangle with angle $A = 45^\circ$, side $a = 16.5$ (the side opposite angle $A$), and side $c = 22$ (the side opposite angle $C$ (which is $B$)). Then by Law of Sines: $\frac{\sin C}{c}=\frac{\sin A}{a}$. So $\sin C=\frac{c\times\sin A}{a}=\frac{22\times\sin 45^\circ}{16.5}$.

Calculate $\sin 45^\circ\approx0.7071$, so $\sin C=\frac{22\times0.7071}{16.5}=\frac{15.5562}{16.5}\approx0.9428$.

Now, the acute angle $C_1=\sin^{-1}(0.9428)\approx70.5^\circ$. Then the obtuse angle $C_2 = 180^\circ - 70.5^\circ=109.5^\circ$. This matches option A.

Answer:

A. $B = 70.5^\circ$ or $B' = 109.5^\circ$