Question

solve each triangle abc that exists.
a = 42.5° a = 8.7 m b = 10.7 m
select the correct choice below and, if necessary, fill in the answer boxes within the choice.
o a. there is only one possible solution for the triangle.
the measurements for the remaining angles b and c and side c are as follows.
b = □°
(round to the nearest tenth as needed.)
c = □°
(round to the nearest tenth as needed.)
c = □
(round to the nearest tenth as needed.)
o b. there are two, possible solutions for the triangle.
the measurements for the solution with the longer side c are as follows.
b₁ = □°
(round to the nearest tenth as needed.)
c₁ = □°
(round to the nearest tenth as needed.)
c₁ = □
(round to the nearest tenth as needed.)
the measurements for the solution with the shorter side c are as follows.
b₂ = □°
(round to the nearest tenth as needed.)
c₂ = □°
(round to the nearest tenth as needed.)
c₂ = □
(round to the nearest tenth as needed.)
o c. there are no possible solutions for this triangle.

Explanation:

Step1: Use the Law of Sines

The Law of Sines states that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. We know $A = 42.5^\circ$, $a = 8.7$ m, and $b = 10.7$ m. First, we solve for $\sin B$:
$$\sin B=\frac{b\sin A}{a}=\frac{10.7\sin(42.5^\circ)}{8.7}$$
Calculate $\sin(42.5^\circ)\approx0.6760$, so:
$$\sin B=\frac{10.7\times0.6760}{8.7}\approx\frac{7.2332}{8.7}\approx0.8314$$

Step2: Find possible values of $B$

Since $\sin B\approx0.8314$, we have two possible solutions for $B$ in the range $0^\circ$ to $180^\circ$:

• $B_1=\arcsin(0.8314)\approx56.3^\circ$
• $B_2 = 180^\circ - 56.3^\circ=123.7^\circ$

We need to check if these angles are valid (i.e., if $A + B<180^\circ$):

• For $B_1 = 56.3^\circ$: $A + B_1=42.5^\circ+56.3^\circ = 98.8^\circ<180^\circ$, valid.
• For $B_2 = 123.7^\circ$: $A + B_2=42.5^\circ+123.7^\circ = 166.2^\circ<180^\circ$, valid. So there are two solutions.

Step3: Calculate $C$ and $c$ for each solution

Solution 1 ($B_1 = 56.3^\circ$):

• $C_1=180^\circ - A - B_1=180^\circ - 42.5^\circ - 56.3^\circ = 81.2^\circ$
• Use the Law of Sines to find $c_1$:

$$c_1=\frac{a\sin C_1}{\sin A}=\frac{8.7\sin(81.2^\circ)}{\sin(42.5^\circ)}$$
$\sin(81.2^\circ)\approx0.9883$, so:
$$c_1=\frac{8.7\times0.9883}{0.6760}\approx\frac{8.5982}{0.6760}\approx12.7\text{ m}$$

Solution 2 ($B_2 = 123.7^\circ$):

• $C_2=180^\circ - A - B_2=180^\circ - 42.5^\circ - 123.7^\circ = 13.8^\circ$
• Use the Law of Sines to find $c_2$:

$$c_2=\frac{a\sin C_2}{\sin A}=\frac{8.7\sin(13.8^\circ)}{\sin(42.5^\circ)}$$
$\sin(13.8^\circ)\approx0.2376$, so:
$$c_2=\frac{8.7\times0.2376}{0.6760}\approx\frac{2.0671}{0.6760}\approx3.1\text{ m}$$

Answer:

• For the solution with the longer side $c$ (Solution 1):

$B_1 = 56.3^\circ$, $C_1 = 81.2^\circ$, $c_1 = 12.7$ m

• For the solution with the shorter side $c$ (Solution 2):

$B_2 = 123.7^\circ$, $C_2 = 13.8^\circ$, $c_2 = 3.1$ m

(So the correct choice is B, as there are two possible solutions.)