In triangle ABC, angle A is 81 degrees, angle B is 32 degrees, and side b is 40.2....
Check the final answer first, then review the worked steps.
Problem
In triangle ABC, angle A is 81 degrees, angle B is 32 degrees, and side b is 40.2. Use the law of sines to find side a. Round your answer to the nearest tenth.
Answer
74.9
Step-by-step solution
1. Find the measure of angle C: The sum of angles in a triangle is $180^\circ$. Therefore, $m\angle C = 180^\circ - m\angle A - m\angle B$.
$$m\angle C = 180^\circ - 81^\circ - 32^\circ$$
$$m\angle C = 180^\circ - 113^\circ$$
$$m\angle C = 67^\circ$$
2. Apply the Law of Sines: The Law of Sines states that for any triangle ABC, the ratio of the length of a side to the sine of its opposite angle is constant. Mathematically, this is expressed as:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
We are given $m\angle A = 81^\circ$, $m\angle B = 32^\circ$, and $b = 40.2$. We need to find side $a$. We can use the portion of the law of sines that relates $a$, $A$, $b$, and $B$:
$$\frac{a}{\sin A} = \frac{b}{\sin B}$$
3. Solve for side a: Substitute the known values into the equation:
$$\frac{a}{\sin 81^\circ} = \frac{40.2}{\sin 32^\circ}$$
To solve for $a$, multiply both sides by $\sin 81^\circ$:
$$a = \frac{40.2 \times \sin 81^\circ}{\sin 32^\circ}$$
4. Calculate the value of a: Using a calculator, find the sine values and perform the calculation:
$$a = \frac{40.2 \times 0.9877}{0.5299}$$
$$a = \frac{39.70554}{0.5299}$$
$$a \approx 74.93006$$
5. Round to the nearest tenth: The problem asks to round the answer to the nearest tenth.
$$a \approx 74.9$$