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In △ABC, m∠A=32°, m∠B=25°, and a=18. Find c to the nearest tenth.

User Las Ten
by
6.3k points

2 Answers

5 votes

Answer:

28.5

Explanation:

I got it correct on founders edtell

User Ahmed Hamdy
by
5.8k points
4 votes

Answer:

28.5

Explanation:

We have been given that in △ABC, m∠A=32°, m∠B=25°, and a=18. We are asked to find the value of c to nearest tenth.

We will use law of sines to solve our given problem.


\frac{a}{\text{sin}(A)}=\frac{b}{\text{sin}(B)}=\frac{c}{\text{sin}(C)}, where a, b and c are opposite sides to angle A, B and C.

First of all, we will find measure of angle C using angle sum property.


m\angle A+m\angle B+m\angle C=180^(\circ)


32^(\circ)+25^(\circ)+m\angle C=180^(\circ)


57^(\circ)+m\angle C=180^(\circ)


57^(\circ)-57^(\circ)+m\angle C=180^(\circ)-57^(\circ)


m\angle C=123^(\circ)

Substituting our values in law of sines, we will get:


\frac{18}{\text{sin}(32^(\circ))}=\frac{c}{\text{sin}(123^(\circ))}

Switch sides:


\frac{c}{\text{sin}(123^(\circ))}=\frac{18}{\text{sin}(32^(\circ))}


\frac{c}{\text{sin}(123^(\circ))}*\text{sin}(123^(\circ))=\frac{18}{\text{sin}(32^(\circ))}*\text{sin}(123^(\circ))


c=(18)/(0.529919264233)*0.838670567945


c=33.9674384664105867*0.838670567945


c=28.487490910261


c\approx 28.5

Therefore, the value of c is 28.5 to the nearest tenth.

User Dashesy
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5.9k points