Question

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

Answer 1

Trig!!!-

We can use trig function to find the answer-

we know that tangent is opp/adjacent

Thus c=180-32-25=123 degrees and

tan 123=18/c

This c=34.7539008597 since it is tan 123=0.51792747158 and we use algebra to solve for c and it is 18/tan 123=c and then c=34.8 when we round to the nearest tenth

Vote my answer for the brainiest pls

Answer 2

Answer:

28.5

Step-by-step 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:

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

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

$c=\frac{18}{0.529919264233}*0.838670567945$

$c=33.9674384664105867*0.838670567945$

$c=28.487490910261$

$c\approx 28.5$

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