Question

In △ABC, m∠A=39 △ A B C , m ∠ A = 39 °, a=11 a = 11 , and b=13 b = 13 . Find c c to the nearest tenth.

Answer 1

Answer:

$c=17.5\ units$

Step-by-step explanation:

step 1

Find the measure of angle B

Applying the law of sines

$\frac{a}{sin(A)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{11}{sin(39\°)}=\frac{13}{sin(B)}$

$sin(B)={sin(39\°)}\frac{13}{11}$

$B=arcsin[sin(39\°)\frac{13}{11}]$

$B=48.1\°$

step 2

Find the measure of angle C

Remember that the sum of the interior angles of a triangle must be equal to 180 degrees

so

$A+B+C=180\°$

substitute the given values

$39\°+48.1\°+C=180\°$

$87.1\°+C=180\°$

$C=180\°-87.1\°=92.9\°$

step 3

Find the length side of c

Applying the law of sines

$\frac{a}{sin(A)}=\frac{c}{sin(C)}$

substitute the given values

$\frac{11}{sin(39\°)}=\frac{c}{sin(92.9\°)}$

$c=\frac{11}{sin(39\°)}sin(92.9\°)$

$c=17.5\ units$