Question

A = 32°, a = 19, b = 14 solve the triangle using the law of cos

Answer 1

Here is your solution.. hope it helps

Answer 2

Answer:

The measures of the triangle are

$A=32\°,B=23\°,C=125\°$

$a=19\ units, b=14\ units, c=29.4\ units$

Step-by-step explanation:

Step 1

Find the value of angle B

Applying the law of sines

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

we have

$a=19\ units, b=14\ units, A=32\°$

Substitute and solve for B

$\frac{19}{sin(32\°)} =\frac{14}{sin(B)}$

$\frac{19}{sin(32\°)} =\frac{14}{sin(B)}\\ \\sin(B)=(14/19)*sin(32\°)$

$B=arcsin((14/19)*sin(32\°))=23\°$

Step 2

Find the measure of angle C

we know that

The sum of the interior angles of triangle is equal to $180\°$

so

$A+B+C=180\°$

we have

$A=32\°$

$B=23\°$

Substitute and solve for C

$32\°+23\°+C=180\°$

$C=180\°-55\°=125\°$

Step 3

Find the length of the side c

Applying the law of cosines

$c^{2} =a^{2}+b^{2}-2abcos(C)$

we have

$a=19\ units, b=14\ units, C=125\°$

substitute

$c^{2} =19^{2}+14^{2}-2(19)(14)cos(125\°)$

$c^{2} =557-532cos(125\°)$

$c =29.4\ units$

The measures of the triangle are

$A=32\°,B=23\°,C=125\°$

$a=19\ units, b=14\ units, c=29.4\ units$