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

The measures of the triangle are

`A=32\°,B=23\°,C=125\°`

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

Explanation:

Step 1

Find the value of angle B

Applying the law of sines

`(a)/(sin(A)) =(b)/(sin(B))`

we have

`a=19\ units, b=14\ units, A=32\°`

Substitute and solve for B

`(19)/(sin(32\°)) =(14)/(sin(B))`

`(19)/(sin(32\°)) =(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`