Question

Solve the triangle.

B = 72°, b = 12, c = 8

Answer 1

The answer is 12^2=x^2+8^2-16x(cos(72°)
x^2-4.9443x=80
(x-2.4721)^2=86.1115
x-2.4721=±9.2796
x=11.7517

B=72°, b=12, c=8, a=11.8
C=arcsin(8/12*sin(72°))=39.3°
A=68.7°

Answer 2

Part 1)
`C=39.3\°`

Part 2)
`A=68.7\°`

Part 3)
`a=11.8\ units`

Explanation:

we know that

Applying the law of sines

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

In this problem we have

`B=72\°`

`b=12\ units`

`c=8\ units`

Step 1

Find the measure angle C

`(b)/(sin(B)) =(c)/(sin(C))`

substitute the values and solve for C

`(12)/(sin(72\°)) =(8)/(sin(C))\\ \\sin(C)=sin(72\°)*(8/12)\\ \\sin(C)= 0.6340\\ \\C=39.3\°`

Step 2

Find the measure of angle A

we know that

The sum of the internal angles of the triangle must be equal to
`180` degrees

so

`m<A+m<B+m<C=180\°`

we have

`m<B=72\°`

`m<C=39.3\°`

`m<A+72\°+39.3\°=180\°`

`m<A=180\°-72\°-39.3\°=68.7\°`

Step 3

Find the measure of side a

Applying the law of cosines

`a^(2)=b^(2)+c^(2)-2(b)(c)cos(A)`

substitute

`a^(2)=12^(2)+8^(2)-2(12)(8)cos(68.7\°)`

`a^(2)=208-(192)cos(68.7\°)`

`a^(2)=138.26`

`a=11.8\ units`