Question

Find the area of a triangle when a = 19.2, A = 53.8°, and C = 65.4 Round to the nearest tenth.

Answer 1

181.3 units^2

Explanation:

Just did it on E2020

I used the law of sines to find the c side then I used the equation A=.5*19.2*21.6*sin(60.8)

Answer 2

`A=181.0\ units^2`

Explanation:

step 1

Find the length side c

Applying the law of sines

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

substitute he given values

`(19.2)/(sin(53.8^o))=(c)/(sin(65.4^o))`

solve for c

`c=(19.2)/(sin(53.8^o))sin(65.4^o)\\\\c=21.6\ units`

step 2

Find the measure of angle B

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

so

`A+B+C=180^o`

substitute the given values

`53.8^o+B+65.4^o=180^o\\B=180^o-119.2^o\\B=60.8^o`

step 3

Find the area of the triangle

we know that

The area of the triangle applying the law of sines is equal to

`A=(1)/(2)(a)(c)sin(B)`

substitute the given vales

`A=(1)/(2)(19.2)(21.6)sin(60.8^o)\\\\A=181.0\ units^2`