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

Answer:

$A=181.0\ units^2$          

Step-by-step explanation:

step 1

Find the length side c

Applying the law of sines

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

substitute he given values

$\frac{19.2}{sin(53.8^o)}=\frac{c}{sin(65.4^o)}$

solve for c

$c=\frac{19.2}{sin(53.8^o)}sin(65.4^o)\\\\c=21.6\ units$

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

$A=\frac{1}{2}(a)(c)sin(B)$

substitute the given vales

$A=\frac{1}{2}(19.2)(21.6)sin(60.8^o)\\\\A=181.0\ units^2$

Answer 2

Answer:

181.3 units^2

Step-by-step 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)