Question

Find the area of triangle ABC with the given parts. Round to the nearest tenth when necessary.A= 26.3°b= 13.2inc= 7.3 inOption 1: 43.2in^2Option 2: 45.2in^2Option 3: 21.3in^2Option 4: 19.3in^2

Answer 1

ANSWER

Option 3: 21.3 in²

EXPLANATION

Let's draw a diagram of the triangle ABC,

When two sides and the included angle are given, we can find the area as half the product of the side lengths and the sine of the included angle,

$A=\frac{1}{2}\cdot b\cdot c\cdot\sin(A)$

Replace the known values and solve,

$A=\frac{1}{2}\cdot13.2in\cdot7.3in\cdot\sin(26.3\degree)\approx21.3in^2$

Hence, the area of triangle ABC is 21.3 in², rounded to the nearest tenth.