A triangle has sides A, B, and C. Sides A and B have lengths of 10 and 8, respectively. The angle between A and C is $(11pi)/24$ and the angle between B and C is $(3pi)/8$ . What is the area of the triangle?

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Answer 1 · 2016-02-05T14:42:22

$"Area" = 20 " units"^2$

Let $alpha$ be the angle opposite to the side $A$ , $beta$ be the angle opposite to the side $B$ and $gamma$ be the angle opposite to the side $C$ .

Thus, you have:

$A = 10$ , $B = 8$ , $beta= (11 pi)/24$ and $alpha = (3 pi)/8$ .

Let's find out the length of the third angle first.

As the sum of all three angles in the triangle must be $180^@ = pi$ , you know that

$gamma = pi - alpha - beta = pi - (11 pi)/24 - (3 pi)/8 = pi/6$

Now you have the sides $A$ and $B$ and $gamma$ , the angle between those two sides. With this information, you can use the formula

$"Area" = 1/2 A * B * sin gamma$

$= 1/2 * 10 * 8 * sin(pi/6)$

$= 5 * 8 * 1/2$

$= 20 " units"^2$

A triangle has sides A, B, and C. Sides A and B have lengths of 10 and 8, respectively. The angle between A and C is (11pi)/24(11pi)/24 and the angle between B and C is (3pi)/8 (3pi)/8. What is the area of the triangle?