A triangle has sides A, B, and C. If the angle between sides A and B is (3pi)/8, the angle between sides B and C is (pi)/2, and the length of B is 12, what is the area of the triangle?

Feb 26, 2016

$A = 72 \left(1 + \sqrt{2}\right)$

Explanation:

Lets take a look at the triangle.

The area of a triangle is given by the formula;

$A = \frac{1}{2} \text{base" xx "height}$

The angle $\frac{\pi}{2}$ is a right angle, so the area of our triangle is;

$A = \frac{1}{2} B \times C$

We are given the length of $B$, and we can solve for $C$ using the tangent formula.

$\tan \theta = \frac{C}{B}$

$\tan \left(\frac{3 \pi}{8}\right) = \frac{C}{12}$

$C = 12 \tan \left(\frac{3 \pi}{8}\right)$

We can solve for $\tan \left(\frac{3 \pi}{8}\right)$ using a calculator or using the half angle formula. Since its not the emphasis of the problem, I'll just include a link to the solution here. The punch line is;

$\tan \left(\frac{3 \pi}{8}\right) = 1 + \sqrt{2}$

So our area function becomes;

$A = \frac{1}{2} B \times C = \frac{1}{2} \left(12\right) \left(12 \left(1 + \sqrt{2}\right)\right)$

$A = 72 \left(1 + \sqrt{2}\right)$