If sides A and B of a triangle have lengths of 6 and 12 respectively, and the angle between them is (pi)/8, then what is the area of the triangle?

May 31, 2018

Area of the triangle is $13.78$ sq.unit.

Explanation:

Angle between Sides $A \mathmr{and} B$ is

$\angle c = \frac{\pi}{8} = \frac{180}{8} = {22.5}^{0}$. Sides are $A = 6 , B = 12$

Area of the triangle is ${A}_{t} = \frac{A \cdot B \cdot \sin c}{2}$ or

${A}_{t} = \frac{6 \cdot 12 \cdot \sin 22.5}{2} \approx 13.78$ sq.unit

Area of the triangle is $13.78$ sq.unit [Ans]

May 31, 2018

$\textrm{a r e a} = \frac{1}{2} a b \sin C = \frac{1}{2} \left(6\right) \left(12\right) \sin \left(\frac{\pi}{8}\right) = 18 \sqrt{2 - \sqrt{2}}$

Explanation:

$\sin \left(\frac{\pi}{8}\right) = \sin \left({45}^{\circ} / 2\right) = \sqrt{\frac{1}{2} \left(1 - \cos {45}^{\circ}\right)} = \sqrt{\frac{1}{2} \left(1 - \frac{\sqrt{2}}{2}\right)} = \frac{1}{2} \sqrt{2 - \sqrt{2}}$

13.8

Explanation:

Let's plot this triangle on an imaginary two-dimensional graph so that:

Your given angle (we can call it $\theta$) is located at the origin, with line A on the $x$ axis and line B slanting upwards and rightwards at the angle $\pi$/$8$ from the origin.

We can now calculate the y-component of line B as $12 \cdot \sin \left(\theta\right)$ which is approximately 4.6 units.

We can now multiply the 'base' of the triangle (A) by the 'height' of the triangle (${b}_{y}$) and we get $6 \cdot 4.6 = 27.6$. Halve that and you get your area.