# A triangle has sides A, B, and C. If the angle between sides A and B is (5pi)/12, the angle between sides B and C is (pi)/2, and the length of B is 13, what is the area of the triangle?

Jul 24, 2016

Reqd. area$= 315.36245 s q . u n i t$

#### Explanation:

We denote, by $\hat{A , B}$, the angle btwn. the sides $A \mathmr{and} B$.

hat(A,B)=5pi/12; hat(B,C)=pi/2 rArrhat(C,A)=pi/12

Now, in the $r i g h t {\angle}^{e d} \Delta$ with $\hat{B , C} = \frac{\pi}{2}$, we have,

$\tan \left(\hat{A , B}\right) = \frac{C}{B} \Rightarrow \tan \left(5 \frac{\pi}{12}\right) = \frac{C}{13} \Rightarrow C = 13 \left(3.7321\right) = 48.5173$

Therefore, the Area of the $r i g h t {\angle}^{e d} \Delta$ [with $\hat{B , C} = \frac{\pi}{2}$],

$= \frac{1}{2} \cdot B \cdot C = \frac{1}{2} \cdot 13 \cdot 48.5173 = 315.36245 s q . u n i t$

But, wait a little! If you don't want to use the Table of Natural Tangents , see the Enjoyment of Maths below to find the value of $\tan 5 \frac{\pi}{12}$ :-

$\tan \left(5 \frac{\pi}{12}\right) = \sin \frac{5 \frac{\pi}{12}}{\cos} \left(5 \frac{\pi}{12}\right) = \sin \frac{5 \frac{\pi}{12}}{\sin} \left(\frac{\pi}{2} - 5 \frac{\pi}{12}\right)$

$= \sin \frac{5 \frac{\pi}{12}}{\sin} \left(\frac{\pi}{12}\right) = \sin \frac{5 \theta}{\sin} \theta , w h e r e , \theta = \frac{\pi}{12}$

$= \frac{\sin 5 \theta - \sin 3 \theta + \sin 3 \theta - \sin \theta + \sin \theta}{\sin} \theta$

$= \frac{2 \cos 4 \theta \cdot \sin \theta + 2 \cos 2 \theta \cdot \sin \theta + \sin \theta}{\sin} \theta$

$= \frac{\cancel{\sin \theta} \left(2 \cos 4 \theta + 2 \cos 2 \theta + 1\right)}{\cancel{\sin}} \theta$

$= 2 \cos \left(4 \frac{\pi}{12}\right) + 2 \cos 2 \frac{\pi}{12} + 1$

$= 2 \cdot \frac{1}{2} + 2 \cdot \frac{\sqrt{3}}{2} + 1$

$= 2 + \sqrt{3}$

$2 + 1.7321$

$3.7321$

Isn't this Enjoyable Maths?!