A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{3}$ $pi/3$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 18, what is the area of the triangle?

Question

1502 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-27T15:58:21

$37.592 \ \text{unit}^2$

Given in $\triangle ABC$ ,

$\angle A=\pi/12$

$\angle C=\pi/3$

$\therefore \angle B=\pi-\angle A-\angle C$

$=\pi-\pi/12-\pi/3$

$={7\pi}/12$

Now, using sine rule in $\triangle ABC$ as follows

$\frac{A}{\sin\angle A}=\frac{B}{\sin\angle B}$

$\frac{A}{\sin(\pi/12)}=\frac{18}{\sin({7\pi}/12)}$

$A=18\frac{\sin(\pi/12)}{\sin({7\pi}/12)}$

$A=18(2-\sqrt3)$

Now, the area $\Delta$ of triangle $\triangle ABC$ with sides $A=18(2-\sqrt3)$ & $18$ & included angle $\angle C=\pi/3$ is given as

$\Delta=1/2AB\sin \angle C$

$=1/2(18)(18(2-\sqrt3))\sin(\pi/3)$

$=162(2-\sqrt3)(\sqrt3/2)$

$=81(2\sqrt3-3)$

$=37.592 \ \text{unit}^2$

A triangle has sides A, B, and C. The angle between sides A and B is pi/3π3pi/3 and the angle between sides B and C is pi/12π12pi/12. If side B has a length of 18, what is the area of the triangle?