A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{2 π}{3}$ $(2pi)/3$ , the angle between sides B and C is $\frac{π}{6}$ $pi/6$ , and the length of B is 2, what is the area of the triangle?

Question

1586 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-17T02:00:50

$A_{b} = 1.732$ $A_b = color(blue)(1.732$

Sum of the three angles of a triangle $= π^{c}$ $= pi^c$

Hence the third angle $ˆ B = π - \frac{2 π}{3} - \frac{π}{6} = \frac{π}{6}$ $hatB = pi - (2pi)/3 - pi/6 = pi/6$

It’s an isosceles triangle with sides a & b equal.

$\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ $a / sin A = b / sin B = c / sin C$

When length 2 corresponds to $∠ B = ∠ (\frac{π}{6})$ $/_B = /_(pi/6)$

$\frac{a}{sin (\frac{π}{6})} = \frac{2}{sin (\frac{π}{6})} = \frac{c}{sin (\frac{2 π}{3})}$ $a/ sin (pi/6) = 2 / sin (pi/6) = c / sin ((2pi)/3)$

$:. a = 2, c = (2 sin ((2pi)/3)) / sin (pi/6) = 3.4641$

Area of triangle $A-b = (1/2) * ac sin B = (1/2) * 2 * 3.4641 * sin (pi/6) = 1.732$

A triangle has sides A,B, and C. If the angle between sides A and B is (2pi)/32π3(2pi)/3, the angle between sides B and C is pi/6π6pi/6, and the length of B is 2, what is the area of the triangle?