A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{π}{2}$ $(pi)/2$ , the angle between sides B and C is $\frac{π}{6}$ $pi/6$ , and the length of B is 18, what is the area of the triangle?

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Answer 1 · 2018-05-26T01:26:22

Area of triangle $A_{t} = 93.51$ $color(crimson)(A_t = 93.51$ sq units

$b = 18, ˆ A = \frac{π}{6}, ˆ C = \frac{π}{2}$ $b = 18, hat A = pi/6, hat C = pi/2$

$ˆ B = π - \frac{π}{6} - \frac{π}{2} = \frac{π}{3}$ $hat B = pi - pi/6 - pi/2 = pi/3$

It’s a right triangle.

As per Law of Sines,

$a = \frac{b sin A}{sin B}$ $a = (b sin A)/ sin B$

$a = \frac{18 \cdot sin (\frac{π}{6})}{sin (\frac{π}{3})} = 10.38$ $a = (18 * sin (pi/6) )/ sin (pi/3) = 10.38$

Area of triangle $A_{t} = (\frac{1}{2}) a b sin C$ $A_t = (1/2) a b sin C$

$A_{t} = (\frac{1}{2}) \cdot 10.39 \cdot 18 \cdot sin (\frac{π}{2}) = 93.51$ $A_t = (1/2) * 10.39 * 18 * sin (pi/2) = 93.51$

A triangle has sides A,B, and C. If the angle between sides A and B is π2(pi)/2, the angle between sides B and C is π6pi/6, and the length of B is 18, what is the area of the triangle?