A triangle has sides A, B, and C. If the angle between sides A and B is $\frac{π}{8}$ $(pi)/8$ , the angle between sides B and C is $\frac{2 π}{3}$ $(2pi)/3$ , and the length of side B is 5, what is the area of the triangle?

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Answer 1 · 2017-07-14T07:10:40

The area of the triangle is $= 9.52 u^{2}$ $=9.52u^2$

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The angles are

$ˆ C = \frac{1}{8} π$ $hat(C)=1/8pi$

$ˆ A = \frac{2}{3} π$ $hat(A)=2/3pi$

This is an isoceles triangle.

$ˆ B = π - (\frac{1}{8} π + \frac{2}{3} π) = \frac{5}{24} π$ $hat(B)=pi-(1/8pi+2/3pi)=5/24pi$

$b = 5$ $b=5$

We apply the sine rule to the triangle

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

$\frac{a}{sin (\frac{2}{3} π)} = \frac{5}{sin (\frac{5}{24} π)}$ $a/sin(2/3pi)=5/sin(5/24pi)$

$a = 5 \cdot \frac{sin (\frac{2}{3} π)}{sin (\frac{5}{24} π)} = 7.11$ $a=5*sin(2/3pi)/sin(5/24pi)=7.11$

The area of the triangle is

$= \frac{1}{2} \cdot a \cdot b sin (ˆ C)$ $=1/2*a*bsin(hat(C))$

$= \frac{1}{2} \cdot 7.11 \cdot 7 \cdot sin (\frac{1}{8} π)$ $=1/2*7.11*7*sin(1/8pi)$

$= 9.52$ $=9.52$

A triangle has sides A, B, and C. If the angle between sides A and B is π8(pi)/8, the angle between sides B and C is 2π3(2pi)/3, and the length of side B is 5, what is the area of the triangle?