A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{4}$ $(pi)/4$ and the angle between sides B and C is $\frac{π}{6}$ $pi/6$ . If side B has a length of 2, what is the area of the triangle?

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{4}$ $(pi)/4$ and the angle between sides B and C is $\frac{π}{6}$ $pi/6$ . If side B has a length of 2, what is the area of the triangle?

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Answer 1 · 2017-07-08T15:08:05

The area of the triangle is $= 0.74 u^{2}$ $=0.74u^2$

Explanation:

The angle between side $A$ $A$ and $C$ $C$ is

$= π - (\frac{1}{4} π + \frac{1}{6} π) = π - \frac{5}{12} π = \frac{7}{12} π$ $=pi-(1/4pi+1/6pi)=pi-5/12pi=7/12pi$

Applying the sine rule to the triangle

$\frac{A}{sin (\frac{1}{6} π)} = \frac{B}{sin (\frac{7}{12} π)} = \frac{2}{sin (\frac{7}{12} π)} = 2.07$ $A/sin(1/6pi)=B/sin(7/12pi)=2/sin(7/12pi)=2.07$

$A = 2.07 \cdot sin (\frac{1}{2} π) = 1.04$ $A=2.07*sin(1/2pi)=1.04$

The area of the triangle is

$= \frac{1}{2} \cdot A \cdot B \cdot sin (\frac{1}{4} π) = \frac{1}{2} \cdot 1.04 \cdot 2 \cdot sin (\frac{1}{4} π) = 0.74 u^{2}$ $=1/2*A*B*sin(1/4pi)=1/2*1.04*2*sin(1/4pi)=0.74u^2$

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{4}$ $(pi)/4$ and the angle between sides B and C is $\frac{π}{6}$ $pi/6$ . If side B has a length of 2, what is the area of the triangle?

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A triangle has sides A, B, and C. The angle between sides A and B is π4(pi)/4 and the angle between sides B and C is π6pi/6. If side B has a length of 2, what is the area of the triangle?

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{4}$ $(pi)/4$ and the angle between sides B and C is $\frac{π}{6}$ $pi/6$ . If side B has a length of 2, what is the area of the triangle?