A triangle has sides A, B, and C. The angle between sides A and B is $\frac{5 π}{12}$ $(5pi)/12$ and the angle between sides B and C is $\frac{π}{6}$ $pi/6$ . If side B has a length of 1, 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 $\frac{5 π}{12}$ $(5pi)/12$ and the angle between sides B and C is $\frac{π}{6}$ $pi/6$ . If side B has a length of 1, what is the area of the triangle?

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Answer 1 · 2016-11-14T12:17:51

The area of the triangle is $\frac{1}{4} s q . u n i t$ $1/4 sq.unit$

Explanation:

The angle between sides $A and B$ $A and B$ is $∠ c = \frac{5 π}{12} = \frac{5 \cdot 180}{12} = 75^{0}$ $/_c=(5pi)/12=(5*180)/12=75^0$ .

The angle between sides $B and C$ $B and C$ is $∠ a = \frac{π}{6} = \frac{180}{6} = 30^{0}$ $/_a=pi/6=180/6=30^0$

The angle between sides $A and C$ $A and C$ is $∠ b = 180 - (75 + 30) = 75^{0}$ $/_b=180-(75+30)=75^0$

$∠ b = ∠ c = 75^{0}$ $/_b= /_c=75^0$ . So it is an isocelles triangle , having opposite sides equal. So $B = C = 1$ $B=C=1$ and their included angle $∠ a = 30^{0}$ $/_a=30^0$

Hence the area of the triangle is $A_{t} = \frac{B \cdot C \cdot sin a}{2} = \frac{1 \cdot 1 \cdot sin 30}{2} = \frac{1}{4} s q . u n i t$ $A_t=(B*C*sin a)/2=(1*1*sin30)/2=1/4 sq.unit$ [Ans]

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{5 π}{12}$ $(5pi)/12$ and the angle between sides B and C is $\frac{π}{6}$ $pi/6$ . If side B has a length of 1, what is the area of the triangle?

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Explanation:

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A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/125π12(5pi)/12 and the angle between sides B and C is pi/6π6pi/6. If side B has a length of 1, what is the area of the triangle?

1 Answer

Explanation:

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{5 π}{12}$ $(5pi)/12$ and the angle between sides B and C is $\frac{π}{6}$ $pi/6$ . If side B has a length of 1, what is the area of the triangle?