A triangle has sides A, B, and C. The angle between sides A and B is $\frac{3 π}{4}$ $(3pi)/4$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 13, 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 $\frac{3 π}{4}$ $(3pi)/4$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 13, what is the area of the triangle?

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Answer 1 · 2016-04-21T11:30:38

Area of the triangle is $30.93$ $30.93$ Sq units

Explanation:

The angle between sides B & C is $∠ A = \frac{π}{12} = \frac{180}{12} = 15^{0}$ $/_A=pi/12=180/12=15^0$
The angle between sides A & B is $∠ C = 3 \cdot \frac{π}{4} = 3 \cdot \frac{180}{4} = 135^{0}$ $/_C=3*pi/4=3*180/4=135^0$
So Angle $∠ B = 180 - (135 + 15) = 30^{0}$ $/_B=180-(135+15)=30^0$ Using sine law we get $\frac{B}{sin B} = \frac{A}{sin A} or \frac{13}{sin 30} = \frac{A}{sin 15} or A = 13 \cdot \frac{sin 15}{sin 30} = 6.73$ $B/sin B=A/sinAor 13/sin30=A/sin15 or A=13*sin15/sin30=6.73$ Area $= \frac{A . B . sin C}{2} = \frac{13 \cdot 6.73 \cdot sin 135}{2} = 30.93$ $= (A.B.sinC)/2 = (13*6.73*sin135)/2=30.93$ sq.units

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{3 π}{4}$ $(3pi)/4$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 13, 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 3π4(3pi)/4 and the angle between sides B and C is π12pi/12. If side B has a length of 13, 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 $\frac{3 π}{4}$ $(3pi)/4$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 13, what is the area of the triangle?