If sides A and B of a triangle have lengths of 9 and 7 respectively, and the angle between them is $\frac{3 π}{8}$ $(3pi)/8$ , then what is the area of the triangle?

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If sides A and B of a triangle have lengths of 9 and 7 respectively, and the angle between them is $\frac{3 π}{8}$ $(3pi)/8$ , then what is the area of the triangle?

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Answer 1 · 2016-02-27T06:15:27

If the angle between two sides A,B be $θ$ $theta$ the area of the triangle will be = $\frac{1}{2} \cdot A \cdot B \cdot sin θ$ $1/2*A*B*sintheta$
here A = 9 unit
B = 7 unit
And $θ = 3 \frac{π}{8}$ $theta=3pi/8$
So area $= \frac{1}{2} \cdot 9 \cdot 7 \cdot sin (3 \frac{π}{8})$ $=1/2*9*7*sin(3pi/8)$ =29.10 squnit

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If sides A and B of a triangle have lengths of 9 and 7 respectively, and the angle between them is $\frac{3 π}{8}$ $(3pi)/8$ , then what is the area of the triangle?

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If sides A and B of a triangle have lengths of 9 and 7 respectively, and the angle between them is $\frac{3 π}{8}$ $(3pi)/8$ , then what is the area of the triangle?