A triangle has sides A, B, and C. Sides A and B have lengths of 7 and 2, respectively. The angle between A and C is $(11pi)/24$ and the angle between B and C is $(3pi)/8$ . What is the area of the triangle?

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Answer 1 · 2018-06-21T12:50:10

$A=7/2$

The third angle is given by
$pi-3/8*pi-11/24*pi=pi/6$
Using that
$A=1/2*a*b*sin(gamma)$ we get

$A=1/2*7*2*sin(pi/6)$ or

$A=7/2$

Answer 2 · 2018-08-10T15:48:10

Hence with given measurements we cannot form a triangle.

$a = 7, b = 2, hat A = (3pi)/8, hat B = (11pi)/24$

Lawof sines. : a / sin A = b / sin B#

$a / sin A = 7 / sin ((3pi)/8) ~~ 7.5767$

$b / sin B = 2 / sin ((3pi)/8) ~~ 2.1648$

From the above we can see,

$a / sin A != b / sin B$

Answer 3 · 2018-08-10T15:54:06

We cannot form a triangle with given measurements.

$a = 7, b = 2, hat A = (3pi)/8 = (9pi)/24, hat B = (12pi)/24$

Greater side will have greater angle opposite to it.

Since $a > b (7 > 2), hat A$ must be $hat B$

But $hat A (9pi)/24 < hat B (11pi)/24$

Since the values do not satisfy the theorem, we cannot form a triangle with given measurements.

A triangle has sides A, B, and C. Sides A and B have lengths of 7 and 2, respectively. The angle between A and C is (11pi)/24(11pi)/24 and the angle between B and C is (3pi)/8 (3pi)/8. What is the area of the triangle?