A triangle has sides A, B, and C. Sides A and B have lengths of 7 and 5, respectively. The angle between A and C is $\frac{7 π}{24}$ $(7pi)/24$ and the angle between B and C is $\frac{5 π}{8}$ $(5pi)/8$ . What is the area of the triangle?

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Answer 1 · 2018-02-17T08:44:41

Since $\frac{a}{sin A} \neq \frac{b}{sin B}$ $a/ sin A != b / sin B$ , cannot form a triangle

Given : $a = 7, b = 5, ˆ A = \frac{5 π}{8}, ˆ B = \frac{7 π}{24}$ $a = 7, b = 5, hatA = (5pi)/8, hatB= (7pi)/24$

$ˆ C = π - (\frac{5 π}{8} + \frac{7 π}{24}) = \frac{π}{12}$ $hatC = pi - ((5pi)/8 + (7pi)/24) = pi/12$

$\frac{a}{sin A} = \frac{7}{sin (\frac{5 π}{8} = 7.5767}$ $a / sin A = 7 / sin ((5pi)/8 = 7.5767$

$\frac{b}{sin B} = \frac{5}{sin (\frac{7 π}{24})} = 6.3024$ $b / sin B = 5 / sin ((7pi)/24) = 6.3024$

Since $\frac{a}{sin A} \neq \frac{b}{sin B}$ $a/ sin A != b / sin B$ , cannot form a triangle

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A triangle has sides A, B, and C. Sides A and B have lengths of 7 and 5, respectively. The angle between A and C is (7pi)/247π24(7pi)/24 and the angle between B and C is (5pi)/85π8 (5pi)/8. What is the area of the triangle?