A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{2}$ $(pi)/2$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 26, what is the area of the triangle?

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Answer 1 · 2018-02-17T04:19:52

$A_{t} = (\frac{1}{2}) b h = 90.5668$ $A_t = (1/2) b h = color (green)(90.5668)$

$ˆ A = \frac{π}{12}, ˆ C = \frac{π}{2}, b = 26$ $hatA = pi/12, hatC = pi/2, b = 26$

$ˆ B = π - (\frac{π}{2} - \frac{π}{12}) = \frac{5 π}{12}$ $hatB = pi - (pi/2 - pi/12) = (5pi)/12$

It’s a right triangle. Hence, area $A_{t} = (\frac{1}{2}) a \cdot b$ $A_t = (1/2) a * b$

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To find side a

$\frac{a}{sin A} = \frac{b}{sin B}$ $a / sin A = b / sin B$

$a = \frac{26 \cdot sin (\frac{π}{12})}{sin (\frac{5 π}{12})} = 6.9667$ $a = (26 * sin (pi/12)) / sin ((5pi)/12) = 6.9667$

$A_{t} = (\frac{1}{2}) \cdot 6.9667 \cdot 26 = 90.5668$ $A_t = (1/2) * 6.9667 * 26 = color (green)(90.5668)$

A triangle has sides A, B, and C. The angle between sides A and B is π2(pi)/2 and the angle between sides B and C is π12pi/12. If side B has a length of 26, what is the area of the triangle?