A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{π}{6}$ $(pi)/6$ , the angle between sides B and C is $\frac{π}{3}$ $pi/3$ , and the length of B is 3, what is the area of the triangle?

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A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{π}{6}$ $(pi)/6$ , the angle between sides B and C is $\frac{π}{3}$ $pi/3$ , and the length of B is 3, what is the area of the triangle?

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Answer 1 · 2018-07-09T14:32:10

$A = \frac{9}{2} \cdot \sqrt{3}$ $A=9/2*sqrt(3)$

Explanation:

Since the third angle is given by

$π - \frac{π}{6} - \frac{π}{3} = \frac{6 π - π - 2 π}{6} = \frac{3 π}{6} = \frac{π}{2}$ $pi-pi/6-pi/3=(6pi-pi-2pi)/6=(3pi)/6=pi/2$
so we get

$tan (\frac{π}{6}) = \frac{3}{a}$ $tan(pi/6)=3/a$ so

$a = \frac{3}{tan (\frac{π}{6})} = \frac{3}{\frac{\sqrt{3}}{3}} = \frac{9}{\sqrt{3}} = 9 \cdot \frac{\sqrt{3}}{3} = 3 \sqrt{3}$ $a=3/tan(pi/6)=3/(sqrt(3)/3)=9/sqrt(3)=9*sqrt(3)/3=3sqrt(3)$

So
$A = \frac{1}{2} \cdot 3 \cdot 3 \cdot \sqrt{3} = \frac{9}{2} \cdot \sqrt{3}$ $A=1/2*3*3*sqrt(3)=9/2*sqrt(3)$

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A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{π}{6}$ $(pi)/6$ , the angle between sides B and C is $\frac{π}{3}$ $pi/3$ , and the length of B is 3, what is the area of the triangle?

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A triangle has sides A,B, and C. If the angle between sides A and B is (pi)/6π6(pi)/6, the angle between sides B and C is pi/3π3pi/3, and the length of B is 3, what is the area of the triangle?

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A triangle has sides A,B, and C. If the angle between sides A and B is $\frac{π}{6}$ $(pi)/6$ , the angle between sides B and C is $\frac{π}{3}$ $pi/3$ , and the length of B is 3, what is the area of the triangle?