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 7, what is the area of the triangle?

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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 7, what is the area of the triangle?

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Answer 1 · 2016-02-04T05:27:09

$S = 49 (1 - \frac{\sqrt{3}}{2}) ≅ 6.5648$ $S = 49(1-sqrt(3)/2)~=6.5648$

Explanation:

It is a right triangle with catheti $A$ $A$ and $B$ $B$ and hypotenuse $C$ $C$ .
Knowing the length of $B = 7$ $B=7$ and an angle between $B$ $B$ and $C$ $C$ equaled $\frac{π}{12}$ $pi/12$ , we can calculate $A$ $A$ as follows.

Since $\frac{A}{B} = tan (\frac{π}{12})$ $A/B = tan(pi/12)$ ,
it follows that $A = B \cdot tan (\frac{π}{12})$ $A = B*tan(pi/12)$

Area of a triangle is
$S = \frac{1}{2} (A \cdot B) = \frac{1}{2} B^{2} tan (\frac{π}{12})$ $S =1/2(A*B) = 1/2B^2tan(pi/12)$

The latter can be simplified using a formula
$tan (ϕ) = \frac{1 - cos (2 ϕ)}{sin (2 ϕ)}$ $tan(phi) = (1-cos(2phi)) / sin(2phi)$
from which follows:
$tan (\frac{π}{12}) = \frac{1 - cos (\frac{π}{6})}{sin (\frac{π}{6})} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3}$ $tan(pi/12) = (1-cos(pi/6)) / sin(pi/6) = [1-sqrt(3)/2]/(1/2) = 2-sqrt(3)$

Therefore, the area of a triangle is
$S = (\frac{1}{2}) 7^{2} (2 - \sqrt{3}) = 49 (1 - \frac{\sqrt{3}}{2})$ $S = (1/2)7^2(2-sqrt(3))=49(1-sqrt(3)/2)$

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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 7, what is the area of the triangle?

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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 7, what is the area of the triangle?