A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{3}$ $pi/3$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 14, 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 $\frac{π}{3}$ $pi/3$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 14, what is the area of the triangle?

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Answer 1 · 2016-03-25T13:21:57

Area $= 22.740979141751$ $=22.740979141751" "$ square units

Explanation:

From the given:
Angles $A = \frac{π}{12} = 15^{\circ}$ $A=pi/12=15^@$ and $C = \frac{π}{3} = 60^{\circ}$ $C=pi/3=60^@$
Therefore $B = 105^{\circ}$ $B=105^@$
but side $b = 14$ $b=14$

We only need one side to solve for the Area.

By the sine law
$\frac{a}{sin A} = \frac{b}{sin B}$ $a/sin A=b/sin B$

$a = \frac{b \cdot sin A}{sin B}$ $a=(b*sin A)/sin B$

$a = \frac{14 \cdot {sin 15}^{\circ}}{{sin 105}^{\circ}}$ $a=(14*sin 15^@)/sin 105^@$

$a = 3.7512886940358$ $a=3.7512886940358$

Now we can use the formula for area:

Area $= \frac{1}{2} \cdot a \cdot b \cdot sin C$ $=1/2*a*b*sin C$
Area $= \frac{1}{2} \cdot (3.7512886940358) \cdot 14 \cdot {sin 60}^{\circ}$ $=1/2*(3.7512886940358)*14*sin 60^@$

Area $= 22.740979141751$ $=22.740979141751" "$ square units

God bless....I hope the explanation is useful.

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{3}$ $pi/3$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 14, what is the area of the triangle?

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Explanation:

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A triangle has sides A, B, and C. The angle between sides A and B is $\frac{π}{3}$ $pi/3$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 14, what is the area of the triangle?