A triangle has sides A, B, and C. The angle between sides A and B is $\frac{5 π}{12}$ $(5pi)/12$ and the angle between sides B and C is $\frac{π}{12}$ $pi/12$ . If side B has a length of 9, what is the area of the triangle?

Question

1607 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-10T04:26:21

$10.8$ $10.8$

We can compute the third angle with the other two angles as the sum of the angles in a triangle is $180^{\circ}$ $180^circ$

$Third angle = 180 - (75 + 15) = 180 - 90 = 90^{\circ}$ $"Third angle"=180-(75+15)=180-90=90^circ$

As the triangle contains a $right angle$ $"right angle"$ ,we can use trigonometry

enter image source here

We need to find one more side to find the area of the triangle

So, we can use

$tan (θ) = \frac{opposite}{hypotenuse}$ $color(orange)(tan(theta)=("opposite") /(" hypotenuse")$

$\to tan (a) = \frac{A}{9}$ $rarrtan(a)=A/9$

$\to tan (15) = \frac{A}{9}$ $rarrtan(15)=A/9$

$\to 0.267 = \frac{A}{9}$ $rarr0.267=A/9$

$\to A = 0.267 \cdot 9$ $rarrA=0.267*9$

$\Rightarrow A = 2.4$ $rArrcolor(green)(A=2.4$

enter image source here

We can calculate the area of the triangle

$Area of triangle = \frac{1}{2} \cdot h \cdot b$ $color(blue)("Area of triangle"=1/2*h*b$

Where,

$h = height = 2.4$ $color(red)(h="height"=2.4$

$b = b a s e = 9$ $color(red)(b=base=9$

$:."Area"=1/2*2.4*9$

$rarr1/cancel2^1*cancel2.4^1.2*9$

$rarr1.2*9$

$rArrcolor(green)(10.8$

A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/125π12(5pi)/12 and the angle between sides B and C is pi/12π12pi/12. If side B has a length of 9, what is the area of the triangle?