Question

A triangle has sides A,B, and C. If the angle between sides A and B is `(pi)/3`, the angle between sides B and C is `pi/6`, and the length of B is 15, what is the area of the triangle?

Answer 1

this is a 30-60-90 angle triangle .
length of hypoteneuse is given that is `B=15`
using trignometry
#A=B×sin30° C=B×cos30°#
Area of `triangle` `ABC =A×C×1/2`
`=1/2×B×sin30°×B×cos30°`
`=1/2×B^2×1/2×sqrt3/2`
`=1/2×15^2×sqrt3/4`
`=1/2×225×sqrt3/8`
`~~48.7139289629`

Answer 2

`225sqrt(3)/8`

Explanation:

We know that `pi/6 "radians"=30^o` and `pi/6 "radians"=60^o`, so we are dealing with a 30-60-90 triangle.

If side B touches both the 30-degree angle and the 60-degree angle, then side B is opposite the 90-degree angle and is the hypotenuse of the triangle.

Since we know that the sides of a 30-60-90 triangle follow the pattern `x`-`xsqrt(3)`-`2x`, we can say that:

`2x=15`

`x=15/2`

`xsqrt(3)=15/2*sqrt(3)`

Now that we know the base and height of the triangle, we can find the area.

`"Area" = 1/2*(x)*(xsqrt(3))`

`=1/2*15/2*15/2*sqrt(3)`

`=225sqrt(3)/8`