Question

A triangle has sides A,B, and C. If the angle between sides A and B is `(7pi)/12`, the angle between sides B and C is `pi/12`, and the length of B is 2, what is the area of the triangle?

Answer 1

`Area=0.5773` square units

Explanation:

First of all let me denote the sides with small letters `a`, `b` and `c`.
Let me name the angle between side `a` and `b` by `/_ C`, angle between side `b` and `c` by `/_ A` and angle between side `c` and `a` by `/_ B`.

Note:- the sign `/_` is read as "angle".
We are given with `/_C` and `/_A`. We can calculate `/_B` by using the fact that the sum of any triangles' interior angels is `pi` radian.
`implies /_A+/_B+/_C=pi`
`implies pi/12+/_B+(7pi)/12=pi`
`implies/_B=pi-(pi/12+(7pi)/12)=pi-(8pi)/12=pi-(2pi)/3=pi/3`
`implies /_B=pi/3`

It is given that side `b=2`.

Using Law of Sines

`(Sin/_B)/b=(sin/_C)/c`

`implies (Sin((pi)/3))/2=sin((7pi)/12)/c`

`implies (0.86602)/2=(0.96592)/c`

`implies 0.43301=0.96592/c`

`implies c=0.96592/0.43301`

`implies c=2.2307`

Therefore, side `c=2.2307`

Area is also given by
`Area=1/2bcSin/_A`

`implies Area=1/2*2*2.2307Sin((pi)/12)=2.2307*0.2588=0.5773` square units
`implies Area=0.5773` square units