A triangle has sides A, B, and C. If the angle between sides A and B is (pi)/6π6, the angle between sides B and C is (5pi)/125π12, and the length of B is 2, what is the area of the triangle?

1 Answer
Jan 3, 2016

Area=1.93184Area=1.93184 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 /_ CC, angle between side "b" and "c" /_ AA and angle between side "c" and "a" by /_ BB.

Note:- the sign /_ is read as "angle".
We are given with /_CC and /_AA. We can calculate /_BB by using the fact that the sum of any triangles' interior angels is pi radian.
implies /_A+/_B+/_C=piA+B+C=π
implies pi/6+/_B+(5pi)/12=piπ6+B+5π12=π
implies/_B=pi-(7pi)/12=(5pi)/12B=π7π12=5π12
implies /_B=(5pi)/12B=5π12

It is given that side b=2.b=2.

Using Law of Sines
(Sin/_B)/b=(sin/_C)/csinBb=sinCc
implies (Sin((5pi)/12))/2=sin((5pi)/12)/csin(5π12)2=sin(5π12)c
implies 1/2=1/c12=1c
implies c=2c=2

Therefore, side c=2c=2

Area is also given by
Area=1/2bcSin/_A=1/2*2*2Sin((7pi)/12)=2*0.96592=1.93184Area=12bcsinA=1222sin(7π12)=20.96592=1.93184square units
implies Area=1.93184Area=1.93184 square units