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?

1 Answer
SagarStudy
Jan 9, 2016

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

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