A triangle has sides A, B, and C. The angle between sides A and B is (5pi)/65π6 and the angle between sides B and C is pi/12π12. If side B has a length of 1, what is the area of the triangle?

1 Answer
Costas C.
Jan 3, 2016

Sum of angles gives an isosceles triangle. Half of the enter side is calculated from coscos and the height from sinsin. Area is found like that of a square (two triangles).

Area=1/4Area=14

Explanation:

The sum of all triangles in degrees is 180^o180o in degrees or π in radians. Therefore:

a+b+c=π

π/12+x+(5π)/6=π

x=π-π/12-(5π)/6

x=(12π)/12-π/12-(10π)/12

x=π/12

We notice that the angles a=b. This means that the triangle is isosceles, which leads to B=A=1. The following image shows how the height opposite of c can be calculated:

Image

For the b angle:

sin15^o=h/A

h=A*sin15

h=sin15

To calculate half of the C:

cos15^o=(C/2)/A

(C/2)=A*cos15^o

(C/2)=cos15^o

Therefore, the area can be calculated via the area of the square formed, as shown in the following image:

1
2
3
4
5

Area=h*(C/2)

Area=sin15*cos15

Since we know that:

sin(2a)=2sinacosa

sinacosa=sin(2a)/2

So, finally:

Area=sin15*cos15

Area=sin(2*15)/2

Area=sin30/2

Area=(1/2)/2

Area=1/4

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