Question

A triangle has two corners with angles of `pi / 12` and `pi / 12`. If one side of the triangle has a length of `5`, what is the largest possible area of the triangle?

Answer 1

Largest area possible `**6.2506**`

Explanation:

Three angles are `pi/12, pi/12, (5pi)/6`
`a/sin a = b/ sin b = c/ sin c`

To obtain largest possible area,
Length 5 should be opposite to the angle with least value.
`5/sin (pi/12) = b/sin (pi/12) = c / sin ((5pi)/6)`
Side b = 5.
Side `c = (5*sin((5pi)/6)) / sin (pi/12)`
Side `c = (5*sin (pi/6))/sin (pi/12) = 5/(2*sin(pi/12))`
`c = 9.6593`
Area
`s= (5+5+9.6593)/2= 9.8297`
`A= sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(9.8297*4.8297*4.8297*0.1704)`

Area `A = 6.2506`

Answer 2

Area = 6.25

Explanation:

Alternate method :
Area of `Delta = (1/2)bh`
Given triangle is isosceles as two angles are `pi/12` each.
`:. h = 5*sin (pi/12)`
`(1/2)b .= 5*cos(pi/12)`
Area `= 5*5*sin(pi/12)*cos(pi/12)`
`=(25/2)*2sin(pi/12)cos(pi/12)`
`=(25/2)sin(pi/6) = (25/2)(1/2)=25/4=6.25`