Question

A triangle has two corners with angles of `pi / 12` and `(7 pi )/ 8`. If one side of the triangle has a length of `8`, what is the largest possible area of the triangle?

Answer 1

Largest possible area of the triangle `Delta = 24.2822`

Explanation:

Given are the two angles `(7pi)/8` and `pi/12` and the length 8

The remaining angle:

`= pi - (pi/12 + (7pi)/8) = (pi)/24`

I am assuming that length AB (8) is opposite the smallest angle.

Using the ASA

Area`=(c^2*sin(A)*sin(B))/(2*sin(C))`

Area`=( 8^2*sin((pi)/12)*sin((7pi)/8))/(2*sin(pi/24))`

Area`=24.2822`

Answer 2

Area of the largest possible triangle is `24.28` sq.unit.

Explanation:

Angle between Sides `A and B` is `/_c= =15^0`

Angle between Sides `B and C` is `/_a= (7pi)/8=157.5^0 :.`

Angle between Sides `C and A` is

`/_b= 180-(157.5+15)=7.5^0` For largest area of the triangle

`8` should be smallest size , which is opposite to the smallest

angle `/_b= 7.5^0 :. B=8` The sine rule states if `A, B and C` are

the lengths of the sides and opposite angles are `a, b ,c` in a

triangle, then `A/sina = B/sinb=C/sinc ; B=8 :. A/sina=B/sinb` or

`A/sin157.5=8/sin7.5:. A = 8* sin157.5/sin7.5 ~~23.45 (2dp)`unit

Now we know sides `A=23.45 , B=8` and their included angle

`/_c = 15^0`. Area of the triangle is `A_t=(A*B*sinc)/2`

`:.A_t=(23.45*8*sin15)/2 ~~ 24.28` sq.unit

Area of the largest possible triangle is `24.28` sq.unit [Ans]