A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{5 π}{8}$ $(5 pi )/ 8$ . If one side of the triangle has a length of $15$ $15$ , what is the largest possible area of the triangle?

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Answer 1 · 2018-07-23T12:55:35

Area of the largest possible triangle is $192.05$ $192.05$ sq.unit.

Angle between sides $A and B$ $A and B$ is $∠ c = \frac{π}{4} = \frac{180}{4} = 45^{0}$ $/_c= pi/4=180/4=45^0$

Angle between sides $B and C$ $BandC$ is $∠ a = \frac{5 π}{8} = \frac{900}{8} = {112.5}^{0}$ $/_a= (5pi)/8=900/8=112.5^0$

Angle between sides $C and A$ $C and A$ is

$∠ b = 180 - (112.5 + 45) = {22.5}^{0}$ $/_b= 180-(112.5+45)=22.5^0$ . For largest area of triangle

$15$ $15$ should be smallest side , which is opposite to the smallest

angle , i.e $B = 15$ $B=15$ The sine rule states if $A, B and C$ $A, B and C$ are the

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

triangle,then, $A/sin a = B/sin b=C/sin c ; B=15 :. A/sin a=B/sin b$

$:. A= B* sin a/sin b :. A= 15 * sin 112.5/sin 22.5~~ 36.21$

Now we know sides $A=36.21 , B=15$ and their included angle

$/_c = 45^0$ . Area of the triangle is $A_t=(A*B*sin c)/2$ or

$A_t=(36.21*15*sin 45)/2 ~~192.05$ sq.unit.

Area of the largest possible triangle is $192.05$ sq.unit [Ans]

A triangle has two corners with angles of pi / 4 π4 pi / 4 and (5 pi )/ 8 5π8 (5 pi )/ 8 . If one side of the triangle has a length of 15 1515 , what is the largest possible area of the triangle?