Two corners of a triangle have angles of $\frac{π}{8}$ $pi / 8$ and $\frac{π}{12}$ $pi / 12$ . If one side of the triangle has a length of $9$ $9$ , what is the longest possible perimeter of the triangle?

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Answer 1 · 2017-12-13T12:56:01

Longest possible perimeter is $43.48$ $43.48$ unit.

Angle between Sides $A and B$ $A and B$ is $∠ c = \frac{π}{8} = \frac{180}{8} = {22.5}^{0}$ $/_c= pi/8=180/8=22.5^0$

Angle between Sides $B and C$ $B and C$ is $/_a= pi/12=180/12=15^0 :.$

Angle between Sides $C and A$ is

$/_b= 180-(22.5+15)=142.5^0$ For largest perimeter of

triangle $9$ should be smallest side , which is opposite to

smallest angle $:.A=9$ unit.

The sine rule states if $A, B and C$ are the lengths of the sides

and opposite angles are $a, b and c$ in a triangle, then:

$A/sinA = B/sinb=C/sinc ; A=9 :. A/sina=B/sinb$ or

$9/sin 15=B/sin142.5 or B= 9* (sin142.5/sin15) ~~ 21.17 (2dp)$

Similarly $A/sina=C/sinc$ or

$9/sin15=C/sin22.55 or C= 9* (sin22.5/sin15) ~~ 13.31 (2dp)$

Perimeter $P=A+B+C =9+21.17+13.31 ~~43.48$ unit.

Longest possible perimeter is $43.48$ unit. [Ans]

Two corners of a triangle have angles of pi / 8 π8pi / 8 and pi / 12 π12 pi / 12 . If one side of the triangle has a length of 9 99 , what is the longest possible perimeter of the triangle?