With the given two angles we can find the 3rd angle by using the concept that sum of all three angles in a triangle is 180^@ or pi180∘orπ:
(3pi)/4 + pi/6 + x = pi3π4+π6+x=π
x = pi - (3pi)/4 - pi/6x=π−3π4−π6
x = pi - (11pi)/12x=π−11π12
x = pi/12x=π12
Hence, the third angle is pi/12π12
Now, let's say
/_A = (3pi)/4, /_B = pi/6 and /_C = pi/12∠A=3π4,∠B=π6and∠C=π12
Using Sine Rule we have,
(Sin /_A)/a = (Sin /_B)/b = (Sin /_C)/csin∠Aa=sin∠Bb=sin∠Cc
where, a, b and c are the length of the sides opposite to /_A, /_B and /_C∠A,∠Band∠C respectively.
Using above set of equations, we have the following:
a = a, b = (Sin /_B)/(Sin /_A)*a, c = (Sin /_C)/(Sin /_A)*aa=a,b=sin∠Bsin∠A⋅a,c=sin∠Csin∠A⋅a
or a = a, b=(Sin (pi/6))/(Sin ((3pi)/4))*a, c=(Sin (pi/12))/(Sin ((3pi)/4))*aora=a,b=sin(π6)sin(3π4)⋅a,c=sin(π12)sin(3π4)⋅a
rArr a = a, b=a/(sqrt2), c=(a*(sqrt(3) - 1))/2⇒a=a,b=a√2,c=a⋅(√3−1)2
Now, to find the longest possible perimeter of the triangle
P = a + b + cP=a+b+c
Assuming, a = 9a=9, we have
a = 9, b = 9/sqrt2 and c = (9*(sqrt(3) - 1))/2a=9,b=9√2andc=9⋅(√3−1)2
rArrP = 9 + 9/(sqrt2) +(9*(sqrt(3) - 1))/2⇒P=9+9√2+9⋅(√3−1)2
or P = (9 (1 + sqrt[2] + sqrt[3]))/2orP=9(1+√2+√3)2
or P~~18.66 orP≈18.66
Assuming, b = 9b=9, we have
a = 9sqrt2, b = 9 and c = (9*(sqrt(3) - 1))/sqrt2a=9√2,b=9andc=9⋅(√3−1)√2
rArrP = 9sqrt2 + 9 +(9*(sqrt(3) - 1))/sqrt2⇒P=9√2+9+9⋅(√3−1)√2
or P = (9 (2 + sqrt[2] + sqrt[6]))/2orP=9(2+√2+√6)2
or P~~26.39 orP≈26.39
Assuming, c = 9c=9, we have
a = 18/(sqrt3 - 1), b = (9sqrt2)/(sqrt3 - 1) and c = 9a=18√3−1,b=9√2√3−1andc=9
rArrP = 18/(sqrt3 - 1) + (9sqrt2)/(sqrt3 - 1) +9⇒P=18√3−1+9√2√3−1+9
or P = (9 (1 + sqrt[2] +sqrt[3]))/(sqrt[3] - 1)orP=9(1+√2+√3)√3−1
or P~~50.98 orP≈50.98
Therefore, Longest possible perimeter of the given triangle is (9 (1 + sqrt[2] +sqrt[3]))/(sqrt[3] - 1)9(1+√2+√3)√3−1
