Question

In ΔABC, if ∠A = 45°, ∠B = 60° then find the ratio of its sides.

Answer 1

By the sine rule,

`a/"sin A" = b/"sin B" = c/"sin C"`

∴ `a/b = "sin A"/"sin B" and b/c = "sin B"/"sin C"`

∴ a : b : c = sin A : sin B : sin C

Given ∠A = 45° and ∠B = 60°

∵ ∠A + ∠B + ∠C = 180°

∴ 45° + 60° + ∠C = 180°

∴ ∠C = 180° – 105° = 75°

Now, sin A = sin 45° = `(1)/sqrt(2)`

sin B = sin 60° = `sqrt(3)/(2)`

And sin C = sin 75° = sin (45° + 30°)

= sin 45° cos 30° + cos 45° sin 30°

= `(1)/(sqrt(2)) xx (sqrt(3))/(2) + (1)/(sqrt(2)) xx (1)/(2)`

= `(sqrt(3))/(2sqrt(2)) + (1)/(2sqrt(2))`

= `(sqrt(3) + 1)/(2sqrt(2))`

∴ The ratio of the sides of ΔABC

= a : b : c

= sin A : sin B : sin C

= `(1)/(sqrt(2)) : (sqrt(3))/(2) : (sqrt(3) + 1)/(2sqrt(2))`

∴ `a : b : c = 2 : sqrt(6) : (sqrt(3) + 1)`.