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