If in ∆ABC, ∠A = 45°, ∠B = 60° and ∠C = 75°, find the ratio of its sides.

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Solution

Let $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Then,

$\frac{a}{\sin 45 °} = \frac{b}{\sin 60 °} = \frac{c}{\sin 75 °} = k \Rightarrow \frac{a}{\frac{1}{\sqrt{2}}} = \frac{b}{\frac{\sqrt{3}}{2}} = \frac{c}{\frac{1}{2 \sqrt{2}} (1 + \sqrt{3})} [∵ \sin 75 ° = \sin (30 ° + 45 °) = \sin 30 ° \cos 45 ° + \sin 45 ° \cos 30 °]$

On multiplying by $2 \sqrt{2}$ , we get:
$a : b : c = 2 : \sqrt{6} : (1 + \sqrt{3})$

Hence, the ratio of the sides is $2 : \sqrt{6} : (1 + \sqrt{3})$ .

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If in ∆ABC, ∠A = 45°, ∠B = 60° and ∠C = 75°, find the ratio of its sides.

Let asinA=bsinB=csinC=k Then, asin45°=bsin60°=csin75°=k⇒a12=b32=c1221+3 ∵sin75°=sin30°+45°=sin30°cos45°+sin45°cos30° On multiplying by 22 , we get: a : b : c=2 : 6 : 1+3 Hence, the ratio of the sides is 2 : 6 : 1+3.