In a - ABC- if - A -45- - B -60- and - C -75- find the ratio of its sides-

By the sine formula, we have a : b : c = sin A : sin B : sin C = sin 45^∘ : sin 60^∘ : sin 75^∘ = 1/√(2) : √(3)/2 : (√(3) + 1)/2 √(2) = 2 : √(6) : ( √(3) + 1). ...

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Standard XII

Mathematics

Higher Order Derivatives

In a ∠ ABC, i...

Question

In a $∠ A B C,$ if $∠ A = 45^{\circ}, ∠ B = 60^{\circ}, a n d ∠ C = 75^{\circ},$ find the ratio of its sides.

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Solution

By the sine formula, we have

$a : b : c = s i n A : s i n B : s i n C$

= $s i n 45^{\circ} : s i n 60^{\circ} : s i n 75^{\circ}$

= $\frac{1}{\sqrt{2}} : \frac{\sqrt{3}}{2} : \frac{(\sqrt{3} + 1)}{2 \sqrt{2}} = 2 : \sqrt{6} : (\sqrt{3} + 1) .$

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In a ∠ABC, if ∠A=45∘,∠B=60∘, and ∠C=75∘, find the ratio of its sides.

By the sine formula, we have a:b:c=sin A:sin B:sin C = sin 45∘:sin 60∘:sin 75∘ = 1√2:√32:(√3+1)2√2=2:√6:(√3+1).

In a $∠ A B C,$ if $∠ A = 45^{\circ}, ∠ B = 60^{\circ}, a n d ∠ C = 75^{\circ},$ find the ratio of its sides.

By the sine formula, we have

$a : b : c = s i n A : s i n B : s i n C$

= $s i n 45^{\circ} : s i n 60^{\circ} : s i n 75^{\circ}$

= $\frac{1}{\sqrt{2}} : \frac{\sqrt{3}}{2} : \frac{(\sqrt{3} + 1)}{2 \sqrt{2}} = 2 : \sqrt{6} : (\sqrt{3} + 1) .$