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Exercise 3.2 · Q33

Q.In △ABC\triangle ABC, if A=45∘A = 45^\circ, B=60∘B = 60^\circ then find the ratio of its sides.

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A=45∘,B=60∘  ⟹  C=180∘−45∘−60∘=75∘A=45^\circ,B=60^\circ\implies C=180^\circ-45^\circ-60^\circ=75^\circ. By the Sine Rule, a:b:c=sin⁡A:sin⁡B:sin⁡C=12:32:sin⁡75∘a:b:c=\sin A:\sin B:\sin C=\dfrac{1}{\sqrt2}:\dfrac{\sqrt3}{2}:\sin75^\circ. Using sin⁡75∘=sin⁡(45∘+30∘)=sin⁡45∘cos⁡30∘+cos⁡45∘sin⁡30∘=12⋅32+12⋅12=3+122=6+24\sin75^\circ=\sin(45^\circ+30^\circ)=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ=\dfrac{1}{\sqrt2}\cdot\dfrac{\sqrt3}{2}+\dfrac{1}{\sqrt2}\cdot\dfrac12=\dfrac{\sqrt3+1}{2\sqrt2}=\dfrac{\sqrt6+\sqrt2}{4}. Multiplying the whole ratio by 222\sqrt2 to clear denominators: $a:b:c=2:\sqrt6:\dfrac …

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