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Exercise 10(b) · Q3

Q.In △ABC\triangle ABC, if A=60∘A=60^\circ, b=8b=8, c=10c=10, find the inradius rr.

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Step 1. Find the missing side using the Cosine Rule:

a2=b2+c2−2bccos⁡A=82+102−2(8)(10)(12)=64+100−80=84 ⟹ a=221.a^2=b^2+c^2-2bc\cos A=8^2+10^2-2(8)(10)\left(\frac12\right)=64+100-80=84\ \Longrightarrow\ a=2\sqrt{21}.

Step 2. Compute the area directly from the given SAS data (no need to wait for aa):

Δ=12bcsin⁡A=12(8)(10)sin⁡60∘=40×32=203.\Delta=\frac12bc\sin A=\frac12(8)(10)\sin60^\circ=40\times\frac{\sqrt3}2=20\sqrt3.

Step 3. Compute the semi-perimeter:

s=a+b+c2=221+8+102=21+9.s=\frac{a+b+c}{2}=\frac{2\sqrt{21}+8+10}{2}=\sqrt{21}+9.

Step 4. Apply r=Δ/sr=\Delta/s and rationalise:

r=20321+9×9−219−21=203(9−21)81−21=203(9−21)60=3(9−21)3.r=\frac{20\sqrt3}{\sqrt{21}+9}\times\frac{9-\sqrt{21}}{9-\sqrt{21}}=\frac{20\sqrt3(9-\sqrt{21})}{81-21}=\frac{20\sqrt3(9-\sqrt{21})}{60}=\frac{\sqrt3(9-\sqrt{21})}{3}.

Step 5. Expand: 93−633=93−373=33−7≈5.196−2.646=2.550\dfrac{9\sqrt3-\sqrt{63}}{3}=\dfrac{9\sqrt3-3\sqrt7}{3}=3\sqrt3-\sqrt7\approx5.196-2.646=2.550.

[!ANSWER] r=33−7≈2.550r=3\sqrt3-\sqrt7\approx2.550.

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