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Exercise 10(a) · Q1

Q.In △ABC\triangle ABC, if b=3b=3, c=4c=4 and A=60∘A=60^\circ, find the side aa.

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Step 1. Since we are given two sides b=3,c=4b=3,c=4 and the included angle A=60∘A=60^\circ (SAS data), the Cosine Rule is the correct tool:

a2=b2+c2−2bccos⁡A.a^2=b^2+c^2-2bc\cos A.

Step 2. Substitute the given values, using cos⁡60∘=12\cos60^\circ=\frac12:

a2=32+42−2(3)(4)(12)=9+16−12.a^2=3^2+4^2-2(3)(4)\left(\frac12\right)=9+16-12.

Step 3. Simplify:

a2=25−12=13.a^2=25-12=13.

Step 4. Take the positive square root (a side length is positive):

a=13≈3.606.a=\sqrt{13}\approx3.606.

[!ANSWER] a=13≈3.606a=\sqrt{13}\approx3.606

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