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Exercise 10(a) · Q2

Q.In △ABC\triangle ABC, if a=5a=5, b=7b=7 and C=60∘C=60^\circ, find the side cc.

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Step 1. Two sides a=5,b=7a=5,b=7 and the included angle C=60∘C=60^\circ are given (SAS), so use

c2=a2+b2−2abcos⁡C.c^2=a^2+b^2-2ab\cos C.

Step 2. Substitute, with cos⁡60∘=12\cos60^\circ=\frac12:

c2=52+72−2(5)(7)(12)=25+49−35.c^2=5^2+7^2-2(5)(7)\left(\frac12\right)=25+49-35.

Step 3. Simplify:

c2=74−35=39.c^2=74-35=39.

Step 4. Take the positive square root:

c=39≈6.245.c=\sqrt{39}\approx6.245.

[!ANSWER] c=39≈6.245c=\sqrt{39}\approx6.245

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