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Cosine Rule

In a triangle...

Question

In a triangle ABC, AB = 3, BC = 5 and AC =7. AP and CP are internal angle bisectors meeting at P. The square of length of AP is

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Solution

The correct option is D 7
P=Incentre
P lies on angular bisector of A, say AD, D divides BC in the ratio c : b = 3:7
$B D = \frac{5 \times 3}{10} = \frac{3}{2}$
Using cosine rule on $Δ$ lies ABC, ABD,
We get $A D = \frac{3 \sqrt{7}}{2}, P$ divides AD in the ratio
b + c : a = 2 : 1
$∴ A P^{2} = 7$

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