Question

triangle ABC is an equilateral triangle point p is base BC such that PC=⅓BC. If AB =12cm. find Ap
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Answer 1

Step-by-step explanation:

It is given that ABC is an equilateral triangle, then AB=BC=AC=6cm and ∠A=∠B=∠C=60°, then according to question, PC=\frac{1}{3}BC

3

1

BC

therefore PC=2 cm.

Now, using the cosine formula in ΔAPC, we have

cos∠C= \frac{AC^{2}+PC^{2}-AP^{2}}{2(AC)(PC)}

2(AC)(PC)

AC

2

+PC

2

−AP

2

cos60°=\frac{6^{2}+2^{2}-AP^{2}}{2(6)(2)}

2(6)(2)

6

2

+2

2

−AP

2\frac{1}{2}=\frac{40-AP^{2}}{24}

2

1 = 24

40−AP 2

AP^{2}=40-12AP

2 =40−12

AP^{2}=28AP

2 =28

AP=2\sqrt{7}cmAP=2

7cm