Question

ABC is an equilateral triangle.Point P is on base BC such that PC=1/3 BC,if AB=6cm.find AP?

Answer 1

here's ur answer.....

Answer 2

Answer:

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=$\frac{1}{3}BC$

therefore PC=2 cm.

Now, using the cosine formula in ΔAPC, we have

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

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

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

$AP^{2}=40-12$

$AP^{2}=28$

$AP=2\sqrt{7}cm$