Question

ВС. .
(iii) A ABC is an equilateral triangle. Point P is on base BC such that PC =1/3BC.
If AB = 12 cm. find AP.​

Answer 1

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

therefore PC=2 cm.

Now, using the cosine formula in ΔAPC, we have

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

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

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

$AP^{2}=40-12$

$AP^{2}=28$

$AP=2\sqrt{7}cm$