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

By the given statement, ABP forms a new triangle where AB = 6cm, BP = 4cm ([2/3]*6) ie BP is ⅔ of BC and Ang ABP = 60°

Therefore We can use
tan 60° = PA/BP
PA = tan 60° * BP
=√3 * 4
= 4√3 cm

Answer 2

ANSWER

Find out the AP

To proof

as given

ABC is an equilateral triangle.

AB=6cm.

as in the equilateral triangle all the sides of the triangles are equal.

thus AB = BC = CA = 6cm

also given

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

$PC = \frac{1}{3} \times6$

PC = 2cm

in the equilateral triangle all the three angles are of the 60°

Now the the diagram is given below

using cosine formula

in ΔAPC

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

AC = 6 cm,PC = 2cm , C = 60°

put the value

$cos 60 ^{\circ} = \frac{36 +4 - AP^{2}}{2\times6\times2}$

$cos60^{\circ} = \frac{1}{2}$

put the value

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

40 -AP² = 12

AP ²= 40 -12

AP ²= 28

AP = 2√ 7

As √ 7 = 2.65 (approx)

AP = 2 × 2.65

    = 5.3 cm ( approx )

Hence proved