In the above figure, PB/PC = 1/2, ∠CBA = 45° and ∠APC = 60°. What is

Using the sine rule, I could get to sin(120 - x) = 2*sin15
We need to solve for x here

Bunuel can you help us with the solution.

Don’t quite understand how to do this one. Please help.

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I didn't know how to solve either. I figured since ∠PAB is 15 (because ∠APB = 120 and 120 + 45 = 165 so 180 - 165 = 15) and since 15 is opposite x, then the angle opposite 2x is 30. This would make the missing angle 90...

Clearly that doesn't work...

chetan2u Do you know how to solve?
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Anyone figure this one out?

can you solve this problem ?? please ..

Hi chetan2u,

Can you please suggest, how to solve this question?

Sorry I missed out on this earlier. CEdward

Yogananda
It is an easy one, when you use sine rule. But let us use methods that could help us in GMAT.

Let PB be x, so PC=2x.
Take \(\triangle APB\), \(\angle BAP=15\), as \(\angle APB=120\) and \(\angle ABP=45\),

In such questions, drawing that magical line is the key to the solution.
Draw a perpendicular from C to O on AP.

STEP I:
Take \(\triangle COP\), which becomes 30-60-90
Therefore, \(CO=\sqrt{3}x\) and \(PO=x\)

STEP II:
Take \(\triangle BOP\).
Here, \(PO=BP=x\), so \(\angle POB= \angle PBO=30\)
Now, draw the perpendicular PN, which will divide the isosceles triangle BOP in two equal parts, eac triangle being 30-60-90.
Here, \(ON=NB=\frac{\sqrt{3}x}{2}\). Thus OB=\(\sqrt{3}x\)

STEP III:
Take \(\triangle AOB\), which is an isosceles triangle with \(\angle OBA= \angle BAO=15\)
Therefore, \(AO=OB=\sqrt{3}x\)

STEP IV:
Finally take \(\triangle AOC\), which has sides \(AO=OC=\sqrt{3}x\), and makes the triangle an isosceles triangle with \(\angle OCA= \angle CAO=45\), with third angle COA as 90.

\(\angle ACP=\angle ACO+ \angle OCP=45+30=75\)


D
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Got it.
Thank you for the explanation chetan2u.

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