Bunuel wrote:
In the above figure, PB:PC = 1:2, ∠CBA = 45° and ∠APC = 60°. What is the measure of ∠ACB ?
A. 60°
B. 65°
C. 70°
D. 75°
E. 80°
Sorry I missed out on this earlier.
CEdwardYoganandaIt 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.Attachment:
1.gif [ 9.14 KiB | Viewed 2500 times ]
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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