KarunMendiratta wrote:
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The attachment Triangle.gif is no longer available
In the figure above, \(\frac{PB}{PC} = \frac{1}{2}\), ∠CBA = 45° and ∠APC = 60°. What is the measure of ∠ACB ?
A. 60°
B. 65°
C. 70°
D. 75°
E. 80°
Explanation:Let PB = \(x\) and PC = \(2x\)
Drop a perpendicular from C on AP and name it as D.
In triangle CDP: DCP = 30
CPD = 60
CDP = 90
Since, it is 30-60-90 triangle: CP = \(2x\), PD = \(x\) and CD = \(\sqrt{3}x\)
Now, let us join BD
In triangle PDB:PD = PB = \(x\)
i.e. PDB = PBD = 30
Also, BD = \(\sqrt{3}x\)
In triangle ADB:DAB = DBA = 15
ADB = 150
i.e. BD = AD = \(\sqrt{3}x\)
Lastly,
In triangle ACD:AD = CD = \(\sqrt{3}x\)
Therefore, CAD = ACD = \(a\) (let)
i.e. \(CAD + ACD + 90 = 180\)
\(2a = 90\)
\(a = 45\)
Thus ACB = \(a + 30 = 45 + 30 = 75\)
Hence, option D
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