Question

If ΔPAB and ΔQBA are similar triangles and AB is the diameter of the circle, what is the ratio of PA to PB?



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Answer 1

Answer:

√3 : 1

Step-by-step explanation:

ΔPAB~ΔQBA (Given)

∴ ∠PAB=∠QBA            (1)

In ΔAOB:

∠AOB=120° (Given)

Let ∠PAB=∠QBA=x    (from (1))

∴ ∠AOB+∠BAO+∠ABO=180°  (Angle sum property)

∴ 120°+x+x=180°      (∵∠BAO=∠PAB, ∠ABO=∠QBA)

120°+2x=180°

2x=180°-120°

2x=60°

∴ x=60°/2

x=30°                                (2)

Also, given that: AB is the diameter

∴ ∠APB=∠BQA=90°        (∵ Angle in a semi-circle is equal to 90°)

In ΔPAB:

∠APB=90°

∠PAB=30° (from (2))

∴ PA/PB=cotA         (∵ cotθ=base/perpendicular)

∴PA/PB=cot30°

∴PA/PB=√3             (∵ cot30°=√3

∵ PA:PB=√3:1

Hence, the ratio of PA to PB is √3 : 1.

HOPE IT HELPS!