In ∆TAP, ∠TAP = 60°, TA = 6 cm, AP = 8 cm. K is the midpoint of AP. A line from K is produced to meet TP at O such that ∠AKO = 120°. Find the length of OK.
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Correct answer: C
Explanation
The correct answer is option 3 i.e. 3 cm.
In ΔTAP & ΔOKP
∠TAP = ∠OKP =60°
∠TPA = ∠OPK
∠ATP = ∠KOP
So, ΔTAP ~ ΔOKP
TA /AP = OK/KP
⇒ 6/8 = OK/4
⇒ OK = 3 cm