Question

hii frnd plzz.. answer this
no spamm!!
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Answer 1

hey there

refer to attachment

Answer 2

Option A...................

Solution :

BD=√p2+q2

∠ABD=∠BDC=α

=> ∠DAB=π−(θ+α)

tanα=p/q

ΔABD

AB/sinθ=BD/sin(π−(θ+α))

=BD/sin(θ+α)

∴AB=BDsinθ/sin(θ+α)

=BD²sinθ/BDsin(θ+α)

=BD²sinθ/BDsinθcosα+BDcosθsinα

=(p2+q2)sinθ/qsinθ+pcosθ.

HOPE THIS HELPS...

STAY HAPPY...