friends answer my question
Attachments:
Answers
Answered by
0
AB║ CD, CB⊥AB (Given)
Let DM ⊥ AB
Then, BC=DM=p and CD=BM=q
In ΔBCD, BD =√(p²+q²)
∠ADB = Θ, ∠ABD =∠BDC =∝,
∴∠ADC=Θ+∝ , ∠BAD=180-(Θ+∝)
AB/SinΘ = BD / Sin(180-(Θ+∝)) = AD/sin∝
AB = BD SinΘ/Sin(Θ+∝)
=BD² SinΘ/ BD Sin(Θ+∝)
=(p²+q²) SinΘ/ BD (SinΘCos∝+CosΘSin∝)
=(p²+q²) SinΘ/ BD [SinΘ. q/BD+CosΘ. p/BD]
=( p²+q²)SinΘ/ BD/BD [qsinΘ+pcosΘ]
= (p²+q²)SinΘ/ p cosΘ+q sinΘ
Similar questions