Question

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Answer 1

Let BD be the surface of water. A is the point of observation. P be the position of cloud and C be the position of the reflection of cloud in the lake.

Draw AQ ⊥ PD. Angle of elevation of cloud from point A is 30°

i.e., ∠PAQ = 30° and depression of the reflection of cloud in the lake is 60°

i.e, ∠QAC = 60°

Let PQ = h m

So, ∠PAQ = 30° and ∠QAC = 60°

QD = AB = 20 m CD = PD = (20 + h)m QC = 20 + h + 20 = (40 + h) m BD = AQ From right angled ∆PAQ,

tan 30° = PD/AQ ⇒ 1/√3 = h/AQ ⇒ AQ = h√3 m From right angled ∆AQC,

tan 60° = QC/AQ ⇒ √3 = (40 + h)/h√3 [∵ From equation (i),

AQ = h√3] h√3 × √3 = 40 + h ⇒ 3h = 40 + h ⇒ 2h = 40 ⇒ h = 40/2 = 20 m From right angled ∆PAQ sin 30° = PQ/AP ⇒ 1/2 = h/Ap ⇒ 1/2 = 20/AP ⇒ AP = 20 × 2 = 40 m

Hence, distance of cloud from A = 40

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Answer 2

Step-by-step explanation:

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