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Question
A man on deck of a ship is 10 m above the water level. He observes that the angle of elevation at the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and height of the cliff.
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Solution
Given:
Height of man above water = 10 m point A.
Let D be the foot base of the cliff at water level and C the top of the cliff.
Let x = horizontal distance AD ship to cliff.
Let h = height of the cliff CD.
Step-wise calculation:
1. From A, angle of depression to D is 30°.
In the right triangle ADN horizontal AD = x, vertical drop AN = 10:
`tan 30^circ = "Opposite"/"Adjacent"`
= `10/x`
⇒ `x = 10/(tan 30^circ)`
= `10/(1/sqrt(3))`
= `10sqrt(3)`
= 17.32 m
2. From A, angle of elevation to C is 45°.
In the right triangle ANC vertical rise CN = h – 10, horizontal AN = x:
`tan 45^circ = (h - 10)/x = 1`
⇒ h – 10 = x
⇒ h = x + 10
= `10sqrt(3) + 10`
= 27.32 m
Distance of the cliff from the ship = `10sqrt(3)` m ≈ 17.32 m.
Height of the cliff = `10(1 + sqrt(3))` m ≈ 27.32 m.
