1. From the top of a cliff 90 metre high, the angles of depression of the top and bottom of a tower are observed to be 30° and 60° respectively. The height of the tower is :
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By: anil on 05 May 2019 01.59 am
Given: CD = 90 meter;
∠KBD = 30° & ∠CAD = 60°
BD × cos30° = BK
AD × cos60° = AC
BD × sin30° = DK
AD × sin60° = DC
BK = AC
BD × cos30° = AD × cos60°
√ 3 × BD = AD ____________ (1)
Now, DC = DK + KC
DC - DK = AB [As, KC = AB]
(AD × sin60°) - (BD × sin30°) = AB
Using equation (1)
($$sqrt{3}$$ ×BD×$$frac{sqrt{3}}{2}$$)−BD2=AB(3×BD×32)−BD2=AB
AB = BD _____________ (2)
Now, AD × sin 60° = DC = 90 meter
AD=$$frac{180}{sqrt{3}}$$
From (1)
BD=AD/$$sqrt{3}$$
=180/$$sqrt{3} imes sqrt{3} $$ = 60meter
From (2) AB = 60 meter
Option B is the correct answer.
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∠KBD = 30° & ∠CAD = 60°
BD × cos30° = BK
AD × cos60° = AC
BD × sin30° = DK
AD × sin60° = DC
BK = AC
BD × cos30° = AD × cos60°
√ 3 × BD = AD ____________ (1)
Now, DC = DK + KC
DC - DK = AB [As, KC = AB]
(AD × sin60°) - (BD × sin30°) = AB
Using equation (1)
($$sqrt{3}$$ ×BD×$$frac{sqrt{3}}{2}$$)−BD2=AB(3×BD×32)−BD2=AB
AB = BD _____________ (2)
Now, AD × sin 60° = DC = 90 meter
AD=$$frac{180}{sqrt{3}}$$
From (1)
BD=AD/$$sqrt{3}$$
=180/$$sqrt{3} imes sqrt{3} $$ = 60meter
From (2) AB = 60 meter
Option B is the correct answer.