RRB NTPC Mains or Stage 2 Model Paper 10 with solutions

Question : 37 of 120

Marks: +1, -0

From a horizontal distance of 50 m, the angles of elevation of the top and the bottom of a vertical cliff face are 45° and 30° respectively. The height of the cliff face (in metres) is :

$50 /√3$
$50/√2$
$50 /{2√3}$
$50[1-1/√3]$

Validate

Solution:

(d) Let the height of cliff face AD be h metres

InΔDCB , tan 30°

=x/50

1/√ 3=x/ 50 or x = 50/√3

... (i)
In ΔACB , tan 45°

{( x+h)}/ 50 or x + h = 50

...(ii)
From (ii) and (i), we get, h

= 50 – 50/√3

= 50 [1 – 1/√3]

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