NDA Mathematics 4 September 2022 | Height and Distance Question 4 | Mathematics

There are two points P and Q due south of a leaning tower, which leans towards north. P is at a distance x and Q is at distance y from the foot of the tower (x > y). The angles of elevation of the top of the tower from P and Q are 15° and 75° respectively.

At what height is the top of the tower above the ground level ?

$\frac{\text{x−y}}{2\sqrt{3}}$

$\frac{\text{x−y}}{4\sqrt{3}}$

$\frac{\text{x−y}}{4}$

$\frac{\text{x−y}}{2}$

NDA Mathematics 4 September 2022

MCQ (Single Correct Answer)

+2.5

-0.83

If θ is the inclination of the tower to the horizontal, then what is cot θ equal to ?

2 + $\frac{\sqrt{3}(\text{x−y})}{\text{x+y}}$

2 − $\frac{\sqrt{3}(\text{x−y})}{\text{x+y}}$

2 + $\frac{\sqrt{3}(\text{x+y})}{\text{x−y}}$

2 − $\frac{\sqrt{3}(\text{x+y})}{\text{x−y}}$

NDA Mathematics 4 September 2022

MCQ (Single Correct Answer)

+2.5

-0.83

What is the length of the tower ?

$\frac{\text{x−y}}{2 \sqrt{3}} \sqrt{1+\left\{2+\frac{\sqrt{3}(\text{x+y})}{\text{x−y}}\right\}^2}$

$\frac{\text{x−y}}{2 \sqrt{3}} \sqrt{1+\left\{2−\frac{\sqrt{3}(\text{x+y})}{\text{x−y}}\right\}^2}$

$\frac{\text{x−y}}{4 \sqrt{3}} \sqrt{1+\left\{2+\frac{\sqrt{3}(\text{x+y})}{\text{x−y}}\right\}^2}$

$\frac{\text{x−y}}{4 \sqrt{3}} \sqrt{1+\left\{2−\frac{\sqrt{3}(\text{x+y})}{\text{x−y}}\right\}^2}$

NDA Mathematics 10 April 2022

MCQ (Single Correct Answer)

+2.5

-0.83

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h. At a point on the plane the angles of elevation of the bottom and top of the flagstaff are θ and 2θ respectively. What is the height of the tower ?

h cos θ

h sin θ

h cos 2θ

h sin 2θ

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