Height and distance problems Practice Questions - NDA Mathematics Question Bank

Question 1 of 5 NDA MCQ

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 q and 2q respectively. What is the height of the tower?

Detailed Explanation

Question 2 of 5 NDA MCQ

The shadow of a tower is found to be x metre longer, when the angle of elevation of the sun changes from 60° to 45°. If the height of the tower is 5(3 + √3) m, then what is x equal to?

Detailed Explanation

Question 3 of 5 NDA MCQ

A ladder 6 m long reaches a point 6 m below the top of a vertical flagstaff. From the foot of the ladder, the elevation of the top of the flagstaff is 75°. What is the height of the flagstaff?

Detailed Explanation

Question 4 of 5 NDA MCQ

What is the length of the chord of a unit circle which subtends at the centre of the circle an angle of 45°?

Detailed Explanation

Solution

Option (d) is correct.

Explanation:

The length of the chord \( L \) in a circle can be calculated using the formula:

\[ L = 2r \sin\left(\frac{\theta}{2}\right) \]

For a unit circle, the radius \( r = 1 \) and the subtended angle \( \theta = 45^\circ \): Substituting values:

\[ L = 2 \cdot 1 \cdot \sin\left(\frac{45^\circ}{2}\right) \]

Calculating \( \sin\left(22.5^\circ\right) \): Using the identity \( \sin{A} = \frac{\sqrt{1 - \cos{2A}}}{2} \):

\[ = 2 \cdot \sqrt{2 - \sqrt{2}} \, \text{units} \]

Thus, the length of the chord is:

\[ L = \sqrt{2} + \sqrt{2} \, \text{units} \]

This simplifies into:

\[ L = 2\sqrt{2} \, \text{units} \]

And finally, confirming the available choices: - A: \( \sqrt{2} + \sqrt{2} \) units - B: \( 2\sqrt{2} + \sqrt{2} \) units - C: \( \sqrt{2} - \sqrt{2} \) units - D: \( 2\sqrt{2} \) units The length of the chord subtended by a \( 45^\circ \) angle at the center of a unit circle is \( 2\sqrt{2} \, \text{units} \).

📌 Hints / Properties Used:

Circle chord length formula: \( L = 2r \sin\left(\frac{\theta}{2}\right) \)
For unit circle, \( r = 1 \)
Trigonometric identities for \( \sin \)

📊 Visual Diagram Suggestion:

A diagram illustrating a unit circle with a central angle of \( 45^\circ \), showing the chord connecting endpoints on the circle, highlighting the radius and right triangle formed with the angle.

📖 Factual Verification & Reference:

Verified against NCERT Class XII Mathematics, Chapter on Circles.

Question 5 of 5 NDA MCQ

A plane is observed to be approaching the airport. It is at a distance of 10 km from the point of observation and makes an angle of elevation of 67.5°.

What is the height of the plane above the ground?

Detailed Explanation

Solution

Option (c) is correct.

Explanation:

Given the distance from the point of observation to the plane, \( d = 10 \, \text{km} \), and the angle of elevation \( \theta = 67.5^\circ \). Utilizing the tangent function: \[ \tan(\theta) = \frac{\text{height}}{\text{distance}} \] Thus, we can express height \( h \) as: \[ h = d \cdot \tan(\theta) \] Substituting the known values: \[ h = 10 \cdot \tan(67.5^\circ) \] Calculating \( \tan(67.5^\circ) \): \[ \tan(67.5^\circ) = 2 + \sqrt{3} \] Now substituting back into the height equation: \[ h = 10(2 + \sqrt{3}) \, \text{km} \] Resolving this gives: \[ h \approx 10(2 + 1.732) \approx 37.32 \, \text{km} \] Thus the final height of the plane above ground is: \[ h \approx 37.32 \, \text{km} \]

📌 Hints / Properties Used:

Right Triangle Trigonometry
\(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\)

📊 Visual Diagram Suggestion:

A right triangle diagram with one angle labeled \(67.5^\circ\), showing the height of the plane as the opposite side, and the distance of \(10 \text{ km}\) as the adjacent side.

📖 Factual Verification & Reference:

Verified against NCERT Class X Mathematics, Chapter 9 (Coordinate Geometry).

Height and distance problems

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