Geometry

Trigonometry

Medium

Two men on either side of a pole of 30 m high observe its top at angles of elevation $\alpha$ and $\beta$ respectively. The distance between the two men is $40\sqrt{3}$ m and the distance between the first man at A and the pole is $30\sqrt{3}$ m.

Figure for CUET Mathematics 2022 10 Aug Shift 1 question 45 (Geometry)

Based on the above information, answer the question:

The value of $\frac{1}{AB^2} + \frac{1}{BC^2}$ is:

$\frac{1}{BD^2}$

$\frac{1}{AD^2}$

$\frac{1}{CD^2}$

$\frac{1}{AC^2}$

✅ Correct Option: 1

Since $\angle ABC = \pi/2$ , triangle ABC is right-angled at B with BD as altitude to hypotenuse AC. The standard result is: $\frac{1}{AB^2} + \frac{1}{BC^2} = \frac{1}{BD^2}$ . (Verification: $AB^2 = 30^2 \cdot 4 = 3600$ , $BC^2 = 30^2 \cdot 4/3 = 1200$ , sum of reciprocals $= 1/3600 + 1/1200 = 4/3600 = 1/900 = 1/30^2 = 1/BD^2$ .)

Question 41

The value of $\alpha$ is

Question 42

$\tan(\alpha + \beta) =$

Question 43

Area of $\triangle ABD$ is

Question 44

$\angle ABC$ is equal to:

The value of 1AB2+1BC2\frac{1}{AB^2} + \frac{1}{BC^2}AB21​+BC21​ is:

Solution

More from this set:

Keyboard Shortcuts

CUET Mathematics Past Year Questions (Topic-Wise)

The value of $\frac{1}{AB^2} + \frac{1}{BC^2}$ is: