Window and Opposite House Problem: Height Proof with Angles of Elevation and Depression | Class 10 Maths

Quick answer: If a window is $h$ metres above the ground and the angles of elevation and depression of the top and bottom of the opposite house are $\alpha$ and $\beta$, the height of the opposite house is $H = h(1 + \tan\alpha\cot\beta)$, which can also be written as $H = \frac{h(\tan\alpha + \tan\beta)}{\tan\beta}$.

Most height-and-distance questions use a single right triangle, like finding the height of a tree from its shadow. Longer board questions combine two right triangles, an observer above the ground, and both an angle of elevation and an angle of depression.

One of the most common of these in CBSE Class 10 Maths, Chapter 9 (Some Applications of Trigonometry), is the window and opposite house problem:

A window of a house is $h$ metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be $\alpha$ and $\beta$ respectively. Prove that the height of the other house is $h(1 + \tan\alpha\cot\beta)$ metres.

In this guide, we label the figure, prove the result step by step, and solve a numerical example.

What You Will Learn

• How to draw and label the figure for an observer at a window $h$ metres above the ground
• Why the horizontal line from the window separates the angle of elevation ($\alpha$) and the angle of depression ($\beta$)
• A complete proof of $H = h(1 + \tan\alpha\cot\beta)$
• The equivalent form $H = \frac{h(\tan\alpha + \tan\beta)}{\tan\beta}$
• A solved example with $h = 15\text{ m}$, $\alpha = 30^\circ$ and $\beta = 45^\circ$
• Common mistakes that cost marks

1. Figure and Notation

• Let $A$ be the window and $B$ the point on the ground directly below it, so $AB = h$.
• Let $CD$ be the opposite house, with top $C$, bottom $D$ and height $CD = H$.
• Let $BD = x$ be the width of the lane.
• Draw the horizontal line $AE$ from the window, parallel to the ground, meeting the opposite house at $E$.

                     C
                   / |
                 /   |  H - h
               / α   |
        A ----------- E
        | \ β        |
      h |   \        |  h
        |     \      |
        B ----------- D
              x
  A = window, CD = opposite house, BD = lane

Facts from the Figure

1. $ABDE$ is a rectangle, so $ED = AB = h$ and $AE = BD = x$.

2. The height of the opposite house splits into two parts:

$H = CD = CE + ED = CE + h \implies CE = H - h$

3. The angles at the window $A$:

• Angle of elevation of the top $C$: $\angle CAE = \alpha$
• Angle of depression of the bottom $D$: $\angle EAD = \beta$
• Since $AE \parallel BD$, the alternate angles are equal: $\angle ADB = \beta$

2. Proof: $H = h(1 + \tan\alpha\cot\beta)$

Step 1: Use the Lower Triangle to Find $x$

In right triangle $ABD$ (right-angled at $B$):

$\tan\beta = \frac{AB}{BD} = \frac{h}{x}$

$x = \frac{h}{\tan\beta} = h\cot\beta \qquad \text{...(1)}$

Step 2: Use the Upper Triangle to Find $H - h$

In right triangle $AEC$ (right-angled at $E$):

$\tan\alpha = \frac{CE}{AE} = \frac{H - h}{x}$

$H - h = x\tan\alpha \qquad \text{...(2)}$

Step 3: Eliminate $x$

Substituting $x = h\cot\beta$ from (1) into (2):

$H - h = h\cot\beta\tan\alpha$

$H = h + h\tan\alpha\cot\beta = h(1 + \tan\alpha\cot\beta)$

Hence proved: the height of the opposite house is $h(1 + \tan\alpha\cot\beta)$ metres.

Step 4: Equivalent Form

Since $\cot\beta = \frac{1}{\tan\beta}$:

$H = h\left(1 + \frac{\tan\alpha}{\tan\beta}\right) = \frac{h(\tan\beta + \tan\alpha)}{\tan\beta}$

Both forms are the same result, and either may appear in the question.

3. Solved Example

Question

A window of a house is $15\text{ m}$ above the ground. From the window, the angles of elevation and depression of the top and bottom of another house on the opposite side of the street are $30^\circ$ and $45^\circ$ respectively. Find the height of the opposite house. (Take $\sqrt{3} = 1.732$)

Solution

Step 1: Write the given values.

• $h = 15\text{ m}$
• $\alpha = 30^\circ$, so $\tan 30^\circ = \frac{1}{\sqrt{3}}$
• $\beta = 45^\circ$, so $\cot 45^\circ = 1$

Step 2: Apply the formula.

$H = h(1 + \tan\alpha\cot\beta) = 15\left(1 + \frac{1}{\sqrt{3}} \times 1\right) = 15\left(\frac{\sqrt{3} + 1}{\sqrt{3}}\right)$

Step 3: Rationalise the denominator.

$H = \frac{15\sqrt{3}(\sqrt{3} + 1)}{3} = 5\sqrt{3}(\sqrt{3} + 1) = 5(3 + \sqrt{3})$

Step 4: Substitute $\sqrt{3} = 1.732$.

$H = 5(3 + 1.732) = 5 \times 4.732 = 23.66\text{ m}$

Answer: The height of the opposite house is $5(3 + \sqrt{3})\text{ m} \approx 23.66\text{ m}$.

Check: The lane width is $x = 15\cot 45^\circ = 15\text{ m}$, and the part above the window is $15\tan 30^\circ = \frac{15}{\sqrt{3}} \approx 8.66\text{ m}$. So $H \approx 15 + 8.66 = 23.66\text{ m}$, which matches.

4. Summary Table

5. Exam Tips and Common Mistakes

Exam Tip: State clearly that $ABDE$ is a rectangle. This is what justifies $ED = h$ and $AE = x$, and examiners look for it.

Exam Tip: In a proof, keep $\tan\alpha$ and $\cot\beta$ as symbols. Do not convert them to decimals.

Common Mistake: Measuring the angle of depression from the vertical wall. Angles of elevation and depression are always measured from the horizontal line through the observer's eye, here the line $AE$.

Frequently Asked Questions

What is the formula for the height of the opposite house?

$H = h(1 + \tan\alpha\cot\beta)$, where $h$ is the height of the window, $\alpha$ is the angle of elevation of the top and $\beta$ is the angle of depression of the bottom of the opposite house.

How do you find the width of the lane?

From the lower triangle, $\tan\beta = \frac{h}{x}$, so the width of the lane is $x = h\cot\beta$.

Why is the angle of depression equal to the angle at the bottom of the opposite house?

The horizontal line from the window is parallel to the ground, so the angle of depression and the angle $\angle ADB$ are alternate angles, and alternate angles are equal.

What is the height of the opposite house if h = 15 m, α = 30° and β = 45°?

$H = 15(1 + \frac{1}{\sqrt{3}}) = 5(3 + \sqrt{3}) \approx 23.66\text{ m}$.