Day 14: Heights & Distances | Secondary Stage Mathematics | Apex Institute of Maths and Sciences

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Day 14: Heights & Distances | Secondary Stage Mathematics | Apex Institute of Maths and Sciences

Day 14: Heights & Distances | Secondary Stage Mathematics

Apex Institute of Maths and Sciences

Level 1: The Quest (Concept)

Welcome, Explorer! Today we use the power of Trigonometry to measure things we cannot reach, like the height of a mountain or the distance of a ship at sea. This is called Heights and Distances.

Two Secret Angles:

🔭 Angle of Elevation: Looking UP at an object above your eye level.

⚓ Angle of Depression: Looking DOWN at an object below your eye level.

Fact: If person A looks up at B, and B looks down at A, the Elevation angle equals the Depression angle!

Level 2: Power-Ups (Tools)

The Magic Formula: In 90% of these problems, we use the Tangent ratio because we deal with Height (Opposite) and Distance (Adjacent).

$$\tan(\theta) = \frac{\text{Height (Opposite)}}{\text{Distance (Adjacent)}}$$
Remember your Power Values:
$\tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577$
$\tan(45^\circ) = 1$
$\tan(60^\circ) = \sqrt{3} \approx 1.732$

Level 3: Mini-Boss Battles

Scenario 1: The Tall Tower
A surveyor stands 50m away from a mobile tower. The angle of elevation to the top is $45^\circ$. Since $\tan(45^\circ) = 1$, the height of the tower is exactly 50m!
Scenario 2: The Lighthouse Guard
A guard looks down from a 100m lighthouse at a boat. If the angle of depression is $30^\circ$, the boat is $100\sqrt{3}$ meters away from the base.

Level 4: Home Quests

Task 1: Shadow Stalker
Measure your own height. Then, measure the length of your shadow on the ground. Use a calculator to find $\tan^{-1}(\frac{\text{Height}}{\text{Shadow}})$ to find the Sun’s angle of elevation!
Task 2: Tree Hunter
Stand at a distance from a tree where your eyes look up at roughly a $45^\circ$ angle to the top. Measure your distance to the tree; that distance is roughly the tree’s height!

Final Boss: Practice Test

1. EASY If the angle of elevation of the sun is $45^\circ$, then the length of the shadow of a pole of height $h$ is:

Magic Solution: Since $\tan(45^\circ) = 1$ and $\tan = \frac{\text{Height}}{\text{Shadow}}$, then $1 = h/\text{Shadow}$, so Shadow $= h$.

2. EASY The line drawn from the eye of an observer to the point in the object viewed by the observer is called:

Magic Solution: The straight line connecting the eye and the object is the ‘Line of Sight’.

3. EASY If the shadow of a tower is $1/\sqrt{3}$ times its height, the angle of elevation is:

Magic Solution: $\tan(\theta) = \text{Height} / (\text{Height}/\sqrt{3}) = \sqrt{3}$. $\tan(60^\circ) = \sqrt{3}$.

4. EASY Angle of depression is formed when the object is _____ the horizontal level.

Magic Solution: Depression means looking ‘down’, so the object must be below your eye level.

5. MODERATE A ladder 15m long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the wall, the height of the wall is:

Magic Solution: Angle with wall is $60^\circ$, so angle with ground is $30^\circ$. $\sin(30^\circ) = \text{Height}/15 \Rightarrow 0.5 = \text{Height}/15 \Rightarrow 7.5$m.

6. MODERATE The angle of elevation of a kite is $60^\circ$. If the string is 100m long, find the height of the kite.

Magic Solution: $\sin(60^\circ) = \text{Height}/100 \Rightarrow (\sqrt{3}/2) \times 100 = 50\sqrt{3}$m.

7. MODERATE If a tower 30m high casts a shadow $10\sqrt{3}$m long, the sun’s elevation is:

Magic Solution: $\tan(\theta) = 30/(10\sqrt{3}) = 3/\sqrt{3} = \sqrt{3}$. Thus $\theta = 60^\circ$.

8. MODERATE Two poles of equal height stand on either side of a 100m road. At a point between them, elevation angles are $30^\circ$ and $60^\circ$. The height of poles is:

Magic Solution: $h = d / (\cot(30) + \cot(60)) = 100 / (\sqrt{3} + 1/\sqrt{3}) = 100 / (4/\sqrt{3}) = 25\sqrt{3}$m.

9. COMPLEX A man on a cliff observes a boat at an angle of depression of $30^\circ$. After 6 mins, the angle becomes $60^\circ$. Time taken to reach the cliff is:

Magic Solution: The distance ratio between the two points is 2:1. If $2x$ takes 6 mins, $x$ takes 3 mins.

10. COMPLEX The angle of elevation of a cloud from a point $h$ meters above a lake is $\alpha$ and the angle of depression of its reflection is $\beta$. The height of the cloud is:

Magic Solution: Using the property of reflection and trigonometry, the height is derived as $h \frac{\tan\beta + \tan\alpha}{\tan\beta – \tan\alpha}$.
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