Line of Sight, Angle of Elevation, Angle of Depression

The concepts of “Line of Sight,” “Angle of Elevation,” and “Angle of Depression” are fundamental in trigonometry and are used to solve real-world problems involving distances and heights. They are primarily based on right-angled triangles.

* Line of Sight: This is the straight line from an observer’s eye to the object being observed. * Angle of Elevation: The angle formed by the line of sight and the horizontal plane when the observer looks *up* at an object. It is always measured from the horizontal upwards. * Angle of Depression: The angle formed by the line of sight and the horizontal plane when the observer looks *down* at an object. It is always measured from the horizontal downwards.

Formulae

The following trigonometric ratios are used to solve problems involving these concepts:

• Sine (sin): $sin(\theta) = \frac{opposite}{hypotenuse}$
• Cosine (cos): $cos(\theta) = \frac{adjacent}{hypotenuse}$
• Tangent (tan): $tan(\theta) = \frac{opposite}{adjacent}$

Where:

• $\theta$ is the angle of elevation or depression.
• “opposite” is the length of the side opposite the angle $\theta$.
• “adjacent” is the length of the side adjacent to the angle $\theta$.
• “hypotenuse” is the length of the side opposite the right angle (the longest side).

Examples

Example-1: A person standing 50 meters away from a building observes the top of the building at an angle of elevation of $30^\circ$. Find the height of the building.

Solution: Let $h$ be the height of the building. The distance from the person to the building (50 meters) is the adjacent side, and the height of the building is the opposite side relative to the angle of elevation. We use the tangent function:

$tan(30^\circ) = \frac{h}{50}$

$h = 50 * tan(30^\circ)$

$h = 50 * \frac{1}{\sqrt{3}} \approx 28.87$ meters

Therefore, the height of the building is approximately 28.87 meters.

Example-2: A ship is observed from the top of a lighthouse, 100 meters high. The angle of depression to the ship is $15^\circ$. Find the distance of the ship from the base of the lighthouse.

Solution: The angle of depression is $15^\circ$. Since the horizontal is parallel to the ground, the angle of elevation from the ship to the top of the lighthouse is also $15^\circ$ (alternate interior angles). Let $d$ be the distance of the ship from the base. The height of the lighthouse (100m) is the opposite side, and $d$ is the adjacent side relative to the angle of elevation.

$tan(15^\circ) = \frac{100}{d}$

$d = \frac{100}{tan(15^\circ)}$

$d \approx \frac{100}{0.2679} \approx 373.21$ meters

Therefore, the distance of the ship from the base of the lighthouse is approximately 373.21 meters.

Real Life Application

The concepts of line of sight, angle of elevation, and angle of depression are widely used in:

• Surveying: Determining distances and heights of land features.
• Navigation: Calculating the position of ships, aircraft, and other objects.
• Architecture and Engineering: Designing buildings, bridges, and other structures.
• Astronomy: Measuring the distances to stars and planets.

Common mistakes by students

• Confusing Angle of Elevation and Depression: Students often incorrectly identify which angle is which or confuse their relationship with the horizontal.
• Incorrectly applying trigonometric ratios: Applying the wrong trigonometric ratio (sin, cos, tan) based on the sides involved (opposite, adjacent, hypotenuse) and the angle given.
• Forgetting Units: Not including units in the final answer.
• Drawing incorrect diagrams: Not sketching a clear diagram, leading to confusion in identifying the sides and angles.

Fun Fact

The use of trigonometric ratios and angles of elevation and depression dates back to ancient civilizations. For example, ancient astronomers used these concepts to measure the Earth’s circumference and the distances to celestial bodies.

Recommended YouTube Videos for Deeper Understanding

Q.1 A bird is sitting on top of a tree. The angle of elevation of the bird from a point on the ground 20 meters away from the base of the tree is $60^\circ$. What is the height of the tree?

Check Solution

Ans: B

Let $h$ be the height of the tree. Then, $\tan(60^\circ) = \frac{h}{20}$. Since $\tan(60^\circ) = \sqrt{3}$, we have $h = 20\sqrt{3}$.

Q.2 From the top of a cliff 100 meters high, the angle of depression of a boat is $30^\circ$. What is the distance of the boat from the base of the cliff?

Check Solution

Ans: B

Let $d$ be the distance of the boat from the base of the cliff. Then, $\tan(30^\circ) = \frac{100}{d}$. Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we have $d = 100\sqrt{3}$.

Q.3 A ladder leans against a wall. The foot of the ladder is 5 meters away from the wall, and the angle of elevation of the ladder is $45^\circ$. How long is the ladder?

Check Solution

Ans: B

Let $L$ be the length of the ladder. Then $\cos(45^\circ) = \frac{5}{L}$. Since $\cos(45^\circ) = \frac{1}{\sqrt{2}}$, we have $L = 5\sqrt{2}$.

Q.4 A person observes the top of a building at an angle of elevation of $45^\circ$. The person walks 50 meters closer to the building, and the angle of elevation becomes $60^\circ$. If the person’s eye level is negligible, what is the height of the building?

Check Solution

Ans: A

Let $h$ be the height of the building, and $x$ be the initial distance from the building. Then, $\frac{h}{x} = \tan(45^\circ) = 1$ and $\frac{h}{x-50} = \tan(60^\circ) = \sqrt{3}$. Therefore, $x = h$ and $h = \sqrt{3}(h-50)$. Solving for $h$, $h(\sqrt{3}-1) = 50\sqrt{3}$. So $h = \frac{50\sqrt{3}}{\sqrt{3}-1} = \frac{50\sqrt{3}(\sqrt{3}+1)}{2} = 25(3+\sqrt{3})$.

Q.5 Two ships are in the sea. The angle of depression of the first ship from the top of a lighthouse is $30^\circ$ and the angle of depression of the second ship from the top of the same lighthouse is $45^\circ$. If the height of the lighthouse is $100$ meters, find the distance between the two ships assuming they are on the same side of the lighthouse.

Check Solution

Ans: A

Let $d_1$ and $d_2$ be the horizontal distances of the first and second ships from the lighthouse respectively. Since $\tan(30^\circ) = \frac{100}{d_1}$ and $\tan(45^\circ) = \frac{100}{d_2}$, we have $d_1 = 100\sqrt{3}$ and $d_2 = 100$. The distance between the ships is $d_1 – d_2 = 100(\sqrt{3} – 1)$.

Next Topic: Heights and Distances: Solving Problems