Mixed and Real-Life Applications

Combining Elevation and Depression

The richest problems mix both angles. A classic: from a point you see the top of a building at an elevation and its bottom (or a window) at a depression — or you stand on one building and measure the top and bottom of another.

Each angle gives one right triangle; they usually share the horizontal distance. Add or subtract the two vertical pieces to get the total height.

Key Point: Elevation + depression from the same spot share the horizontal distance. Total height = (part above eye) + (part below eye).

[Board Important] The horizontal distance is the common link. Find it from the easier triangle first, then use it in the other.

Two Buildings / Cliff and Object

If you stand on top of a building of height $a$ and look at another building across a distance $d$:

• Depression of its foot $\alpha$: the vertical drop is $a$, so $\tan\alpha = \dfrac{a}{d}$.
• Elevation of its top $\beta$: the rise above your eye is some $b$, so $\tan\beta = \dfrac{b}{d}$.
• The other building's total height is $a + b$.

Key Point: Your own building height appears as the depression's vertical side; the extra rise comes from the elevation.

[JEE/NEET Tip] Solve the depression equation first (it usually contains the known height), get $d$, then plug into the elevation equation.

Rivers, Balloons and Aeroplanes

Real-life set-ups are dressed-up versions of the same triangles:

• River width: elevation of a tree/tower on the far bank gives the horizontal distance across.
• Balloon / aeroplane: height stays constant while horizontal distance changes — a two-triangle problem.
• Two banks of a river: observer between or on one side, leading to elevations on both sides.

Key Point: Strip away the story and find the right triangle(s); the physics is always height vs horizontal distance.

[Board Important] Always convert the words into a labelled figure; the marking scheme rewards the correct diagram and the correct ratio.

Choosing the Right Ratio Quickly

A fast checklist once your triangle is drawn:

• Know vertical & horizontal, want the angle ⇒ $\tan$.
• Know the angle & horizontal, want vertical ⇒ vertical $=$ horizontal $\times \tan$.
• Know the angle & slant, want vertical ⇒ vertical $=$ slant $\times \sin$.
• Know the angle & vertical, want horizontal ⇒ horizontal $=$ vertical $/\tan$.

Key Point: Match the two sides (one known, one wanted) to the ratio that uses exactly those two.

[Board Important] Don't introduce the hypotenuse unless the problem gives or asks for a slant length — most height/distance problems stay with $\tan$.

Solved Examples

Example 1: Elevation and depression together

From the top of a 7 m building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.

Solution:

1. Depression of foot 45°: $\tan 45° = \dfrac{7}{d} \Rightarrow d = 7$ m.
2. Elevation of top 60°: rise above eye $b$, $\tan 60° = \dfrac{b}{7} \Rightarrow b = 7\sqrt3$.
3. Tower height $= 7 + 7\sqrt3 = 7(1 + \sqrt3)$.

Final Answer: $7(1 + \sqrt3) \approx 19.12$ m.

Takeaway: Depression gives $d$; elevation gives the rise; add the building height.

Example 2: Two buildings

From the top of a 20 m building, the angle of depression of the foot of another building is 30° and the elevation of its top is 45°. Find the second building's height.

Solution:

1. $\tan 30° = \dfrac{20}{d} \Rightarrow d = 20\sqrt3$.
2. $\tan 45° = \dfrac{b}{d} = 1 \Rightarrow b = d = 20\sqrt3$.
3. Height $= 20 + 20\sqrt3 = 20(1 + \sqrt3)$.

Final Answer: $20(1 + \sqrt3) \approx 54.64$ m.

Takeaway: Total height = own height + extra rise above eye level.

Example 3: River width

A tree 15 m tall stands on the far bank of a river. From the near bank its top has elevation 30°. Find the river's width.

Solution:

1. $\tan 30° = \dfrac{15}{d} \Rightarrow d = 15\sqrt3$.

Final Answer: $15\sqrt3 \approx 25.98$ m.

Takeaway: Width is the horizontal distance from the elevation triangle.

Example 4: Balloon between observations

A balloon at constant height is seen at elevation 60° from a point; after moving the observer 30 m away (directly), the elevation is 30°. Find the balloon's height.

Solution:

1. Near: $\tan 60° = \dfrac{h}{x} \Rightarrow x = \dfrac{h}{\sqrt3}$.
2. Far: $\tan 30° = \dfrac{h}{x + 30} \Rightarrow x + 30 = h\sqrt3$.
3. $\dfrac{h}{\sqrt3} + 30 = h\sqrt3 \Rightarrow 30 = \dfrac{2h}{\sqrt3} \Rightarrow h = 15\sqrt3$.

Final Answer: $15\sqrt3 \approx 25.98$ m.

Takeaway: Constant-height balloon ⇒ two-triangle problem.

Example 5: Tower on a building

A vertical tower stands on a 50 m building. From a point on the ground, the elevation of the bottom of the tower (building top) is 45° and of the top of the tower is 60°. Find the tower's height.

Solution:

1. $\tan 45° = \dfrac{50}{d} \Rightarrow d = 50$.
2. $\tan 60° = \dfrac{50 + t}{d} = \dfrac{50 + t}{50} = \sqrt3 \Rightarrow 50 + t = 50\sqrt3$.
3. $t = 50\sqrt3 - 50 = 50(\sqrt3 - 1)$.

Final Answer: $50(\sqrt3 - 1) \approx 36.6$ m.

Takeaway: Use the lower angle to fix $d$, then the upper angle for the total height.

Example 6: Aeroplane height

The angle of elevation of an aeroplane from a point on the ground is 60°. After 15 seconds of horizontal flight, the elevation drops to 30°. If the plane flies at a constant height of $1500\sqrt3$ m, find its speed.

Solution:

1. Near: $x_1 = \dfrac{1500\sqrt3}{\tan 60°} = \dfrac{1500\sqrt3}{\sqrt3} = 1500$.
2. Far: $x_2 = \dfrac{1500\sqrt3}{\tan 30°} = 1500\sqrt3\cdot\sqrt3 = 4500$.
3. Distance $= 4500 - 1500 = 3000$ m in 15 s.
4. Speed $= \dfrac{3000}{15} = 200$ m/s.

Final Answer: 200 m/s.

Takeaway: Constant-height flight is a two-triangle problem; speed = distance ÷ time.

Example 7: Both banks of a river

From a point between two poles on opposite banks, the elevations of their tops are 60° and 30°. The poles are equal in height $h$ and the banks are 40 m apart. Find $h$ and the point's position.

Solution:

1. Let the point be $x$ from the first pole, $40 - x$ from the second. $\tan 60° = \dfrac{h}{x}$, $\tan 30° = \dfrac{h}{40 - x}$.
2. $h = x\sqrt3$ and $h = \dfrac{40 - x}{\sqrt3}$. So $x\sqrt3 = \dfrac{40 - x}{\sqrt3} \Rightarrow 3x = 40 - x \Rightarrow x = 10$.
3. $h = 10\sqrt3$.

Final Answer: $h = 10\sqrt3 \approx 17.32$ m; the point is 10 m from the first pole.

Takeaway: Equate the two expressions for $h$ to locate the point.

Example 8: Shadow lengthens

The shadow of a tower is $40$ m longer when the sun's elevation is 30° than when it is 60°. Find the tower's height.

Solution:

1. At 30°: shadow $= \dfrac{h}{\tan 30°} = h\sqrt3$. At 60°: shadow $= \dfrac{h}{\tan 60°} = \dfrac{h}{\sqrt3}$.
2. $h\sqrt3 - \dfrac{h}{\sqrt3} = 40 \Rightarrow \dfrac{2h}{\sqrt3} = 40 \Rightarrow h = 20\sqrt3$.

Final Answer: $20\sqrt3 \approx 34.64$ m.

Takeaway: Lower sun ⇒ longer shadow; the difference gives the height.

Example 9: Two boats, depression

From the top of a $40\sqrt3$ m tower, the depressions of two boats due east are 30° and 60°. Find the distance between them.

Solution:

1. Near (60°): $x_1 = \dfrac{40\sqrt3}{\sqrt3} = 40$. Far (30°): $x_2 = \dfrac{40\sqrt3}{1/\sqrt3} = 40\sqrt3\cdot\sqrt3 = 120$.
2. Distance $= 120 - 40 = 80$ m.

Final Answer: 80 m.

Takeaway: The height $40\sqrt3$ is chosen so the answers are whole numbers.

Example 10: Pole on a building, depression

From the foot of a tower, the elevation of the top of a building is 30°; from the foot of the building, the elevation of the top of the tower is 60°. If the building is 10 m tall, find the tower's height.

Solution:

1. Let $d$ be the distance between feet, tower height $H$.
2. From tower foot: $\tan 30° = \dfrac{10}{d} \Rightarrow d = 10\sqrt3$.
3. From building foot: $\tan 60° = \dfrac{H}{d} \Rightarrow H = d\sqrt3 = 10\sqrt3\cdot\sqrt3 = 30$.

Final Answer: Tower height $= 30$ m.

Takeaway: Two cross elevations share the same horizontal distance $d$.