1. Basic Angle of Elevation

• Formula: If a vertical object has height (h) and the observer is (d) units away:

$$\tan\theta=\frac{h}{d}$$

Hence:

$$h=d\tan\theta$$

• Example: An observer is 20 m from a tower and the angle of elevation is (60^\circ). Find the height of the tower. Solution:

$$h=20\tan60^\circ$$ $$=20\sqrt3$$ $$\boxed{20\sqrt3\text{ m}}$$

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2. Angle of Elevation When Observer Moves Away

• Formula: If the observer moves (x) m away:

$$h=d\tan\theta_1=(d+x)\tan\theta_2$$

• Example: The angle of elevation of a tower changes from (60^\circ) to (30^\circ) when the observer moves 20 m away. Find the tower height. Solution: Let the initial distance be (d).

At (60^\circ):

$$h=d\tan60^\circ=d\sqrt3$$

After moving 20 m away:

$$h=(d+20)\tan30^\circ=\frac{d+20}{\sqrt3}$$

Equating:

$$d\sqrt3=\frac{d+20}{\sqrt3}$$ $$3d=d+20$$ $$2d=20\Rightarrow d=10$$

Therefore:

$$h=10\sqrt3$$ $$\boxed{10\sqrt3\text{ m}}$$

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3. Angle of Elevation When Observer Moves Towards the Object

• Formula: If the observer moves (x) m towards the object:

$$h=d\tan\theta_1=(d-x)\tan\theta_2$$

• Example: From a point, the angle of elevation of a tower is (30^\circ). After moving 30 m towards it, the angle becomes (45^\circ). Find the tower height. Solution: Let the original distance be (d).

Initially:

$$h=d\tan30^\circ=\frac d{\sqrt3}$$

After moving 30 m:

$$h=(d-30)\tan45^\circ=d-30$$

Thus:

$$\frac d{\sqrt3}=d-30$$ $$d(\sqrt3-1)=30\sqrt3$$ $$d=\frac{30\sqrt3}{\sqrt3-1} =15(3+\sqrt3)$$

Hence:

$$h=d-30 =15(1+\sqrt3)$$ $$\boxed{15(1+\sqrt3)\text{ m}}$$

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4. Angle of Depression

• Formula: The angle of depression from a higher point equals the angle of elevation from the lower point:

$$\text{Angle of depression}=\text{Angle of elevation}$$

Then use:

$$\tan\theta=\frac{\text{vertical height}}{\text{horizontal distance}}$$

• Example: From the top of a 40 m building, the angle of depression of a car is (45^\circ). Find its horizontal distance from the building. Solution:

$$\tan45^\circ=\frac{40}{d}$$ $$1=\frac{40}{d}$$ $$\boxed{d=40\text{ m}}$$

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5. Shadow Length & Elevation of Sun

• Formula:

$$\tan\theta=\frac{\text{height of object}}{\text{shadow length}}$$

• Example: A 6 m pole casts a shadow of (2\sqrt3) m. Find the sun’s angle of elevation. Solution:

$$\tan\theta=\frac6{2\sqrt3} =\sqrt3$$

Since:

$$\tan60^\circ=\sqrt3$$

Therefore:

$$\boxed{\theta=60^\circ}$$

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6. Shadow Length from Sun’s Elevation

• Formula:

$$\text{Shadow length}=\frac{\text{Height}}{\tan\theta}$$

• Example: A 10 m pole is standing when the sun’s elevation is (45^\circ). Find its shadow length. Solution:

$$L=\frac{10}{\tan45^\circ}$$ $$=\frac{10}{1} =\boxed{10\text{ m}}$$

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7. Two Points of Observation on the Same Side

• Formula: If a tower is observed from two points on the same straight line:

$$h=d_1\tan\theta_1=d_2\tan\theta_2$$

If the points are (x) m apart and the farther point is at distance (d):

$$h=d\tan\theta_1=(d-x)\tan\theta_2$$

• Example: The angle of elevation of a tower is (30^\circ) from point A and (60^\circ) from point B, which is 20 m closer to the tower. Find the height. Solution: Let distance of B from tower be (d).

From B:

$$h=d\tan60^\circ=d\sqrt3$$

From A:

$$h=(d+20)\tan30^\circ =\frac{d+20}{\sqrt3}$$

Equating:

$$d\sqrt3=\frac{d+20}{\sqrt3}$$ $$3d=d+20$$ $$d=10$$

Therefore:

$$h=10\sqrt3$$ $$\boxed{10\sqrt3\text{ m}}$$

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8. Two Points of Observation on Opposite Sides

• Formula: If two observers are on opposite sides of the base of a tower:

$$h=d_1\tan\theta_1=d_2\tan\theta_2$$

If the distance between observers is (D):

$$D=d_1+d_2$$

• Example: Two observers on opposite sides of a tower are 40 m apart. Their angles of elevation are (30^\circ) and (60^\circ). Find the tower height. Solution: Let distances from the tower be (d_1,d_2).

$$h=d_1\sqrt3$$

and:

$$h=\frac{d_2}{\sqrt3}$$

Thus:

$$d_2=3d_1$$

Since:

$$d_1+d_2=40$$ $$4d_1=40\Rightarrow d_1=10$$

Therefore:

$$h=10\sqrt3$$ $$\boxed{10\sqrt3\text{ m}}$$

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Advanced Variants

9. Two Towers with Complementary Angles

• Formula: If two towers of heights (h_1,h_2) stand at distances (d) from an observer and their angles of elevation are complementary:

$$\tan\theta=\frac{h_1}{d}$$ $$\tan(90^\circ-\theta)=\frac{h_2}{d}$$

Since:

$$\tan\theta\cot\theta=1,$$

we get:

$$d^2=h_1h_2$$

• Example: Two towers of heights 30 m and 50 m stand on the same side of an observer. The angles of elevation from the observer are complementary, and the observer is equidistant from both towers. Find the distance from the observer to each tower. Solution:

$$d^2=30\times50=1500$$ $$d=10\sqrt{15}$$

Therefore:

$$\boxed{10\sqrt{15}\text{ m}}$$

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10. Distance Between Two Towers with Complementary Angles

• Formula: If the observer is midway between two towers and the tower heights are (h_1,h_2):

$$d=\sqrt{h_1h_2}$$

Distance between towers:

$$D=2d=2\sqrt{h_1h_2}$$

• Example: Two towers are 30 m and 50 m high. An observer at their midpoint sees their tops at complementary angles. Find the distance between the towers. Solution:

$$d=\sqrt{30\times50} =10\sqrt{15}$$

Hence:

$$D=2d =\boxed{20\sqrt{15}\text{ m}}$$

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11. Moving Observer — Direct Height Calculation

• Formula: If an observer moves (x) m towards a tower and the angles change from (\theta_1) to (\theta_2):

$$\frac{h}{\tan\theta_1} ---------------------- \frac{h}{\tan\theta_2} =x$$

Therefore:

$$h= \frac{x\tan\theta_1\tan\theta_2} {\tan\theta_2-\tan\theta_1}$$

• Example: An observer moves 20 m towards a tower. The angle of elevation changes from (30^\circ) to (45^\circ). Find the height. Solution:

$$h= \frac{20\times\tan30^\circ\times\tan45^\circ} {\tan45^\circ-\tan30^\circ}$$

[

\frac{20\times(1/\sqrt3)\times1} {1-1/\sqrt3} ]

$$=\boxed{10(1+\sqrt3)\text{ m}}$$

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12. Observer Moves Away — Direct Height Calculation

• Formula: If an observer moves (x) m away and the angle changes from (\theta_1) to (\theta_2):

$$\frac{h}{\tan\theta_2} ---------------------- \frac{h}{\tan\theta_1} =x$$

Hence:

$$h= \frac{x\tan\theta_1\tan\theta_2} {\tan\theta_1-\tan\theta_2}$$

• Example: The angle of elevation of a tower changes from (60^\circ) to (30^\circ) when an observer moves 20 m away. Find the height. Solution:

$$h= \frac{20(\sqrt3)(1/\sqrt3)} {\sqrt3-1/\sqrt3}$$ $$=\frac{20}{2/\sqrt3} =\boxed{10\sqrt3\text{ m}}$$