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📐 Key Concepts: Right Triangle Trigonometry

For a right triangle with angle $\theta$, opposite side $o$, adjacent side $a$, and hypotenuse $h$:

$$\sin\theta = \frac{o}{h}, \quad \cos\theta = \frac{a}{h}, \quad \tan\theta = \frac{o}{a}$$

Angles of Elevation and Depression: When a point is above the horizontal, the angle looking up is the angle of elevation; looking down is the angle of depression. Both are measured from the horizontal.

Two-observation method: If two angles of elevation $\alpha$ and $\beta$ are observed from two points separated by distance $d$:

$$h = \frac{d\tan\alpha\tan\beta}{\tan\alpha - \tan\beta} \quad (\alpha < \beta)$$

Problem 1: Height of the Tower (Right Triangles)

A man finds the angle of elevation of the top of a tower to be 30°. He walks 65 m nearer the tower and finds its angle of elevation to be 60°. What is the height of the tower?

Right Triangles – Problem 1 – Diagram Right Triangles – Problem 1 – Diagram Right Triangles – Problem 1 – Diagram
Solution Solution Solution Solution

Let $h$ = height of the tower, $x$ = distance from the tower when the angle is 60°.

$$\tan60° = \frac{h}{x} \Rightarrow x = \frac{h}{\sqrt{3}}$$
$$\tan30° = \frac{h}{x + 65} \Rightarrow x + 65 = h\sqrt{3}$$

Substituting $x = h/\sqrt{3}$ into the second equation:

$$\frac{h}{\sqrt{3}} + 65 = h\sqrt{3} \Rightarrow 65 = h\sqrt{3} - \frac{h}{\sqrt{3}} = h\!\left(\frac{3-1}{\sqrt{3}}\right) = \frac{2h}{\sqrt{3}}$$
$$h = \frac{65\sqrt{3}}{2} \approx \boxed{56.3 \text{ m}}$$

Problem 2: Tilted Piece of Wood

A rectangular piece of wood 4 cm × 12 cm is tilted at an angle of 45°. Find the vertical distance between the lower corner and the upper corner.

Tilted Wood – Diagram Tilted Wood – Diagram Tilted Wood – Diagram
Solution Solution Solution Solution

The piece is tilted at 45°. Decompose the rectangle's corner-to-corner vector into vertical and horizontal components. The vertical rise from the lowest corner (A) to the highest corner (D) equals:

$$\Delta V = L\sin\theta + W\cos\theta$$

where $L = 12$ cm (length), $W = 4$ cm (width), and $\theta = 45°$:

$$\Delta V = 12\sin45° + 4\cos45° = \frac{12}{\sqrt{2}} + \frac{4}{\sqrt{2}} = \frac{16}{\sqrt{2}} = 8\sqrt{2} \approx \boxed{11.31 \text{ cm}}$$

Problem 3: Ladder Against a Wall

A ladder 10 m long leans against a vertical wall. If the foot of the ladder makes an angle of 75° with the ground, find: (a) the height the ladder reaches on the wall, and (b) the horizontal distance from the wall to the foot of the ladder.

Let the ladder length be the hypotenuse $h = 10$ m and the angle with the ground be $\theta = 75°$.

Part (a) — Height on wall (opposite side):

$$\text{Height} = h\sin\theta = 10\sin 75° = 10(0.9659) \approx \boxed{9.66 \text{ m}}$$

Part (b) — Distance from wall (adjacent side):

$$\text{Distance} = h\cos\theta = 10\cos 75° = 10(0.2588) \approx \boxed{2.59 \text{ m}}$$

Alternatively: $\tan 75° = \frac{9.659}{x}$, giving $x = \frac{9.659}{\tan 75°} = \frac{9.659}{3.732} \approx 2.59$ m.

Problem 4: Shadow and Angle of the Sun

A vertical pole casts a shadow 8 m long when the angle of elevation of the sun is 38°. A man standing 2 m from the base of the pole observes the top of the pole. Find: (a) the height of the pole, and (b) the angle of elevation of the top of the pole as seen by the man.

Part (a) — Height of the pole

The shadow length $s = 8$ m and the sun's elevation angle $\theta = 38°$. The pole and its shadow form a right triangle:

$$h = s\tan\theta = 8\tan 38° = 8(0.7813) \approx \boxed{6.25 \text{ m}}$$

Part (b) — Elevation angle from the man

The man stands 2 m from the base, so horizontal distance = 2 m, vertical height = 6.25 m:

$$\alpha = \arctan\!\left(\frac{h}{d}\right) = \arctan\!\left(\frac{6.25}{2}\right) = \arctan(3.125) \approx \boxed{72.3°}$$