Trigonometric Identities and Applications - Solve heights and distances problems using angles of elevation and depression (30 degree, 45 degree, 60 degree)

A tower stands vertically on the ground. From a point on the ground, which is $15$ m away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^{\circ}$. Find the height of the tower.

Let $AB$ be the tower of height $h$ and $C$ be the point on the ground. Given: Distance $BC = 15$ m and $\angle ACB = 60^{\circ}$. In right $\triangle ABC$: $\tan 60^{\circ} = \frac{AB}{BC}$ $\sqrt{3} = \frac{h}{15}$ $h = 15\sqrt{3} \text{ m}$ If we take $\sqrt{3} \approx 1.732$, then $h = 15 \times 1.732 = 25.98$ m.

The problem provides the adjacent side (distance from the foot) and asks for the opposite side (height). Therefore, the tangent ratio is the most direct formula to use.

From the top of a $7$ m high building, the angle of elevation of the top of a cable tower is $60^{\circ}$ and the angle of depression of its foot is $45^{\circ}$. Determine the height of the tower.

Let $AB = 7$ m be the building and $CD$ be the tower. Let $AE$ be the horizontal line from the top of the building to the tower. Then $AB = EC = 7$ m. In right $\triangle ABC$, the angle of depression is $45^{\circ}$, so $\angle ACB = 45^{\circ}$ (alternate angles). $\tan 45^{\circ} = \frac{AB}{BC} \Rightarrow 1 = \frac{7}{BC} \Rightarrow BC = 7 \text{ m}$ Since $AE = BC$, $AE = 7$ m. In right $\triangle AED$: $\tan 60^{\circ} = \frac{DE}{AE} \Rightarrow \sqrt{3} = \frac{DE}{7} \Rightarrow DE = 7\sqrt{3} \text{ m}$ Total height of tower $CD = CE + DE = 7 + 7\sqrt{3} = 7(1 + \sqrt{3}) \text{ m}$.

This solution breaks the tower into two parts: the portion equal to the building height and the portion above it. We use the angle of depression to find the distance between the two structures, which then serves as the base for the second triangle to find the remaining height.

Let $AB$ be the chimney and $CD$ be the observer. Here, $CD = 1.5$ m and the distance $CB = 28.5$ m. Draw $DE \parallel CB$. Then $DE = CB = 28.5$ m and $EB = CD = 1.5$ m. In right $\triangle ADE$: $\tan 45^{\circ} = \frac{AE}{DE}$ $1 = \frac{AE}{28.5}$ $AE = 28.5$ m. Height of chimney $h = AE + EB$ $h = 28.5 + 1.5 = 30$ m.

Since the observer has a height, we subtract it from the total height of the chimney to form a right triangle at the eye level. Using $\tan 45^{\circ}$, we find the upper portion of the chimney and then add the observer's height back.

Let $h$ be the height of the tower $AB$. Let $BC = x$ be the shadow when altitude is $60^{\circ}$. The shadow becomes $BD = x + 40$ when altitude is $30^{\circ}$. In $\triangle ABC$: $\tan 60^{\circ} = \frac{h}{x} \implies \sqrt{3} = \frac{h}{x} \implies x = \frac{h}{\sqrt{3}}$ ... (1) In $\triangle ABD$: $\tan 30^{\circ} = \frac{h}{x + 40}$ $\frac{1}{\sqrt{3}} = \frac{h}{x + 40} \implies x + 40 = h\sqrt{3}$ ... (2) Substitute (1) into (2): $\frac{h}{\sqrt{3}} + 40 = h\sqrt{3}$ $40 = h\sqrt{3} - \frac{h}{\sqrt{3}}$ $40 = \frac{3h - h}{\sqrt{3}} = \frac{2h}{\sqrt{3}}$ $h = \frac{40\sqrt{3}}{2} = 20\sqrt{3}$ m.

We use two right triangles sharing the same height. By expressing the base of both triangles in terms of the height and using the difference between the shadow lengths (40m), we solve for the height $h$.