Some Applications of Trigonometry: questions with answers

Let $\theta$ be the requested angle of elevation, measured from the horizontal. Whether it is called the angle of depression at the top or the angle of elevation at the car, the angle is measured from the HORIZONTAL. By alternate angles the two are equal. In the right triangle the tower is opposite that angle and the horizontal distance is adjacent: $\tan \theta = \dfrac{ 20 }{ 20\sqrt{3} } = \frac{1}{\sqrt{3}}$, so the angle is $30^\circ$. The angle at the top between the tower and the line of sight is $60^\circ$, its complement, and is not the angle asked.

The broken part is the hypotenuse of the right triangle whose base is the ground distance $15$ m and whose angle at the ground is $30^\circ$: $\cos 30^\circ = \dfrac{ 15 }{\text{broken}}$, so broken $= \dfrac{ 15 }{\cos 30^\circ} = \frac{30}{\sqrt{3}}$ m. $\frac{15}{\sqrt{3}}$ is the standing part (from $\tan$), $15\sqrt{3}$ is the whole tree, and $15$ is the base, which is shorter than the hypotenuse.

The standing part is opposite the $30^\circ$ angle: $\tan 30^\circ = \dfrac{\text{standing}}{ 18 }$, standing $= \frac{18}{\sqrt{3}}$ m. The broken part is the hypotenuse: $\cos 30^\circ = \dfrac{ 18 }{\text{broken}}$, broken $= \frac{36}{\sqrt{3}}$ m. The tree's height is the standing part PLUS the broken part: $\frac{18}{\sqrt{3}} + \frac{36}{\sqrt{3}} = 18\sqrt{3}$ m. $\frac{18}{\sqrt{3}}$ alone is only the stump, $\frac{36}{\sqrt{3}}$ alone is only the fallen piece, and $36$ takes both pieces to be as long as the ground distance, which neither of them is.

The right triangle starts at the observer's eyes, $1.5$ m above the ground. Its horizontal leg is $30\sqrt{3}$ m and its vertical leg is the part of the building ABOVE eye level: $\tan 30^\circ \times 30\sqrt{3} = 30$ m. The building also has the $1.5$ m below eye level, so its height is $30 + 1.5 = 31.5$ m. $30$ is only the part above the eyes; $28.5$ subtracts the observer's height, as if the eyes were below the foot of the building; $33$ adds the observer's height twice.

Draw the horizontal from the window across the street. The right triangle formed has base $10$ m, and its vertical leg is the part of the building above the window: $10 \tan 60^\circ = 10\sqrt{3}$ m. The building continues $15$ m below the window, down to the ground, so its height is $15 + 10\sqrt{3}$ m. $10\sqrt{3}$ alone leaves out the part below the window; $10\sqrt{3} - 15$ subtracts the window height as if the ground were above the window; $15\sqrt{3} + 10$ multiplies the wrong length by $\sqrt{3}$.

Draw the horizontal through the eye at $15$ m. Lower triangle: the foot is $15$ m below the horizontal and the depression angle is $30^\circ$, so the horizontal distance is $15 \cot 30^\circ = 15\sqrt{3}$ m. Upper triangle on that base with elevation $60^\circ$: part above $= 15\sqrt{3} \times \sqrt{3} = 45$ m. Total height $= 15 + 45 = 60$ m. Swapping the angles ($60^\circ$ to the foot, $30^\circ$ to the top) gives $20$; $45$ leaves out the $15$ m below the eye; $30$ subtracts it.

At $60^\circ$, the horizontal distance from the observer is $90 \cot 60^\circ = 30\sqrt{3}$ m. At $30^\circ$, it is $90 \cot 30^\circ = 90\sqrt{3}$ m. Both ground positions lie on the same side of the observer. Their separation is $90\sqrt{3} - 30\sqrt{3} = 60\sqrt{3}$ m. There is no reversal, so this separation is also the distance travelled. $90\sqrt{3}$ is only the final horizontal distance from the observer. $120\sqrt{3}$ adds the two horizontal distances instead of finding their difference. $30\sqrt{3}$ is only the initial horizontal distance from the observer.

Let the height be $h$ and the shorter shadow (altitude $60^\circ$) be $x$: $h = x \tan 60^\circ = x\sqrt{3}$. The longer shadow (altitude $30^\circ$) is $x + 48$: $h = (x + 48) \tan 30^\circ = \dfrac{x + 48 }{\sqrt{3}}$. So $3x = x + 48$, $x = 24$ m and $h = 24\sqrt{3}$ m. The $48$ m is the DIFFERENCE of the two shadows; using it as the whole $30^\circ$ shadow gives $\frac{48}{\sqrt{3}}$, and as the $60^\circ$ shadow gives $48\sqrt{3}$; $24$ is the shorter shadow itself.

Let the height of the tower be $h$. At $30^\circ$ the car is $h\sqrt{3}$ from the foot; at $60^\circ$ it is $\dfrac{h}{\sqrt{3}}$ from the foot. In $6$ minutes it covered $h\sqrt{3} - \dfrac{h}{\sqrt{3}} = \dfrac{2h}{\sqrt{3}}$, twice the $\dfrac{h}{\sqrt{3}}$ that remains. At uniform speed, the remaining distance takes half the time: $3$ minutes. The angle changes by $30^\circ$ in each stretch, but equal changes of angle do NOT take equal times. The distances are in the ratio $2:1$, so, at the same speed, the times are $6$ minutes and $3$ minutes.

The two feet are $x$ m apart, and $x$ is the base of BOTH triangles. From the foot of the building, the tower's top is at $60^\circ$: $x = 60 \cot 60^\circ = 20\sqrt{3}$ m. From the foot of the tower, the building's top is at $30^\circ$: building $= x \tan 30^\circ = 20\sqrt{3} \times \dfrac{1}{\sqrt{3}} = 20$ m. $20\sqrt{3}$ is the distance between the feet, not a height; $60\sqrt{3}=60\tan60^\circ$ treats the tower’s height as the base. $180=60\sqrt{3}\times\sqrt{3}$ uses $\tan60^\circ$ at both steps, again starting with the tower’s height as the base.

Smaller triangle: $\tan 30^\circ = \dfrac{ 15 }{PB}$, so $PB = 15\sqrt{3}$ m. Larger triangle on the same base: height to the top of the transmission mast is $15\sqrt{3} \times \tan 60^\circ = 45$ m. The transmission mast alone is $45 - 15 = 30$ m. $45$ is the whole height from the ground; $\frac{30}{\sqrt{3}}=15\sqrt{3}-\frac{ 15 }{\sqrt{3}}$ incorrectly treats the building’s height as the height for both angles and subtracts the resulting horizontal bases; $15\sqrt{3}$ is the base $PB$.

Let the stand be $h$ m and the point be $x$ m from its foot. Smaller triangle: $\tan 45^\circ = \dfrac{h}{x}$, so $x = h$. Larger triangle, same base: $\tan 60^\circ = \dfrac{h + 8 }{x}$, so $h + 8 = h\sqrt{3}$, $h(\sqrt{3} - 1) = 8$, $h = \dfrac{ 8 }{\sqrt{3} - 1} = 4(\sqrt{3} + 1)$ m. $4(3 + \sqrt{3})$ is $h + 8$, the top of the statue, not the stand. $8 - \frac{8}{\sqrt{3}}$ applies both angles to the statue's $8$ m, which is the leg of neither triangle. $4(\sqrt{3} - 1)$ flips the sign when solving for $h$.

The lighthouse $90$ m is the vertical leg of both triangles, one on each side of it. Ship at $30^\circ$: $90 \cot 30^\circ = 90\sqrt{3}$ m from the foot. Ship at $60^\circ$: $90 \cot 60^\circ = 30\sqrt{3}$ m from the foot. Opposite sides, so the distance between the ships is the sum, $90\sqrt{3} + 30\sqrt{3} = 120\sqrt{3}$ m. $60\sqrt{3}$ is their difference, which is correct only if the ships were on the same side. $180 + 60\sqrt{3}$ adds the two slant lines of sight, but those do not lie along the sea. $90\sqrt{3}$ is the distance of one ship only.

With the poles $h$ m high and the point $x$ m from the pole seen at $60^\circ$: $h = x\sqrt{3}$ and $h = \dfrac{ 84 - x}{\sqrt{3}}$, so $3x = 84 - x$ and $x = 21$ m; the other pole is $84 - 21 = 63$ m away. The point is nearer to the pole it sees at the LARGER angle. It is not midway: at $42$ m from each, both angles would be equal. Reversing the distances puts the $60^\circ$ pole farther, and thirds do not satisfy the $3x = 84 - x$ relation.

Let the poles be $h$ m and the point be $x$ m from the pole seen at $60^\circ$, so it is $100 - x$ m from the other. $h = x \tan 60^\circ = x\sqrt{3}$ and $h = (100 - x) \tan 30^\circ = \dfrac{ 100 - x}{\sqrt{3}}$. So $3x = 100 - x$, $x = 25$ m and $h = 25\sqrt{3} = 25\sqrt{3}$ m. The point is NOT midway: at $50$ m from each pole the two angles would be equal, and $50\sqrt{3}$ is what the midway assumption gives with $60^\circ$. $25$ is the distance $x$, not the height; $100\sqrt{3}$ uses the whole road as the base of one triangle.