Two-angle problems

Once you can handle one observer and one object, the natural next step is two angles in the same diagram. These problems are the bread and butter of the board exam's $4$ - and $5$ -mark questions. They look intimidating because they involve two unknowns, but the recipe is identical to the single-angle case , applied twice , followed by a simple algebraic substitution.

The most common configurations are: (i) one object viewed from two points along a horizontal line; (ii) two objects of unequal height viewed from one point; (iii) one observer looking down at two objects (two depressions); and (iv) a tower with a flagstaff on top, where the flagstaff and the tower subtend two different angles.

Set-up template

Whenever you see two angles in a problem, define two unknowns at the start. Almost always: $h$ for the height (or a related vertical), and $x$ for the unknown horizontal distance. Then write two equations from the two right triangles.

For configuration (i), one tower of height $h$ viewed from two points $A$ and $B$ on the same side, with $AB = d$ :

$\tan\alpha = \frac{h}{x+d}, \qquad \tan\beta = \frac{h}{x},$

where $\alpha$ is the elevation from the farther point and $\beta$ from the closer point. Two equations, two unknowns ( $h$ and $x$ ). Eliminate $h$ :

$h = (x+d)\tan\alpha = x \tan\beta \Rightarrow x = \frac{d\tan\alpha}{\tan\beta - \tan\alpha}.$

A workhorse derivation

Many board questions reduce to: "From two points $d$ metres apart on the same side of a tower, the angles of elevation of the top are $\alpha$ and $\beta$ ( $\alpha < \beta$ ). Find the height of the tower."

From the formulas above,

$h = \frac{d \tan\alpha \tan\beta}{\tan\beta - \tan\alpha}.$

If $\alpha = 30^\circ$ and $\beta = 60^\circ$ , then $\tan\alpha = 1/\sqrt 3$ , $\tan\beta = \sqrt 3$ , so

$h = \frac{d \cdot (1/\sqrt 3) \cdot \sqrt 3}{\sqrt 3 - 1/\sqrt 3} = \frac{d}{(3-1)/\sqrt 3} = \frac{d\sqrt 3}{2}.$

Memorise the $30^\circ$ – $60^\circ$ result: $h = d\sqrt 3 / 2$ . It saves time in the exam.

Tower with flagstaff

A tower of height $H$ has a flagstaff of height $f$ on top. From a point at distance $b$ on the ground, the bottom of the flagstaff (top of tower) subtends angle $\alpha$ and the top of the flagstaff subtends angle $\beta$ . Then

$\tan\alpha = \frac{H}{b}, \qquad \tan\beta = \frac{H+f}{b}.$

Subtracting: $f = b(\tan\beta - \tan\alpha)$ . If $b$ is known, the flagstaff length is one calculation away. If $f$ is known and $H$ unknown, you have two equations in $H$ and $b$ .

Worked examples

Example 1. From two points on the ground, on the same side of a tower and in line with it, the angles of elevation of the top of the tower are $30^\circ$ and $60^\circ$ . The two points are $50$ m apart. Find the height of the tower.

Using $h = d\sqrt 3/2$ with $d = 50$ : $h = 50\sqrt 3/2 = 25\sqrt 3 \approx 43.3$ m.

Example 2. The angles of depression of two ships from the top of a $100$ -m lighthouse are $30^\circ$ and $45^\circ$ . The ships are on the same side of the lighthouse. Find the distance between the two ships.

Let the distances of the ships from the lighthouse be $x_1$ (nearer, angle $45^\circ$ ) and $x_2$ (farther, angle $30^\circ$ ).

$\tan 45^\circ = 100/x_1 \Rightarrow x_1 = 100$ . $\tan 30^\circ = 100/x_2 \Rightarrow x_2 = 100\sqrt 3$ .

Distance between ships $= x_2 - x_1 = 100\sqrt 3 - 100 = 100(\sqrt 3 - 1) \approx 73.2$ m.

Example 3. A vertical tower stands on the ground with a flagstaff on top. From a point on the ground $9$ m from the foot, the angle of elevation of the top of the tower is $30^\circ$ and that of the top of the flagstaff is $60^\circ$ . Find the height of the flagstaff.

Tower height $H$ : $\tan 30^\circ = H/9 \Rightarrow H = 9/\sqrt 3 = 3\sqrt 3$ .

Flagstaff top: $\tan 60^\circ = (H+f)/9 = \sqrt 3 \Rightarrow H + f = 9\sqrt 3$ .

So $f = 9\sqrt 3 - 3\sqrt 3 = 6\sqrt 3 \approx 10.39$ m.

Example 4. A man on the top of a $30$ -m tower observes a car moving towards the foot of the tower at a uniform speed. The angle of depression changes from $30^\circ$ to $60^\circ$ in $6$ seconds. Find the speed of the car (in m/s).

Distances of the car from the foot: $x_1 = 30/\tan 30^\circ = 30\sqrt 3$ (initially) and $x_2 = 30/\tan 60^\circ = 30/\sqrt 3 = 10\sqrt 3$ (after $6$ s).

Distance covered $= 30\sqrt 3 - 10\sqrt 3 = 20\sqrt 3$ m in $6$ s.

Speed $= 20\sqrt 3 / 6 = 10\sqrt 3/3 \approx 5.77$ m/s.

Try it yourself

From two points $20$ m apart on the same side of a tower, the elevations of the top are $30^\circ$ and $60^\circ$ . Find the height.
The angles of depression of two stones, in line with the foot of a $50$ -m tower, are $45^\circ$ and $60^\circ$ . Find the distance between them.
A tower stands on a base. From $10$ m away on the ground, the top of the tower and the top of a $4$ -m statue on top subtend angles $30^\circ$ and $45^\circ$ . Find the tower's height.
A man sees an aeroplane flying horizontally. Its angle of elevation changes from $60^\circ$ to $30^\circ$ in $15$ seconds. The plane is at constant height $1500$ m. Find its speed.
From the top of a building, two cars on the road on the same side are observed at $30^\circ$ and $45^\circ$ . The building is $20$ m tall. Distance between cars?
A tower and a flagstaff together stand $50$ m tall. From a point $50$ m away, the foot of the flagstaff subtends $30^\circ$ . Find the flagstaff height.
From two points $A$ and $B$ on opposite sides of a tower, the angles of elevation of the top are $45^\circ$ and $60^\circ$ . If $AB = 40$ m, find the height.
From the top of a cliff $60$ m high, the angles of depression of the top and bottom of a tower are $30^\circ$ and $60^\circ$ . Find the tower's height.
The angle of elevation of a cloud above a lake from a point $h$ m above the lake is $\alpha$ , and the angle of depression of its reflection is $\beta$ . Show that the cloud's height above the lake is $h \cdot (\tan\beta + \tan\alpha)/(\tan\beta - \tan\alpha)$ .
A boat moves in a straight line away from a $100$ -m cliff. The angle of depression changes from $60^\circ$ to $30^\circ$ in $5$ s. Find the boat's speed.

Pitfalls / Insight

(a) In two-angle problems on the same side, the larger angle goes with the closer observer. Sketch it and confirm. (b) When solving simultaneously, eliminate the variable you don't want to find , usually it is the horizontal distance. (c) For "approaching" problems (car, plane, boat), set up two distances and subtract; speed = distance covered / time.

A pleasant identity worth remembering: when the two elevations are $30^\circ$ and $60^\circ$ and the points are $d$ apart, the closer point is at horizontal $d/2$ and the tower is $d\sqrt 3/2$ tall.