Chapter 9 ยท Some Applications of Trigonometry

9.2 Heights and Distances

Two looks are better than one. When the foot of a tower is out of reach, measure the angle of elevation from two places, or from a rooftop look both up and down โ€” two right triangles share a side and the unknown height drops out.

A tower stands on the far bank of a river. You cannot cross, so you cannot measure the distance to its foot, and in Section 9.1 you needed that distance. What can you do? Measure the angle of elevation of the top from where you stand, walk a known distance straight towards the tower, and measure the angle again. The tower height, and the distance you never knew, can both be found, because the two right triangles share the same vertical side.

The same idea works from a rooftop: look up at the top of a neighbouring tower and look down at its foot. The two angles give two triangles with a common horizontal side. This section contains all the heights-and-distances situations of the board syllabus: at most two right triangles, and only the angles $30^\circ$, $45^\circ$ and $60^\circ$. We use $\sqrt3\approx1.732$ and $\sqrt2\approx1.414$ and round answers to two decimal places.

Two observations of the same top

Let $AB$ be a vertical tower of height $h$ with foot $B$ and top $A$. Take two points $P$ and $Q$ on the level ground in a straight line with $B$, a distance $d=PQ$ apart. Two situations arise: $P$ and $Q$ on the same side of the tower, or on opposite sides. The nearer point always sees the top at the larger angle (Interactive 1).

๐Ÿ“ Theorem 9.3 โ€” Two observation points
Let a vertical tower of height $h$ stand on level ground with foot $B$. Let $P$, $Q$ be two points of the ground on a straight line through $B$, with $PQ=d$. The angle of elevation of the top is $\beta$ from $P$ and $\alpha$ from $Q$.
(a) Same side ($P$ nearer, so $\beta\gt\alpha$): $\;PB=\dfrac{h}{\tan\beta},\; QB=\dfrac{h}{\tan\alpha}$, and $$h=\frac{d\,\tan\alpha\,\tan\beta}{\tan\beta-\tan\alpha}.$$ (b) Opposite sides ($B$ between $P$ and $Q$): $$h=\frac{d\,\tan\alpha\,\tan\beta}{\tan\alpha+\tan\beta}.$$

Proof. In the right triangle at $B$ with the point $P$: $\tan\beta=\dfrac{h}{PB}$, so $PB=\dfrac{h}{\tan\beta}$. In the same way $QB=\dfrac{h}{\tan\alpha}$.

In an examination, do not quote the formula: write the two right triangles and the two equations, as the examples below do; use the formula to check your answer.

๐Ÿงญ Problem-solving strategy โ€” two triangles, one unknown height
  1. Draw one figure showing the vertical object, both observation points and both lines of sight. Mark each angle at its own point.
  2. Name the vertical side $h$ and the nearer horizontal distance $x$. The farther distance is then $x+d$ (same side); on opposite sides the two distances add up to $d$.
  3. Write one equation for each triangle: $\tan(\text{angle})=\dfrac{h}{\text{distance}}$. Use exact values like $\tan60^\circ=\sqrt3$ and $\tan30^\circ=\dfrac1{\sqrt3}$.
  4. Either eliminate $h$ (set the two expressions for $h$ equal) and solve for $x$, or eliminate $x$. Then find $h$.
  5. Check by substituting into both triangles: they must give the same $h$.
โœ๏ธ Example 1 โ€” Walking towards a tower (30ยฐ and 60ยฐ)
A tower stands on level ground. From a point $Q$ the angle of elevation of its top is $30^\circ$. Walking $20$ m straight towards the tower, to a point $P$, the angle becomes $60^\circ$. Find the height of the tower and the distance of $P$ from its foot. (Use $\sqrt3\approx1.732$.)
Figure Tower $AB$, $PB=x$, $QB=x+20$, $AB=h$, $\angle APB=60^\circ$, $\angle AQB=30^\circ$.
Triangle at P $\tan60^\circ=\dfrac hx$, so $h=\sqrt3\,x$.  (1)
Triangle at Q $\tan30^\circ=\dfrac h{x+20}$, so $h=\dfrac{x+20}{\sqrt3}$.  (2)
Solve From (1) and (2): $\sqrt3\,x=\dfrac{x+20}{\sqrt3}$. Multiply by $\sqrt3$: $3x=x+20$, so $x=10$ m. Then $h=10\sqrt3\approx17.32$ m.
Check At $Q$: $\dfrac{h}{x+20}=\dfrac{17.32}{30}=0.577=\tan30^\circ$ โœ“. Formula: $h=\dfrac{\sqrt3}{2}\times20=17.32$ โœ“.
โœ๏ธ Example 2 โ€” Angles 45ยฐ and 60ยฐ
From a point $Q$ on the ground, the angle of elevation of the top of a tower is $45^\circ$. From a point $P$, $20$ m nearer to the tower and on the same line, it is $60^\circ$. Find the height of the tower. (Use $\sqrt3\approx1.732$.)
Figure $PB=x$, $QB=x+20$, $\angle APB=60^\circ$, $\angle AQB=45^\circ$.
Equations $\tan60^\circ=\dfrac hx\Rightarrow h=\sqrt3\,x$. $\;\tan45^\circ=\dfrac h{x+20}\Rightarrow h=x+20$.
Solve $\sqrt3\,x=x+20$, so $x(\sqrt3-1)=20$ and $x=\dfrac{20}{\sqrt3-1}=\dfrac{20(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}=\dfrac{20(\sqrt3+1)}{2}=10(\sqrt3+1)\approx27.32$ m.
Height $h=x+20=47.32$ m. (Or $h=\sqrt3\,x=1.732\times27.32=47.32$ m โœ“.)
Check Formula: $h=\dfrac{20\times1\times1.732}{1.732-1}=\dfrac{34.64}{0.732}=47.32$ m โœ“.
Interactive 1 โ€” Two Looks at the Same Tower: Solved Live

Pick the two angles of elevation and set the distance $d=PQ$ between the observers (slider, or press Play). The nearer point $P$ sees the top at the larger angle $\beta$ (red); the farther point $Q$ sees it at $\alpha$ (blue). The page solves the two triangles and reports the tower height $h$. Check that both triangles agree on $h$.

20.0
$PB$ near (m)โ€”
$QB$ far (m)โ€”
tower height $h$ (m)โ€”
$PB\tan\beta$ and $QB\tan\alpha$ (m)โ€”
โœ๏ธ Example 3 โ€” Points on opposite sides of the tower
A vertical tower is $30$ m high. Two observers $P$ and $Q$ stand on opposite sides of it, in line with its foot. They see the top at elevations $30^\circ$ (from $P$) and $60^\circ$ (from $Q$). Find the distance $PQ$. (Use $\sqrt3\approx1.732$.)
Figure Tower $AB=30$ m, $B$ between $P$ and $Q$, $\angle APB=30^\circ$, $\angle AQB=60^\circ$.
Triangle at P $\tan30^\circ=\dfrac{30}{PB}$, so $PB=30\sqrt3\approx51.96$ m.
Triangle at Q $\tan60^\circ=\dfrac{30}{QB}$, so $QB=\dfrac{30}{\sqrt3}=10\sqrt3\approx17.32$ m.
Add $PQ=PB+BQ=40\sqrt3\approx69.28$ m.
Check Formula (b) reversed: $d=h\left(\dfrac1{\tan\alpha}+\dfrac1{\tan\beta}\right)=30(1.732+0.577)=69.28$ โœ“.

An object on top of another: two angles from one point

A pole on a building, a flag on a tower, a statue on a pedestal: from a single point on the ground you can see the bottom of the object (the top of the support) and the top of it, at two different angles of elevation. Both right triangles have the same horizontal side $x$.

๐Ÿ“ Theorem 9.4 โ€” A pole on a support
A vertical pole of length $p$ stands on top of a vertical support of height $b$ on level ground. From a point on the ground at horizontal distance $x$ from the foot of the support, the angle of elevation of the bottom of the pole is $\alpha$ and of its top is $\beta$ (with $\beta\gt\alpha$). Then $$b=x\tan\alpha,\qquad b+p=x\tan\beta,\qquad\text{so}\quad p=x\,(\tan\beta-\tan\alpha).$$

Here we know $p$ and the two angles, and find $x$ and $b$: $x=\dfrac{p}{\tan\beta-\tan\alpha}$. In another type of problem the observer stands on a building and looks up at the top and down at the foot of a tower. Then the horizontal distance $x$ is again shared, and the tower height is the part above the observer plus the part below.

โœ๏ธ Example 4 โ€” From a rooftop
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 $45^\circ$. Find the height of the tower. (Use $\sqrt3\approx1.732$.)
Figure Building $PQ=7$ m ($P$ top, $Q$ foot); tower $AB$ with foot $B$ on the same ground. Draw $PC$ horizontal, meeting the tower at $C$. Then $CB=PQ=7$ m and $PC=QB=x$.
Depression The angle of depression of $B$ is $45^\circ$, so $\angle PBQ=45^\circ$ (Theorem 9.1) and $\tan45^\circ=\dfrac{7}{x}$, so $x=7$ m.
Elevation In the triangle $APC$: $\tan60^\circ=\dfrac{AC}{x}$, so $AC=7\sqrt3\approx12.12$ m.
Tower $AB=AC+CB=7\sqrt3+7=7(\sqrt3+1)\approx7\times2.732=19.12$ m.
Check $\dfrac{AB}{x}$ should be $\tan60^\circ+\tan45^\circ$ because the two parts add up: $\dfrac{19.12}{7}=2.731\approx1.732+1$ โœ“.
โœ๏ธ Example 5 โ€” A flagpole on a building
A flagpole $10$ m long stands on top of a building. From a point on the ground the angles of elevation of the bottom and the top of the flagpole are $30^\circ$ and $45^\circ$. Find the height of the building and the distance of the point from its foot. (Use $\sqrt3\approx1.732$.)
Figure Building height $b$, distance $x$, flagpole $10$ m, so the top is at height $b+10$.
Equations $\tan30^\circ=\dfrac bx\Rightarrow b=\dfrac{x}{\sqrt3}$. $\;\tan45^\circ=\dfrac{b+10}{x}\Rightarrow b+10=x$.
Solve $\dfrac x{\sqrt3}+10=x$, so $x\left(1-\dfrac1{\sqrt3}\right)=10$, $x=\dfrac{10\sqrt3}{\sqrt3-1}=\dfrac{10\sqrt3(\sqrt3+1)}{2}=5(3+\sqrt3)\approx23.66$ m.
Height $b=x-10=13.66$ m. (Also $b=\dfrac{x}{\sqrt3}=\dfrac{23.66}{1.732}=13.66$ โœ“.)
Check Theorem 9.4: $p=x(\tan45^\circ-\tan30^\circ)=23.66\times(1-0.577)=10.0$ โœ“.

Looking down at two objects

From a lighthouse, cliff or building, two boats or cars on the ground give two angles of depression. By Theorem 9.1 each depression angle is the angle of elevation at the object, so each object is at distance $\dfrac{H}{\tan(\text{angle})}$ from the foot, where $H$ is the height of the observer. The two objects are $H\left(\dfrac1{\tan30^\circ}-\dfrac1{\tan45^\circ}\right)$ apart on the same side, or $H\left(\dfrac1{\tan30^\circ}+\dfrac1{\tan45^\circ}\right)$ apart on opposite sides. Interactive 2 shows both cases.

โœ๏ธ Example 6 โ€” Two ships seen from a lighthouse
From the top of a $75$ m high lighthouse, the angles of depression of two ships on the same side of the lighthouse, in line with its foot, are $45^\circ$ and $30^\circ$. Find the distance between the ships. (Use $\sqrt3\approx1.732$.)
Figure Lighthouse $AB=75$ m; ships at $C$ ($45^\circ$, nearer) and $D$ ($30^\circ$, farther). By Theorem 9.1, $\angle ACB=45^\circ$ and $\angle ADB=30^\circ$.
Near ship $\tan45^\circ=\dfrac{75}{BC}$, so $BC=75$ m.
Far ship $\tan30^\circ=\dfrac{75}{BD}$, so $BD=75\sqrt3\approx129.9$ m.
Distance Same side: $CD=BD-BC=75(\sqrt3-1)\approx75\times0.732=54.9$ m.
Check $129.9-75=54.9$ โœ“. (If the ships were on opposite sides, the distance would be $129.9+75=204.9$ m.)
Interactive 2 โ€” Two Boats Seen From a Lighthouse

Change the height $H$ of the lighthouse (slider, or Play). Boat $C$ is seen at a depression of $45^\circ$ and boat $D$ at $30^\circ$. Choose whether $D$ is on the same side as $C$ or on the opposite side. The distance between the boats is always a fixed multiple of $H$ โ€” for the same side, $H(\sqrt3-1)\approx0.732H$.

60
$BC=H\cot45^\circ$ (m)โ€”
$BD=H\cot30^\circ$ (m)โ€”
distance $CD$ (m)โ€”
$CD\div H$โ€”

Exercises

I
Two observation points. Use $\sqrt3\approx1.732$, give answers to two decimal places and ignore the observer's height unless stated.
  • 1 From a point $Q$ the angle of elevation of the top of a tower is $30^\circ$; moving $40$ m straight towards the tower, to $P$, it is $60^\circ$. Find the height of the tower and $PB$, the distance of $P$ from its foot.
  • 2 The angles of elevation of the top of a tower from two points on the same line through its foot, $30$ m apart, are $45^\circ$ (farther) and $60^\circ$ (nearer). Find the height of the tower.
  • 3 The elevations of the top of a tower from two points on one side of it, $60$ m apart, are $30^\circ$ (farther) and $45^\circ$ (nearer). Find the height.
  • 4 A man walks $10$ m towards a tower and the angle of elevation of its top changes from $45^\circ$ to $60^\circ$. How tall is the tower?
  • 5 An aeroplane flies horizontally at a height of $1500\sqrt3$ m. Its angle of elevation from a point on the ground is $60^\circ$; $15$ seconds later it is $30^\circ$ (the plane has moved away from the observer). Find its speed in m/s and in km/h.
  • 6 The shadow of a tower is $40$ m longer when the sun's elevation is $30^\circ$ than when it is $60^\circ$. Find the height of the tower.
  • 7 A tree stands on the opposite bank of a river. From a point opposite the tree on this bank the angle of elevation of its top is $60^\circ$; going $16$ m directly away, it is $30^\circ$. Find the height of the tree and the width of the river.
  • 8 A tower $50$ m high is seen from two points on opposite sides of it, in line with its foot, at elevations $30^\circ$ and $45^\circ$. Find the distance between the two points.
1 $h=\sqrt3x$, $h=\dfrac{x+40}{\sqrt3}$: $3x=x+40$, $x=20$ m, $h=20\sqrt3\approx34.64$ m. $PB=20$ m.
2 $h=\sqrt3x=x+30$: $x=\dfrac{30}{\sqrt3-1}=15(\sqrt3+1)\approx40.98$, $h=x+30\approx70.98$ m.
3 $h=x$ (nearer, $45^\circ$) and $h=\dfrac{x+60}{\sqrt3}$: $\sqrt3x=x+60$, $x=\dfrac{60}{\sqrt3-1}=30(\sqrt3+1)\approx81.96$ m $=h$.
4 Nearer point is at $60^\circ$: $h=\sqrt3x=x+10$, $x=\dfrac{10}{\sqrt3-1}=5(\sqrt3+1)\approx13.66$, $h=x+10\approx23.66$ m.
5 $H=1500\sqrt3$. First distance $=\dfrac{H}{\sqrt3}=1500$ m; second $=H\sqrt3=4500$ m. Distance flown $=3000$ m in $15$ s: $200$ m/s $=200\times3.6=720$ km/h.
6 Shadow at $30^\circ$ is $h\sqrt3$; at $60^\circ$ it is $\dfrac h{\sqrt3}$. $h\sqrt3-\dfrac h{\sqrt3}=\dfrac{2h}{\sqrt3}=40$, so $h=20\sqrt3\approx34.64$ m.
7 Width $=x$: $h=\sqrt3x=\dfrac{x+16}{\sqrt3}$, so $3x=x+16$, $x=8$ m. Tree $h=8\sqrt3\approx13.86$ m; river width $8$ m.
8 $PB=\dfrac{50}{\tan30^\circ}=50\sqrt3\approx86.60$ m; $QB=\dfrac{50}{\tan45^\circ}=50$ m. $PQ\approx136.60$ m.
II
Two angles from one point, and objects on a support.
  • 1 From the top of a $12$ m building the angle of elevation of the top of a tower is $60^\circ$ and the angle of depression of its foot is $30^\circ$. Find the height of the tower.
  • 2 From the top of a $20$ m building the elevation of the top of a tower is $30^\circ$ and the depression of its foot is $60^\circ$. Find the height of the tower.
  • 3 A $5$ m flagstaff stands on a tower. From a point on the ground the elevations of the bottom and the top of the flagstaff are $30^\circ$ and $45^\circ$. Find the height of the tower.
  • 4 A $2$ m statue stands on a pedestal. From a point on the ground the elevation of the top of the pedestal is $30^\circ$ and of the top of the statue is $45^\circ$. Find the height of the pedestal.
  • 5 From the top of a $60$ m tower the angles of depression of the top and the foot of a building are $30^\circ$ and $60^\circ$. Find the height of the building.
  • 6 A $20$ m communication tower stands on a building. From a point on the ground the elevations of the bottom and the top of the tower are $45^\circ$ and $60^\circ$. Find the height of the building.
  • 7 From the top of a $24$ m building the angle of depression of the foot of a chimney is $60^\circ$ and the angle of elevation of its top is $30^\circ$. Find the height of the chimney.
1 Horizontal distance: $\tan30^\circ=\dfrac{12}{x}$, $x=12\sqrt3$. Part above the building: $x\tan60^\circ=12\sqrt3\times\sqrt3=36$ m. Tower $=36+12=48$ m.
2 $\tan60^\circ=\dfrac{20}{x}$, $x=\dfrac{20}{\sqrt3}$. Above: $x\tan30^\circ=\dfrac{20}{\sqrt3}\cdot\dfrac1{\sqrt3}=\dfrac{20}3$. Tower $=20+\dfrac{20}{3}=\dfrac{80}{3}\approx26.67$ m.
3 $b=\dfrac{x}{\sqrt3}$, $b+5=x$: $x=\dfrac{5}{1-1/\sqrt3}=\dfrac{5\sqrt3}{\sqrt3-1}$, so $b=\dfrac{x}{\sqrt3}=\dfrac{5}{\sqrt3-1}=\dfrac{5(\sqrt3+1)}{2}\approx6.83$ m.
4 $b=\dfrac{x}{\sqrt3}$, $b+2=x$, so $b=\dfrac{2}{\sqrt3-1}=\sqrt3+1\approx2.73$ m.
5 Foot at $60^\circ$: $x=\dfrac{60}{\sqrt3}=20\sqrt3$. Top at $30^\circ$: the drop is $x\tan30^\circ=20$ m. Building $=60-20=40$ m.
6 $b=x\tan45^\circ=x$ and $b+20=x\sqrt3$: $x=\dfrac{20}{\sqrt3-1}=10(\sqrt3+1)\approx27.32$ m, so $b\approx27.32$ m.
7 $\tan60^\circ=\dfrac{24}{x}$, $x=8\sqrt3$. Above the observer: $x\tan30^\circ=8$ m. Chimney $=24+8=32$ m.
III
Applications and a proof.
  • 1 A $10$ m ladder is placed against a wall on one side of a street, making $60^\circ$ with the ground, and, with the same foot, against the opposite wall on the other side, making $45^\circ$. Find the width of the street and the heights reached on the two walls. (Use $\sqrt2\approx1.414$.)
  • 2 A rocket rises vertically. Two observers $8$ km apart, on opposite sides of the launch point, in line with it, see it at elevations $60^\circ$ and $30^\circ$. Find the rocket's height at that moment.
  • 3 A pilot flying at $1.5$ km above the ground sees two milestones on opposite sides of the aeroplane's track, in line with it, at angles of depression $30^\circ$ and $60^\circ$. Find the distance between the milestones.
  • 4 From a $100$ m lighthouse, two boats on the same side are at depressions $45^\circ$ and $30^\circ$. How far apart are they?
  • 5 From the top of a $90$ m cliff, two boats are seen on the same side, at depressions $30^\circ$ and $60^\circ$. Find the distance between them.
  • 6 A girl whose eyes are $1.6$ m above the ground sees the top of a flagpole at elevation $30^\circ$. She walks $20$ m towards it and the elevation is $60^\circ$. Find the height of the flagpole.
  • 7 Prove: if the elevations of the top of a tower from two points $P$ (near) and $Q$ (far) on the same side are $60^\circ$ and $30^\circ$, then the tower's height is $\dfrac{\sqrt3}{2}\,PQ$.
1 Foot to the first wall: $10\cos60^\circ=5$ m; foot to the second: $10\cos45^\circ=5\sqrt2\approx7.07$ m. Width $\approx12.07$ m. Heights: $10\sin60^\circ=5\sqrt3\approx8.66$ m and $10\sin45^\circ=5\sqrt2\approx7.07$ m.
2 $\dfrac{h}{\tan60^\circ}+\dfrac{h}{\tan30^\circ}=8$: $\dfrac{h}{\sqrt3}+h\sqrt3=\dfrac{4h}{\sqrt3}=8$, so $h=2\sqrt3\approx3.46$ km.
3 By Theorem 9.1 the angles at the milestones are $30^\circ$ and $60^\circ$. Distances: $1.5\sqrt3$ and $\dfrac{1.5}{\sqrt3}$. Sum $=1.5\left(\sqrt3+\dfrac1{\sqrt3}\right)=1.5\times\dfrac{4}{\sqrt3}=2\sqrt3\approx3.46$ km.
4 $100\sqrt3-100=100(\sqrt3-1)\approx73.2$ m.
5 $90\sqrt3-\dfrac{90}{\sqrt3}=90\sqrt3-30\sqrt3=60\sqrt3\approx103.92$ m.
6 Above the eye: $h'=\sqrt3x=\dfrac{x+20}{\sqrt3}$, so $x=10$, $h'=10\sqrt3\approx17.32$ m. Flagpole $=17.32+1.6=18.92$ m.
7 Let $PB=x$ and $d=PQ$. Then $h=\sqrt3x$ and $h=\dfrac{x+d}{\sqrt3}$, so $3x=x+d$, $x=\dfrac d2$ and $h=\dfrac{\sqrt3}{2}d$. $\blacksquare$
IV
Challenge.
  • 1 An aeroplane flies horizontally $1$ km above the ground. The angle of elevation of the aeroplane from a point on the ground changes from $60^\circ$ to $30^\circ$ in $10$ seconds (the plane moving away). Find its speed in km/h.
  • 2 Two poles of equal height stand on either side of a road $80$ m wide. A point on the road between the poles sees the tops at elevations $60^\circ$ and $30^\circ$. Find the height of the poles and the distances of the point from the poles.
  • 3 Use Theorem 9.4 to find the length of a pole standing on a building if, from a point $30$ m from the building, the elevations of the bottom and top of the pole are $30^\circ$ and $60^\circ$.
1 Distances: $\dfrac1{\sqrt3}\approx0.5774$ km and $\sqrt3\approx1.7321$ km. Flown: $\sqrt3-\dfrac1{\sqrt3}=\dfrac{2}{\sqrt3}\approx1.1547$ km in $10$ s: $0.11547$ km/s $\approx415.7$ km/h.
2 $\dfrac h{\tan60^\circ}+\dfrac h{\tan30^\circ}=80$: $\dfrac{4h}{\sqrt3}=80$, $h=20\sqrt3\approx34.64$ m. Distances: $\dfrac{h}{\sqrt3}=20$ m from the pole seen at $60^\circ$ and $h\sqrt3=60$ m from the pole seen at $30^\circ$.
3 $p=30(\tan60^\circ-\tan30^\circ)=30\left(\sqrt3-\dfrac1{\sqrt3}\right)=30\times\dfrac{2}{\sqrt3}=20\sqrt3\approx34.64$ m.
๐ŸŽฏ Exam corner Board ยท 1 markBoard ยท 2 marksBoard ยท 3 marksBoard ยท 5 marksBoard ยท Case study

How this is examined: a heights-and-distances question is worth $3$ or $5$ marks, with credit for the figure, each triangle's equation, the solved unknowns and the final answer with units. Case studies set the problem in a real scene (a flagpole, drone, lighthouse). MCQs reward quick recognition of which angle goes with the nearer point. Write every equation with exact values first ($\sqrt3$, $\tfrac1{\sqrt3}$), and give the decimal at the end.

E1. Board ยท 1 mark The angle of elevation of the top of a tower is $30^\circ$ from a point $Q$, and $60^\circ$ from a point $P$ that is $24$ m closer to the tower on the same line. The height of the tower is  (A) $12$ m  (B) $12\sqrt3$ m  (C) $24\sqrt3$ m  (D) $36$ m

Solve $h=\sqrt3x=\dfrac{x+24}{\sqrt3}$, so $3x=x+24$, $x=12$ and $h=12\sqrt3\approx20.78$ m.
Trap (C) $24\sqrt3$ uses the distance walked ($24$) as if it were the distance to the tower, which would be right only for a single triangle. (A) $12$ is the distance $PB$, not the height.
Answer (B).

E2. Board ยท 1 mark Assertion (A): From two points on the same side of a tower, the angles of elevation of its top are $30^\circ$ and $45^\circ$. The point where the angle is $45^\circ$ is nearer to the tower. Reason (R): $\tan\theta=\dfrac{h}{x}$, so for a fixed height $h$ a larger angle means a smaller distance $x$.
(A) Both A and R are true and R is the correct explanation of A.  (B) Both A and R are true but R is not the correct explanation of A.  (C) A is true but R is false.  (D) A is false but R is true.

Check R $\tan\theta$ increases with $\theta$ for $0^\circ\lt\theta\lt90^\circ$; with $h$ fixed, $x=\dfrac h{\tan\theta}$ decreases as $\theta$ grows. True.
Check A $x=h$ at $45^\circ$ and $x=h\sqrt3\approx1.732h$ at $30^\circ$, so the $45^\circ$ point is nearer. True, and R explains it.
Answer (A).

E3. Board ยท 2 marks From the top of a $50$ m high lighthouse the angles of depression of two boats on the same side are $30^\circ$ and $60^\circ$. Find the distance between the boats. (Use $\sqrt3\approx1.732$.)

Distances (1 mark) Far boat ($30^\circ$): $\dfrac{50}{\tan30^\circ}=50\sqrt3$ m. Near boat ($60^\circ$): $\dfrac{50}{\tan60^\circ}=\dfrac{50}{\sqrt3}$ m.
Difference (1 mark) $50\sqrt3-\dfrac{50}{\sqrt3}=\dfrac{150-50}{\sqrt3}=\dfrac{100}{\sqrt3}\approx57.74$ m.

E4. Board ยท 3 marks From the top of a $20$ 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 $45^\circ$. Find the height of the tower. (Use $\sqrt3\approx1.732$.)

Figure (1 mark) Building $PQ=20$ m; tower $AB$; $PC$ horizontal with $C$ on the tower, $PC=QB=x$, $CB=20$ m.
Depression (1 mark) $\angle PBQ=45^\circ$ (alternate angles), so $\tan45^\circ=\dfrac{20}{x}$ and $x=20$ m.
Elevation (1 mark) $AC=x\tan60^\circ=20\sqrt3\approx34.64$ m. Tower $AB=34.64+20=54.64$ m.

E5. Board ยท 5 marks The angle of elevation of the top of a tower from a point $Q$ on the ground is $30^\circ$. After walking $30$ m towards the tower along level ground, to a point $P$, the angle of elevation is $60^\circ$. Find the height of the tower and the distance of $P$ from its foot. (Use $\sqrt3\approx1.732$.)

Figure (1 mark) Tower $AB=h$; $PB=x$; $QB=x+30$; $\angle APB=60^\circ$, $\angle AQB=30^\circ$.
Equations (2 marks) $\tan60^\circ=\dfrac hx\Rightarrow h=\sqrt3x$. $\tan30^\circ=\dfrac h{x+30}\Rightarrow h=\dfrac{x+30}{\sqrt3}$.
Solve (1 mark) $\sqrt3x=\dfrac{x+30}{\sqrt3}$, $3x=x+30$, $x=15$ m.
Answer (1 mark) $h=15\sqrt3\approx25.98$ m, and $PB=15$ m.

E6. Board ยท Case study At a school sports day a flagpole $6$ m long is fixed on top of the pavilion. Rohan stands on the ground in front of the pavilion; he sees the bottom of the flagpole at an angle of elevation of $30^\circ$ and its top at $45^\circ$. Let the pavilion be $b$ m high and let Rohan be $x$ m from its foot. (Use $\sqrt3\approx1.732$.)
(i) Write two equations connecting $b$ and $x$. (2 marks)
(ii) Find the height of the pavilion. (1 mark)
(iii) Find Rohan's distance from the foot of the pavilion. (1 mark)

(i) $\tan30^\circ=\dfrac bx$, i.e. $b=\dfrac{x}{\sqrt3}$; and $\tan45^\circ=\dfrac{b+6}{x}$, i.e. $b+6=x$. (1 + 1 marks)
(ii) $\dfrac{x}{\sqrt3}+6=x$, so $x=\dfrac{6\sqrt3}{\sqrt3-1}=3\sqrt3(\sqrt3+1)=9+3\sqrt3\approx14.20$ m; $b=x-6=3(\sqrt3+1)\approx8.20$ m. (1 mark)
(iii) $x=9+3\sqrt3\approx14.20$ m. (1 mark)
๐Ÿ”ฌ Researchโ€“Teaching NexusMini project ยท optional, about 45 minutes

Research question. Can you find the height of a tower or the width of a pond or ditch you cannot cross using a clinometer app, and a measured walk of known length โ€” and how large is the error?

Why it matters. Surveyors triangulate: they measure one baseline on easy ground and two angles, and from that compute distances to things they cannot reach, such as mountain peaks, rivers and ships. The Great Trigonometrical Survey of India, which measured the Himalayan peaks, was built on exactly this two-angle idea, just with a bigger instrument.

What to do.

  1. Choose a tall object (tree, tower, building) with a clear straight line of level ground leading to its foot. Measure your eye height $e$.
  2. From point $Q$ read the angle of elevation $\alpha$ of the top. Walk straight towards the object by a measured distance $d$ (for example $10$ m) to point $P$ and read the angle $\beta$ again.
  3. Calculate the height above your eye with the formula $h'=\dfrac{d\tan\alpha\tan\beta}{\tan\beta-\tan\alpha}$ (use a calculator for tan), then add $e$.
  4. Measure the distance to the foot of the object with the tape (if you can reach it) and compare with $\dfrac{h'}{\tan\beta}$.
  5. Repeat with $d=5$ m and $d=15$ m. Which baseline gives the most consistent answer?

Tools: A phone clinometer app, a long tape or a rope marked in metres, a notebook and calculator

Hand in: A neat figure with all measured values, a table of $d$, $\alpha$, $\beta$, $h$ for each trial, and one paragraph on the source of the biggest error.

Going further. Try two observation points on opposite sides of a tree, or measure the width of a ditch using a pole of known height standing on the other bank.

Teacher/parent note: if the two angles are very close together (a short baseline), the denominator $\tan\beta-\tan\alpha$ is small and the answer becomes unreliable โ€” this is a good discussion point about measurement error. Angles in the real survey will not be $30^\circ$, $45^\circ$ or $60^\circ$, so a calculator is allowed for the project.

๐Ÿงช Self-check quiz

0/5 correct

Q1. The elevation of the top of a tower is $30^\circ$ from a point $Q$ and $60^\circ$ from a point $P$ that is $12$ m closer, on the same line. The height of the tower is

Q2. From the top of a $40$ m cliff a car on level ground is seen at a depression of $45^\circ$. It moves straight towards the cliff until the depression is $60^\circ$. The distance it moved is

Q3. Two points on the same side of a tower see its top at elevations $30^\circ$ and $45^\circ$. The point where the elevation is $45^\circ$ is

Q4. A $30$ m tower is seen at an elevation of $45^\circ$ from two points on opposite sides of it, in line with its foot. The distance between the two points is

Q5. A $10$ m pole stands on a building. From a point on the ground the elevations of the bottom and top of the pole are $45^\circ$ and $60^\circ$. The height of the building is

โ† 9.1 Line of Sight, Angles of Elevation and DepressionChapter 10 โ€” Circles โ†’