Q: The angles of depression of two ships A and B as observed from the top of a light house 60 m high are $60^\circ$ and $45^\circ$ respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.

60° 45° A D B C (60 m) Step 1 Let $CD = 60\text{ m}$ be the lighthouse. $A$ and $B$ are on opposite sides of base $D$. Angle of depression $60^\circ \implies \angle CAD = 60^\circ$. Angle of depression $45^\circ \implies \angle CBD = 45^\circ$. Step 2: Distance AD In right $\triangle CDA$ ($\angle D = 90^\circ$): $$\tan 60^\circ = \frac{CD}{AD} \implies \sqrt{3} = \frac{60}{AD} \implies AD = \frac{60}{\sqrt{3}} = 20\sqrt{3}\text{ m}$$ [1.5 Marks] Step 3: Distance BD In right $\triangle CDB$ ($\angle D = 90^\circ$): $$\tan 45^\circ = \frac{CD}{BD} \implies 1 = \frac{60}{BD} \implies BD = 60\text{ m}$$ [1 Mark] Step 4: Total Distance AB $$AB = AD + BD = 20\sqrt{3} + 60 = 20(1.732) + 60 = 34.64 + 60 = 94.64\text{ m}$$ To the nearest whole number: $\mathbf{95\text{ metres}}$ [1.5 Marks] Answer: The distance between the two ships is 95 metres .

Question 1

A flagpole is erected at the top of a building. The angle of elevation of the top and foot of the flagpole from a point 100 m away, on the same level as that of the foot of the building, are 33° and 31° respectively. Find the height of the flagpole. Give your answer correct to the nearest metre. (Us

Accepted Answer

A B C D 31° 33° 100 m Vertical flagpole CD on building BC with elevation angles 31° and 33° from 100 m away Geometry Let building height BC = h, flagpole CD = x, ground distance AB = 100 m. In right-angled ΔABC: tan 31° = BC / AB = h / 100 ⇒ h = 100 × tan 31° = 100 × 0.6009 = 60.09 m . [2 Marks] Total Height In right-angled ΔABD: tan 33° = BD / AB = (h + x) / 100 ⇒ h + x = 100 × tan 33° = 100 × 0.6494 = 64.94 m . [2 Marks] Height of Flagpole x = (h + x) − h = 64.94 − 60.09 = 4.85 m. Rounding to the nearest metre: x = 5 metres . [1 Mark]

Question 2

AB is vertical tower 100 m away from foot of 30-storied building CD. Angles of depression from C and E (midpoint of CD) are 35° and 14° respectively. Find height of: (a) tower AB, (b) building CD.

Accepted Answer

In ΔAPC: CP = 100 tan 35° = 100(0.7002) = 70.02 m. [1.5 Marks]
In ΔAPE: EP = 100 tan 14° = 100(0.2493) = 24.93 m. [1.5 Marks]
CE = CP − EP = 70.02 − 24.93 = 45.09 m.
Since E is midpoint of CD: Height of building CD = 2 × 45.09 = 90.18 m. [1 Mark]
Tower AB = PD = ED − EP = 45.09 − 24.93 = 20.16 m. [1 Mark]

Question 3

The angle of elevation of the top of a 100 m high tree from two points A and B on opposite sides of the tree are 52° and 45° respectively. Find the distance AB , to the nearest metre.

Accepted Answer

In right Δ CDA : tan 52° = 100 / AD ⇒ AD = 100 / 1.2799 ≈ 78.13 m. [2 Marks] In right Δ CDB : tan 45° = 100 / BD ⇒ BD = 100 m. [1.5 Marks] AB = AD + BD = 78.13 + 100 = 178.13 m ≈ 178 m . [1.5 Marks]

Question 4

From the top of a tower 100 m high a man observes the angles of depression of two ships $A$ and $B$, on opposite sides of the tower as $45^\circ$ and $38^\circ$ respectively. If the foot of the tower and the ships are in the same horizontal line, find the distance between the two ships $A$ and $B$ to the nearest metre.
(Use Mathematical Tables for this question.)

Accepted Answer

Let $CD$ be the tower of height 100 m, with foot $C$.
By alternate interior angles, angle of elevation at ship $A$ is $\angle CAD = 45^\circ$, and at ship $B$ is $\angle CBD = 38^\circ$.

Distance AC In right-angled $\triangle ACD$:
$\tan 45^\circ = \frac{CD}{AC} \implies 1 = \frac{100}{AC} \implies AC = 100\text{ m}$.1.5 Marks

Distance BC In right-angled $\triangle BCD$:
$\tan 38^\circ = \frac{CD}{BC} \implies BC = \frac{100}{\tan 38^\circ}$.
From natural tangents tables: $\tan 38^\circ \approx 0.7813$.
$BC = \frac{100}{0.7813} \approx 127.99\text{ m}$.1.5 Marks

Total Distance AB
$AB = AC + BC = 100 + 127.99 = 227.99\text{ m} \approx \mathbf{228\text{ m}}$ (to the nearest metre).1 Mark

Question 5

From the top of a cliff, the angle of depression of the top and bottom of a tower are observed to be $45^\circ$ and $60^\circ$ respectively. If the height of the tower is 20 m, find:
(i) the height of the cliff
(ii) the distance between the cliff and the tower.

Accepted Answer

45° 60° 45° 60° A B C D E 20 m Distance d Figure: Cliff $AB$ (height $H$) and tower $CD$ (height $20	ext{ m}$) separated by distance $d$ Step 1 Let cliff height be $H$ ($AB = H$) and distance between them be $d$ ($BD = EC = d$). Height of tower $CD = EB = 20	ext{ m} \implies AE = H - 20$. In right $	riangle AEC$: $	an 45^\circ = \frac{AE}{EC} = \frac{H - 20}{d} \implies 1 = \frac{H - 20}{d} \implies \mathbf{H = d + 20}$. [1.5 Marks] Step 2 In right $	riangle ABD$: $	an 60^\circ = \frac{AB}{BD} = \frac{H}{d} \implies \sqrt{3} = \frac{H}{d} \implies \mathbf{H = d\sqrt{3}}$. [1 Mark] Step 3 Equating expressions for $H$: $$d\sqrt{3} = d + 20 \implies d(\sqrt{3} - 1) = 20 \implies d = \frac{20}{\sqrt{3} - 1} = \frac{20(\sqrt{3} + 1)}{2} = 10(1.732 + 1) = \mathbf{27.32	ext{ m}}$$ Then cliff height $H = d + 20 = 27.32 + 20 = \mathbf{47.32	ext{ m}}$. Answer: (i) Height of cliff $= 47.32	ext{ m}$, (ii) Distance $= 27.32	ext{ m}$ . [1.5 Marks]

Question 6

A man observes the angle of elevation of the top of the tower to be $45^\circ$. He walks towards it in a horizontal line through its base. On covering $20	ext{ m}$ the angle of elevation changes to $60^\circ$. Find the height of the tower correct to 2 significant figures.

Accepted Answer

C D B A (Top) 45° 60° 20 m h ≈ 47 m Vector Diagram: Elevation angles 45° and 60° to tower top with 20 m baseline approach Step 1 Let $AB = h$ be the height of the vertical tower and $B$ be its base. Let observer start at $C$ ($\angle ACB = 45^\circ$) and walk $20	ext{ m}$ to $D$ ($\angle ADB = 60^\circ$). Let $DB = x$, then $CB = x + 20$. [1 Mark] Step 2 In right triangle $\Delta ABC$: $	an 45^\circ = \frac{AB}{CB} \implies 1 = \frac{h}{x + 20} \implies x + 20 = h \implies x = h - 20$. [1 Mark] Step 3 In right triangle $\Delta ABD$: $	an 60^\circ = \frac{AB}{DB} \implies \sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3}$. Substitute $x = h - 20$: $h = (h - 20)\sqrt{3} \implies h = h\sqrt{3} - 20\sqrt{3} \implies h(\sqrt{3} - 1) = 20\sqrt{3}$. $h = \frac{20\sqrt{3}}{\sqrt{3} - 1} = \frac{20\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = 10\sqrt{3}(\sqrt{3} + 1) = 10(3 + \sqrt{3})$. [1 Mark] Step 4 Evaluate numerically with $\sqrt{3} \approx 1.732$: $h = 10(3 + 1.732) = 10(4.732) = 47.32	ext{ m}$. Rounding to 2 significant figures: $\mathbf{h \approx 47	ext{ m}}$. Height of tower = 47 m . [1 Mark]

Question 7

As observed from the top of a light house, $80	ext{ m}$ above sea level, the angle of depression of a ship, sailing directly towards it, changes from $35^\circ$ to $55^\circ$. Draw a free hand sketch representing the given information. Find the distance travelled by the ship during the period of ob

Accepted Answer

A (Lighthouse Top) B (Base) C D 35° 55° 35° 55° 80 m Distance CD ≈ 58 m Vector Diagram: Elevation/depression geometry of ship approaching 80 m lighthouse Step 1 (a) Freehand sketch with angles of depression $35^\circ$ and $55^\circ$ converting to alternate angles of elevation. [1.5 Marks] Step 2 (b) In right triangle $\Delta ABD$ ($\angle ADB = 55^\circ$): $	an 55^\circ = \frac{AB}{BD} \implies BD = \frac{80}{	an 55^\circ}$. Using mathematical tables: $	an 55^\circ \approx 1.4281 \implies BD = \frac{80}{1.4281} \approx 56.02	ext{ m}$. [1.5 Marks] Step 3 In right triangle $\Delta ABC$ ($\angle ACB = 35^\circ$): $	an 35^\circ = \frac{AB}{BC} \implies BC = \frac{80}{	an 35^\circ}$. Using mathematical tables: $	an 35^\circ \approx 0.7002 \implies BC = \frac{80}{0.7002} \approx 114.25	ext{ m}$. [1 Mark] Step 4 Distance travelled by ship ($CD$): $CD = BC - BD = 114.25 - 56.02 = 58.23	ext{ m}$. Rounding to nearest meter: $\mathbf{CD \approx 58	ext{ m}}$. [1 Mark]

Question 8

The angle of elevation from a point P of the top of a tower QR, 50m high is $60^\circ$ and that of the tower PT from a point Q is $30^\circ$. Find the height of the tower PT, correct to the nearest metre.

Accepted Answer

Step 1 Let distance between the two towers $PQ = d$, and height of tower $PT = h$.
In right $\triangle PQR$ ($\angle PQR = 90^\circ$):
$$\tan 60^\circ = \frac{QR}{PQ} = \frac{50}{d}$$ $$\sqrt{3} = \frac{50}{d} \implies d = \frac{50}{\sqrt{3}}\text{ m}$$ [1.5 Marks]
Step 2 In right $\triangle QPT$ ($\angle QPT = 90^\circ$):
$$\tan 30^\circ = \frac{PT}{PQ} = \frac{h}{d}$$ $$\frac{1}{\sqrt{3}} = \frac{h}{d} \implies h = \frac{d}{\sqrt{3}}$$ [1.5 Marks]
Step 3 Substitute $d = \frac{50}{\sqrt{3}}$:
$$h = \frac{\frac{50}{\sqrt{3}}}{\sqrt{3}} = \frac{50}{3} = 16.67\text{ m}$$ [0.5 Mark]
Rounding to the nearest metre:
$$h \approx \mathbf{17\text{ metres}}$$ [0.5 Mark]
Answer: The height of the tower $PT$ is 17 metres.

Question 9

The angles of depression of two ships A and B as observed from the top of a light house 60 m high are $60^\circ$ and $45^\circ$ respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.

Accepted Answer

60° 45° A D B C (60 m) Step 1 Let $CD = 60	ext{ m}$ be the lighthouse. $A$ and $B$ are on opposite sides of base $D$. Angle of depression $60^\circ \implies \angle CAD = 60^\circ$. Angle of depression $45^\circ \implies \angle CBD = 45^\circ$. Step 2: Distance AD In right $	riangle CDA$ ($\angle D = 90^\circ$): $$	an 60^\circ = \frac{CD}{AD} \implies \sqrt{3} = \frac{60}{AD} \implies AD = \frac{60}{\sqrt{3}} = 20\sqrt{3}	ext{ m}$$ [1.5 Marks] Step 3: Distance BD In right $	riangle CDB$ ($\angle D = 90^\circ$): $$	an 45^\circ = \frac{CD}{BD} \implies 1 = \frac{60}{BD} \implies BD = 60	ext{ m}$$ [1 Mark] Step 4: Total Distance AB $$AB = AD + BD = 20\sqrt{3} + 60 = 20(1.732) + 60 = 34.64 + 60 = 94.64	ext{ m}$$ To the nearest whole number: $\mathbf{95	ext{ metres}}$ [1.5 Marks] Answer: The distance between the two ships is 95 metres .

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