A vertical tower TC of height h metres stands on a horizontal plane with its base at C. Po… — JAMB UTME

A vertical tower $TC$ of height $h$ metres stands on a horizontal plane with its base at $C$ . Point $A$ on the plane is due West of $C$ , and point $B$ is due South of $A$ , such that the distance $AB = 40\text{ m}$ . If the angles of elevation of the top $T$ of the tower from $A$ and $B$ are $60^\circ$ and $30^\circ$ respectively, what is the height of the tower?

$10\sqrt{6}\text{ m}$ Answer
B
$4\sqrt{30}\text{ m}$
C
$20\sqrt{2}\text{ m}$
D
$20\sqrt{3}\text{ m}$

Answer

The height of the tower is

10\sqrt{6}\text{ m}

Using the trigonometric tangent ratios, the horizontal distances from the base of the tower are

AC = \frac{h}{\sqrt{3}}

and

BC = h\sqrt{3}

. Because point

A

is West of

C

and point

B

is South of

A

, the angle

\angle CAB = 90^\circ

. Applying Pythagoras' theorem

BC^2 = AC^2 + AB^2

yields

3h^2 = \frac{h^2}{3} + 1600

, which simplifies to

h = 10\sqrt{6}\text{ m}

Step-by-Step Solution

Express the horizontal distances

AC

and

BC

in terms of height

h

In vertical right-angled triangle

TAC

\tan(60^\circ) = \frac{h}{AC} \implies AC = \frac{h}{\sqrt{3}}

. In vertical right-angled triangle

TBC

, \tan(30^\circ) = \frac{h}{BC} \implies BC = h\sqrt{3}$.

Relate the vertical height to the horizontal ground distances using tangent ratios.

Identify the geometry of the ground plane triangle

\triangle CAB

Since

A

is due West of

C

and

B

is due South of

A

, the line segments

CA

(East-West) and

AB

(North-South) are perpendicular. Thus,

\triangle CAB

is right-angled at

A

Cardinal directions (West and South from

A

) are perpendicular to each other.

Apply Pythagoras' theorem to

\triangle CAB

BC^2 = AC^2 + AB^2 \implies (h\sqrt{3})^2 = \left(\frac{h}{\sqrt{3}}\right)^2 + 40^2 \implies 3h^2 = \frac{h^2}{3} + 1600

Hypotenuse

BC

relates the two legs

AC

and

AB

on the horizontal plane.

Solve the equation for

h

3h^2 - \frac{h^2}{3} = 1600 \implies \frac{8h^2}{3} = 1600 \implies 8h^2 = 4800 \implies h^2 = 600 \implies h = \sqrt{600} = 10\sqrt{6}\text{ m}

Simplify algebraic terms to determine

h

in surd form.

Key Concept

Combining angles of elevation in 3D vertical planes with horizontal plane coordinate geometry and Pythagoras' theorem.

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