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Q12:
From the top of a $7$ m high building, the angle of elevation of the top of a cable tower is $60°$ and the angle of depression of its foot is $45°$. Determine the height of the tower.

Solution :

Given:

1. Height of the building ($AB$) = $7$ m.

2. Angle of elevation of the top of the tower ($EC$) from the top of the building ($A$) = $60^\circ$.

3. Angle of depression of the foot of the tower ($D$) from the top of the building ($A$) = $45^\circ$.

To find:

The total height of the cable tower ($CD$).

7m A B C D 60° 45°

Step 1: Define the variables and geometric relationships.

Let $AB$ be the building of height $7$ m. Let $CD$ be the cable tower. Let $A$ be the top of the building and $B$ be its foot. Let $C$ be the top of the tower and $D$ be its foot. Let $E$ be a point on $CD$ such that $AE$ is horizontal. Thus, $AE = BD$ and $ED = AB = 7$ m.

Step 2: Analyze triangle $\triangle ABD$.

In the right-angled triangle $\triangle ABD$, the angle of depression from $A$ to $D$ is $45^\circ$. Therefore, the angle of elevation from $D$ to $A$ is also $45^\circ$ (alternate interior angles).

Using the trigonometric ratio tangent:

$\tan(45^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{BD}$

Since $\tan(45^\circ) = 1$:

$1 = \frac{7}{BD}$

$BD = 7$ m

[Since $AE$ is parallel to $BD$, $AE = BD = 7$ m].

Step 3: Analyze triangle $\triangle AEC$.

In the right-angled triangle $\triangle AEC$, the angle of elevation is $60^\circ$.

$\tan(60^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{CE}{AE}$

We know $\tan(60^\circ) = \sqrt{3}$ and $AE = 7$ m:

$\sqrt{3} = \frac{CE}{7}$

$CE = 7\sqrt{3}$ m

Step 4: Calculate the total height of the tower.

The total height of the tower $CD = CE + ED$.

We know $ED = AB = 7$ m.

$CD = 7\sqrt{3} + 7$

$CD = 7(\sqrt{3} + 1)$ m

Using the approximation $\sqrt{3} \approx 1.732$:

$CD \approx 7(1.732 + 1) = 7(2.732) = 19.124$ m.

Final Answer: The height of the tower is $7(\sqrt{3} + 1)$ m or approximately $19.12$ m.


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