Problem Statement
From the top of a $7\text{ m}$ 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 $45^\circ$. Determine the height of the tower.
Verified Solution & Marking Scheme
Construct geometric model
Let $AB = 7\text{ m}$ be the building. Let $CD = H$ be the cable tower with base $D$.
Draw horizontal $AE \perp CD$ with $E$ on $CD$.
Then $ED = AB = 7\text{ m}$ and $CE = H - 7$.
Angle of elevation of top $C$ from $A$ is $\angle CAE = 60^\circ$.
Angle of depression of foot $D$ from $A$ is $\angle EAD = 45^\circ \implies \angle ADB = 45^\circ$.
Find distance between building and tower BD
In right $\triangle ABD$:
$\tan 45^\circ = \frac{AB}{BD} \implies 1 = \frac{7}{BD} \implies BD = 7\text{ m}$
Since $AE = BD$, $AE = 7\text{ m}$.
Find height of upper section CE
In right $\triangle AEC$:
$\tan 60^\circ = \frac{CE}{AE} \implies \sqrt{3} = \frac{CE}{7} \implies CE = 7\sqrt{3}\text{ m}$
Calculate total height of the tower CD
$CD = CE + ED = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1)\text{ m}$
If $\sqrt{3} \approx 1.732$, $CD = 7(2.732) \approx 19.124\text{ m}$.