A $1.6$ m tall observer is $20$ m away from a building. The angle of elevation of the top of the building from the observer's eyes is $60^\circ$. Find the height of the building.
Step-by-Step Solution
Key Concept: Find the height of the building above eye level first, then add the observer's own height.
Let the height above eye level be $x$: $\tan60^\circ=\dfrac{x}{20}\Rightarrow\sqrt3=\dfrac{x}{20}$. [1.0 Mark]
$x=20\sqrt3$ m. [1.0 Mark]
Total height of building $=x+1.6=(20\sqrt3+1.6)$ m $\approx35.2+1.6=36.8$ m (using $\sqrt3\approx1.76$ approx, or leave in surd form as $(20\sqrt3+1.6)$ m). [1.0 Mark]
Correct Answer: