Question

The angle of elevation of the top of a 10-metre high building from a point P on the earth is 30°. There is a flag on top of the building. The angle of elevation of the top of the flag from point P is 45°. Find the length of the flag.

Hint:

For height and distance problems, always form two right-angled triangles and apply $\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$. Subtract the smaller height from the larger to find the required vertical distance.

Answer 1

Step 1: Let the height of the flag be $h$ metres.
Then, total height of the building and flag = $(10 + h)$ m. Step 2: Draw a right-angled triangle.
Let the distance of point P from the base of the building be $x$ m. From the figure (mentally imagined): - $\triangle PAB$ represents the building top, and - $\triangle PAC$ represents the top of the flag. Step 3: Use trigonometric ratios.
From $\triangle PAB$: \[ \tan 30° = \frac{\text{height of building}}{\text{distance}} = \frac{10}{x} \Rightarrow x = \frac{10}{\tan 30°} \Rightarrow x = \frac{10}{\frac{1}{\sqrt{3}}} = 10\sqrt{3} \]
Step 4: From $\triangle PAC$: \[ \tan 45° = \frac{\text{total height}}{\text{distance}} = \frac{10 + h}{x} \] Substitute $x = 10\sqrt{3}$ and $\tan 45° = 1$: \[ 1 = \frac{10 + h}{10\sqrt{3}} \Rightarrow 10 + h = 10\sqrt{3} \] \[ h = 10(\sqrt{3} - 1) \]
Step 5: Approximate value (if needed).
Since $\sqrt{3} \approx 1.732$, \[ h = 10(1.732 - 1) = 10(0.732) = 7.32 \, \text{m} \]
Step 6: Conclusion.
Hence, the length of the flag is \[ \boxed{h = 10(\sqrt{3} - 1)\ \text{metres} \ (\approx 7.32 \text{ m})} \]