From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take 3 = 1.732)
Step-by-Step Solution
Key Concept: Use the definition of tangent in a right‑angled triangle: \(\tan\theta = \dfrac{\text{opposite side}}{\text{adjacent side}}\). Apply it first to the building (height 10 m) and then to the total height (building + flagstaff). Solve the two equations to obtain the distance and the flagstaff length.
1. Let the horizontal distance from point \(P\) to the foot of the building be \(x\) metres.
2. For the top of the building (height = 10 m) the angle of elevation is \(30^{\circ}\).
$$\tan 30^{\circ}=\frac{10}{x}$$
Using \(\tan 30^{\circ}=\frac{1}{\sqrt{3}}\) and \(\sqrt{3}=1.732\),
$$\frac{1}{1.732}=\frac{10}{x}\quad\Rightarrow\quad x=10\times 1.732=10\sqrt{3}\;\text{m}$$
Hence, \(x = 10\sqrt{3}\) m \(\approx 17.32\) m.
3. Let the length of the flagstaff be \(h\) metres. The total height of the flagstaff top is \(10+h\) metres.
4. The angle of elevation to the top of the flagstaff is \(45^{\circ}\).
$$\tan 45^{\circ}=\frac{10+h}{x}$$
Since \(\tan 45^{\circ}=1\),
$$1=\frac{10+h}{x}\quad\Rightarrow\quad 10+h = x$$
5. Substitute the value of \(x\) from step 2:
$$10+h = 10\sqrt{3}\quad\Rightarrow\quad h = 10\sqrt{3}-10 = 10(\sqrt{3}-1)\;\text{m}$$
Numerically, \(h = 10(1.732-1)=10\times0.732=7.32\) m.
6. Therefore,
- Distance of the building from \(P\): \(x = 10\sqrt{3}\) m \(\approx 17.32\) m.
- Length of the flagstaff: \(h = 10(\sqrt{3}-1)\) m \(\approx 7.32\) m.
Correct Answer: Distance of the building from P = $10\sqrt{3}$ m \(\approx 17.32\) m; Length of the flagstaff = $10(\sqrt{3}-1)$ m \(\approx 7.32\) m.