Chapter 9 · Question 5

From a point $P$ on the ground, the angle of elevation of the top of a $10\text{ m}$ tall building is $30^\circ$ . A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from $P$ is $45^\circ$ . Find the length of the flagstaff and the distance of the building from point $P$ . (Take $\sqrt{3} = 1.732$ )

Answer Revealed

Direct Answer:

Let building height

AB = 10\text{ m}

, flagstaff

BD = x\text{ m}

. In

\triangle PAB

\tan 30^\circ = \frac{10}{AP}

, so

AP = 10\sqrt{3} \approx 17.32\text{ m}

. In

\triangle PAD

\tan 45^\circ = \frac{10 + x}{AP} = 1

, so

10 + x = 10\sqrt{3}

, giving

x = 10(\sqrt{3} - 1) \approx 7.32\text{ m}

. Distance

AP = 10\sqrt{3} \approx 17.32\text{ m}

The building is about

17.32\text{ m}

away from point

P

, and the flagstaff is about

7.32\text{ m}

long.

Let

AB = 10\text{ m}

(building height) and

BD = x\text{ m}

(flagstaff). The total height from ground to flag top is

AD = (10 + x)\text{ m}

. Step 1 (Find distance $AP$ ): In right

\triangle PAB

\tan 30^\circ = \frac{AB}{AP}

\frac{1}{\sqrt{3}} = \frac{10}{AP}

, so

AP = 10\sqrt{3} \approx 17.32\text{ m}

. Step 2 (Find flagstaff length): In right

\triangle PAD

\tan 45^\circ = \frac{AD}{AP}

1 = \frac{10 + x}{10\sqrt{3}}

, so

10 + x = 10\sqrt{3}

, giving

x = 10(\sqrt{3} - 1) \approx 10 \times 0.732 = 7.32\text{ m}

. Answer: Distance of building from point

P

10\sqrt{3}\text{ m}

(about

17.32\text{ m}

), and the length of the flagstaff is

10(\sqrt{3} - 1)\text{ m}

(about

7.32\text{ m}

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