Question

The angle of elevation of the top of a 10m high building from a point P on the ground is 30°. A flag is hoisted at the top of the building and the angle
of elevation of the top of the flag staff from P is 45°. Find the length of the flag staff and the distance of the building from point P. ​

Answer 1

Given :-

• Height of building=10m
• Angle of elevation from the top of building to point P is 30°
• And angle of elevation from Flag staff from P is 45°

To Find :-

• We have to find the length of flag staff
• And the distance of building from point P

Solution

• Let RQ be the building and ST be the flag staff
• Then ,

$\begin{cases}\sf{RQ=10m}\\\sf{\angle QPR=30^{\circ}}\\\sf{\angle QPS=45^{\circ}}\end{cases}$

• Let SR = h metres

From right ∆PQR , We have

$\dashrightarrow\sf\ \dfrac{RQ}{PQ}= tan30^{\circ}\\ \\ \\ \dashrightarrow\sf\ \dfrac{10}{PQ}=\dfrac{1}{\sqrt{3}}\ \ \ \ \ \Big\lgroup\ tan30^{\circ}=\dfrac{1}{\sqrt{3}}\Big\rgroup\\ \\ \\ \dashrightarrow\sf\ \ 10\sqrt{3}=PQ\\ \\ \\ \dashrightarrow\underline{\boxed{\purple{\sf\ \ PQ=10\sqrt{3}}}}$

★From right ∆PQS, we have

$\dashrightarrow\sf\ \dfrac{SQ}{PQ}=tan45^{\circ}\\ \\ \\ \dashrightarrow\sf\ \dfrac{10+h}{10\sqrt{3}}=1\ \ \ \ \ \Big\lgroup\ tan45^{\circ}=1\Big\rgroup\\ \\ \\ \dashrightarrow\sf\ 10+h=10\sqrt{3}\\ \\ \\ \dashrightarrow\sf\ h= 10\sqrt{3}-10\\ \\ \\ \dashrightarrow\sf\ h= 10(\sqrt{3}-1)\\ \\ \\ \dashrightarrow\sf\ h= 10\big(1.732-1\big)\ \ \ \ \Big\lgroup \sqrt{3}=1.732\Big\rgroup\\ \\ \\ \dashrightarrow\sf\ h= 10\times 0.732\\ \\ \\ \dashrightarrow\underline{\boxed{\red{\sf\ \ h=7.32m}}}$

$\therefore\underline{\sf Height \ of\ flag\ staff= 7.32\ m}\\ \\ \therefore \underline{\sf Distance\ of\ P\ from\ building=10\sqrt{3}m}$

Answer 2

$\huge{\underline{\mathrm{\blue{Answer-}}}} </p><p></p><p>$

Distance of the building from point P is 17.32 m.

Length of flagstaff = 7.32 m

$\huge{\underline{\mathrm{\blue{Explanation-}}}} </p><p>$

In ∆PAB,

Tan30° =

$\sf{\dfrac{AB}{AP}} </p><p>$

$⟹ \sf{\dfrac{1}{\sqrt{3}}\:=\:\dfrac{10}{AP}} </p><p>$

⟹ AP = 10√3

⟹ AP = 10 × 1.73

⟹ AP = 17.32 m

∴ Distance of the building from point P is 17.32 m.

Now, let's suppose DB = x m.

AD = (10+x) m.

In ∆PAD,

Tan45° =

$\sf{\dfrac{AD}{AP}} </p><p>$

$⟹ 1 = \sf{\dfrac{10+x}{10\sqrt{3}}} </p><p>$

⟹ x = 10(√3-1)

⟹ x = 7.32 m

∴ Length of flagstaff = 7.32 m