Question

Two persons on same side of tall building notice angle of elevation of top of building to be 30° and 60°. If height of building is 72m find distance between two persons √3=1.73

Answer 1

▶ Question :-

→ Two persons on same side of tall building notice angle of elevation of top of building to be 30° and 60°. If height of building is 72m, find distance between two persons . [ √3=1.73 ]


$\huge \orange{ \mid{ \underline{ \overline{ \sf Solution :- }} \mid}}$


→ Let AB be the height of the building = 72 m .

→ And, suppose two person standing at point C and D making an angle of elevation to the top of the building is 60° and 30° respectively.

→ Let the distance between two person CD be x m.

→ And, CB = y m.


▶ Now,

In right ∆ABC,

$\sf \because \tan 60 \degree = \frac{AB}{BC}. \\ \\ \sf \implies \sqrt{3} = \frac{72}{y} . \\ \\ \sf \implies y = \frac{72}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } . \\ \\ \sf \implies y = \frac{72 \sqrt{3} }{3} . \\ \\ \sf \therefore y = 24 \sqrt{3} m.$

And,

In right ∆ABD

$\sf \because \tan 30 \degree = \frac{AB}{BD} . \\ \\ \sf \implies \frac{1}{ \sqrt{3} } = \frac{72}{x + y} . \\ \\ \sf \implies \frac{1}{ \sqrt{3} } = \frac{72}{x + 24 \sqrt{3} } . \\ \\ \sf \implies x + 24 \sqrt{3} = 72 \sqrt{3} . \\ \\ \sf \implies x = 72 \sqrt{3} - 24 \sqrt{3} . \\ \\ \boxed{ \pink{ \sf \therefore x = 48 \sqrt{3} m.}} \\ \\ or \\ \\ \boxed{ \pink{ \sf \therefore x = 83.04 m.}}$


✔✔ Hence, the distance between two person is 83.04 m ✅✅ .


$\huge \boxed{ \boxed{ \red{\mathcal{THANKS}}}}$

Answer 2

$\huge \bf \red{ANSWER}$

$\bf{QUESTION}$
Two persons on same side of tall building notice angle of elevation of top of building to be 30° and 60°. If height of building is 72m find distance between two persons √3=1.73.

$\bf{step \: by \: step \: explanation}$

$\bf{Given - }$

height of building= 72m

angles are

30° and 60°

figure is in attachment

Let BC=x

and

CD=y

=> In triangle ABC

$tan60 \degree = \frac{ab}{bc}$

$\sqrt{3} = \frac{72}{x}$

$x = \frac{72}{ \sqrt{3} }$

$x = \frac{72}{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} }$

$x = \frac{72 \sqrt{3} }{3}$

now

In triangle ABD

$tan30 \degree = \frac{ab}{bd}$

$\frac{1}{ \sqrt{3} } = \frac{72}{bd}$
$bd = 72 \sqrt{3}$
hence

distance between two persons=CD

CD=BD-BC

$cd = {72 \sqrt{3}} - \frac{72 \sqrt{3} }{3}$
$cd = 72 \sqrt{3} - 24 \sqrt{3}$
$\bf{cd = 48 \sqrt{3}}$

$\sqrt{3}=1.73$

Hence

$48\sqrt{3}=83.04m$

Hence solved✔✔

$\bf{thanks}$