Question

Class-10 Math NCERT ex 9.1 Questions no 6


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Answer 1

Note :-

• Refer to the above attachment !

Given :-

• height of the boy is 1.5 m

• height of the building is 30 m

• first angle of elevation is 30°

• second angle of elevation is 60°

To Find :-

• the distance walked the by the boy towards the building (DF).

Remember :-

⇒ tan 30° = $\large{\sf{\frac{1}{√3}}}$

⇒ tan 60° = $\large{\sf{√3}}$

Solution :-

• DE (height of the boy) = 1.5 m

• AC (height of the building) = 30 m

⇒ AB = AC - DE

⇒ AB = 30 m - 1.5 m

∴ AB = 28.5 m

• ∠ADB = 30°

• ∠AFB = 60°

★ we know :-

⇒ tan 30° = $\large{\sf{\frac{1}{√3}=\frac{28.5}{DB}}}$

∴ DB = 28.5√3

⇒ tan 60° = $\large{\sf{\frac{√3}{1}=\frac{28.5}{FB}}}$

∴ FB = $\large{\sf{\frac{28.5}{√3}}}$

so, then :-

• DF (distance walked walked by boy) = DB - FB

⇒DF = $\large{\sf{28.5√3 -\frac{28.5}{√3}}}$

⇒DF = $\large{\sf{\frac{(28.5×3)-28.5}{√3}}}$

⇒DF = $\large{\sf{\frac{57}{√3}}}$

∴ DF = 32.91 m (approx.)

Answer :-

• Therefore, the distance walked by the boy is 32.91 m

Answer 2

Easy Solution:

AB= 30-1.5=28.5 cm

In ΔABD:−tan30∘= $\frac{AB}{BD}$

⇒BD=(28.5) $\sqrt{3}$

In ΔABC:−tan60 ∘= $\frac{AB}{BC}$

⇒BC= $\frac{(28.5)}{ \sqrt{3} }$

CD= BD-BC

=$28.5( \sqrt{3} - \frac{1}{ \sqrt{3} })$

$= 28.5( \frac{2}{ \sqrt{3}} ) = 32.91m$

Therefore,The distance walked by boy is 32.91m.