AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A balloon moving in a straight line passes vertically above two points $A$ and $B$ , on a horizontal plane, $1000 ft$ apart. When above $A$ , it has an altitude of 60 $°$ as seen from $B$ and when above $B$ it has an altitude of $45 °$ as seen from $A$ . The distance, from $A$ , of the point at which it will touch the plane is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

2266

2466

2566

2366

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

It is given a balloon moving in a straight line passes vertically above two points

A

and

B

, on a horizontal plane

1000 ft

apart.
According to the given data, the figure has been drawn below.
Question Image

It is given that,

AB = 1000

ft.
Now, in

△ ABE

tan θ = P e r p e n d i c u l a r b a s e

\tan {60 °}^{} = \frac{AE}{AB}

\Rightarrow AE = ABtan 60 °

= 1000 \sqrt{3}

∵ tan 60 ° = 3

Similarly, in

△ BAD

tan θ = P e r p e n d i c u l a r b a s e

\tan 45 ° = \frac{DB}{AB}

\Rightarrow DB = ABtan 45 °

∵ tan 45 ° = 1

= 1000 (1)

= 1000

△ ACE

and

△ DBC

are similar so we have,

\frac{AE}{AC} = \frac{DB}{BC}

\Rightarrow \frac{1000 \sqrt{3}}{1000 + d} = \frac{1000}{d}

\Rightarrow \frac{\sqrt{3}}{1000 + d} = \frac{1}{d}

\Rightarrow \sqrt{3} = \frac{d + 1000}{d}

\Rightarrow \sqrt{3} = 1 + \frac{1000}{d}

\Rightarrow \sqrt{3} - 1 = \frac{1000}{d}

\Rightarrow d = \frac{1000}{\sqrt{3} - 1}

Now, substituting value of

\sqrt{3}

=1.732,

\Rightarrow d

= 1366 ft

Distance from point C to Point A

= 1000 + d

\Rightarrow 1000 ft + 1366 ft

\Rightarrow 2366 ft

So the correct option is 4.

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!