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 $1.2 m$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2 m$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $6 0^{\circ}$ . After some time, the angle of elevation reduces to $3 0^{\circ}$ (See figure). Find the distance travelled by the balloon during the interval.

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

58 \sqrt{4} m

8 \sqrt{3} m

5 \sqrt{3} m

58 \sqrt{3} m

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Given,
Height of girl,

AG = 1.2 m

EF = 88.2 m

∠ AEB = 6 0^{\circ}

∠ DAC = 3 0^{\circ}

In the figure,

AG ∥ BF

AG = BF = 1.2 m

Description: C:UsersadminDesktopierkhtrjhgtjg.PNG

Description: C:UsersadminDesktopCapture.PNG

Distance travelled by the balloon

= BC

By the figure,

EF = BE + BF

∴ BE = EF - BF

Put the values,

BE = 88.2 - 1.2 = 87 m

BE = DC = 87 m

Again, by the figure,

∠ ABE ∠ ACD

are right angle

∴ ∠ ABE = ∠ ACD = 9 0^{\circ}

Also,

∠ EAB = 6 0^{\circ}

BE = 87 m

Thus,

\tan A = \frac{BE}{AB}

\tan 6 0 = \frac{87}{AB}

\Rightarrow AB = \frac{87}{\tan 6 0^{\circ}}

\Rightarrow AB = \frac{87}{\sqrt{3}} m

△ DAC

DC = 87 m

∠ DAC = 3 0^{\circ}

Thus,

\tan 3 0 = \frac{DC}{AC}

\Rightarrow \tan 3 0^{\circ} = \frac{87}{AC}

∴ AC = \frac{87}{\tan 30}

AC = \frac{87}{\frac{1}{\sqrt{3}}} (∵ \tan 3 0 = \frac{1}{\sqrt{3}})

AC = 87 \sqrt{3} m

It can be seen in the figure that,

AC = AB + BC

∴ BC = AC - AB

BC = 87 \sqrt{3} - \frac{87}{\sqrt{3}}

BC = \frac{87 \times 3 - 87}{\sqrt{3}} = \frac{174}{\sqrt{3}}

BC = \frac{174 \sqrt{3}}{\sqrt{3} \times \sqrt{3}}

BC = \frac{174 \sqrt{3}}{3}

BC = 58 \sqrt{3} m

Thus, distance travelled by the balloon

= 58 \sqrt{3} m

Hence option 4 is correct.

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!