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Samiti

· started a discussion

· 1 Months ago

Explain (Question:97093.0)

Pls explain

Question:
A round balloon of radius r subtends an angle \(\alpha\) at the eye of an observer while the angle of elevation of its center is β. The height of the center of the balloon is:
Options:
A)  r sin β. cosec \(\cfrac{\alpha}{2}\)
B) r cos β. Cosec \(\cfrac{\alpha}{2}\)
C) r cosec \(\alpha\). sin β
D) r2 sin \(\cfrac{\beta}{2}\). cos \(\cfrac{\alpha}{2}\)
Solution:

Let O be the center of balloon of radius r. The observer’s eye is at C.

∠ACB = \(\alpha\) and   ∠OCD = β clearly, CA and CB are tangents to the circle. So

∠ACO = ∠BCO = \(\cfrac{\alpha}{2}\)

In right angled ∆OBC,


Knowledge Expert

· commented

· 1 Months ago

Dear student,
Please read the solution properly.
Keep learning,
Team TR

Knowledge Expert

· commented

· 1 Months ago

Dear student,
Please read the solution properly.
Keep learning,
Team TR
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