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Trignometry

Blazing goIITian

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6 Sep 2008 17:37:20 IST

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482

Heigth and distance:

None

At wach end of a horizonatl base of length 2a it is found that the angular heigth of a certain peak is $\theta$ and that at the

midpoint it is $\phi$ . Prove that the vertical heigth of the peak is $\frac{a~\sin~\theta\sin~\phi}{\sqrt{\sin(\phi+\theta)\sin(\phi-\theta)}}$

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®µD®A

Blazing goIITian

Joined: 12 Apr 2008

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7 Sep 2008 08:43:50 IST

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Please any body solve this one.

®µD®A

Blazing goIITian

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7 Sep 2008 19:36:04 IST

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Atleast give a hint. :(

Shreya

Blazing goIITian

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8 Sep 2008 13:05:01 IST

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wach end means?

Gaurav |spideyunlimited| Ragtah

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8 Sep 2008 13:43:22 IST

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'wach' is a typo for 'each' I guess.

®µD®A

Blazing goIITian

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8 Sep 2008 14:09:05 IST

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OOps sorry. That is a typo for each .

Gaurav |spideyunlimited| Ragtah

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8 Sep 2008 14:20:39 IST

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Let AB be the distance between midpoint of rod and base of mountain and let the height of the mountain be H

tan (phi) = H / AB
AB = H / tan(phi)

Let CB be distance between one end of the rod to the base of the mountain.
Using pythagoras theorem:
(CB)^2 = H^2 / tan^2 (phi) + a^2

now Tan (theta) = H / CB
so equating CB from both equations:
a^2 + H^2 / tan^2 (phi) = H^2 / tan^2 (theta)
Write in terms of sin and cos.

Solving, you get:
H = a sin(theta).sin(phi) / root [ sin(phi + theta).sin(phi - theta) ]

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