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Cutting into rings , the area of a small ring at an angle $\theta$ would be $2 \pi R^2 sin\theta d\theta$.

Hence area of visible surface for an angle $\phi$(Shown as $\frac{\pi}{4}$ in Pranav A.'s image) = $\displaystyle \int_{0}^{\phi}2 \pi R^2 sin\theta d\theta$

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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TopNewestHi, it should be $6400Km$

Here is the solution,

Cutting into rings , the area of a small ring at an angle $\theta$ would be $2 \pi R^2 sin\theta d\theta$.

Hence area of visible surface for an angle $\phi$(Shown as $\frac{\pi}{4}$ in Pranav A.'s image) = $\displaystyle \int_{0}^{\phi}2 \pi R^2 sin\theta d\theta$

= $2 \pi R^2(1 - cos\phi) = \frac{1}{4}4 \pi R^2$(given)

Hence, $\phi = \frac{\pi}{3}$.

Hence, height from surface = $\frac{R}{cos\phi} - R = R = 6400Km$

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Great!, first, when i posted the question resonance was showing answer to be $3200Km$, but now they have changed their answer.

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Image

Since $\cos(\pi/4)=R/(R+h)$, h can be calculated. R is the radius of earth.

Is it possible for you to share the answer?

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I would like to request the members to stop voting up my reply. Jatin's answer is correct and nicely explained.

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Why people voted down?,they should not have the right to vote down something other than an abuse .

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Can you tell me what would be the expected cutoff.

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I don't have any idea , previous year it was 56

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Hello Jatin, I am just curious to know that you're in which class?

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Its ok.

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