Calculate distance from a point 100m from a plane given two angles

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The discussion centers on calculating the distance from a point 100 meters above a horizontal plane, given two angles of 8 degrees and 21 degrees. Participants note that the original question lacks sufficient information and clarity, particularly regarding the concept of a "plane of 100 meters." It is suggested that the inquiry may actually involve finding the height of a triangle with a base of 100 meters and the specified angles. A diagram is recommended for better understanding, and there is a hint that this might be related to a schoolwork problem. Clearer parameters are necessary to provide a meaningful solution.
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I am looking for a formula.

From a horizontal plane of 100 meters; If angle on the left is 8 degrees and the angle on the right is 21 degrees at what distance from the centre of the horizontal plane will these two angles converge?
 
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Your question above does not contain enough information for an answer. We will need more parameters to provide any meaningful answer.
 
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This is not differential geometry.

There is no such thing as a 'plane of 100 metres', and planes do not have a centre.

I think the question you meant to ask is "a triangle has a base of 100 metres and sides at angles 8 degrees and 21 degrees from the base; what is the height of the triangle?". If this is not what you meant or you are unsure, draw a diagram.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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