Question

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I need help with a) ii. & iii.
For i, the answer is: h tan (a)

Answer 1

if you draw OB, you should be able to find the sides of the right triangle OBX easily. R is the hypotenuse of that triangle.

Answer 2

then what is OX?

Answer 3

what is AX?
what is AO?

Answer 4

If you read the question:

AX = h
I do not know what AO is, but the point O represents the centre of the circumference of the circle.

Answer 5

I read the question. I just think you can figure it out.

Answer 6

I have tried it recently, but it's not the correct answer. I end up getting \[R = h \sqrt{1+\tan(a)^{2}}\].

The correct answer is: \[R = \frac{h}{2} (1+ \tan^2(a)\] (from the book)

Answer 7

\[\frac{BX}{h} = \tan \alpha\]
\[BX = h \tan \alpha\]
\[OX = h - R = \frac{BX}{tan \alpha} - R \]
\[OX^2 + BX^2 = R^2\]

Answer 8

I get the answer in the book.

Answer 9

\[R^2 = (h-R)^2 + h^2 \tan^2 \alpha = h^2 - 2R + R^2 + h^2 \tan^2 \alpha\]
\[2R = h^2(1 + \tan^2 \alpha)\]

Answer 10

I have h^2. Hmm.. Can you work it from here?

Answer 11

I dropped a factor of h
should have -2hR in multiplying out (h - R)^2

Answer 12

Thank you I got it!

Answer 13

I have another problem could you help me? I don't understand it.