how he is using the d

plz do in simple steps

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- July 9th 2011, 01:35 AMmoonnightingalekindly explain the step
how he is using the d

plz do in simple steps

file attached - July 9th 2011, 01:42 AMSironRe: kindly explain the step
If you write:

as

Can you complete? ...

For the other term it's similar. - July 9th 2011, 01:44 AMGokuRe: kindly explain the step
First take the d^2 as a common factor for both terms and then work from there..

i.e (ht + hr)^2 +d^2 = d^2 [(ht+hr)^2 / d^2 + 1] - July 9th 2011, 01:45 AMmoonnightingaleRe: kindly explain the step
other term is delta

- July 9th 2011, 01:48 AMmoonnightingaleRe: kindly explain the step
- July 9th 2011, 01:51 AMSironRe: kindly explain the step
Because you have to take the square root.

- July 9th 2011, 01:51 AMGokuRe: kindly explain the step
take the

of d^2 [(ht+hr)^2 / d^2 + 1], this whole term is underneath a square root...__square root__

(d^2 [(ht+hr)^2 / d^2 + 1])^1/2 = d [(ht+hr)^2 / d^2 + 1]^1/2 - July 9th 2011, 01:53 AMmoonnightingaleRe: kindly explain the step
no it is not working

can u show me these simple steps

thanks a lot - July 9th 2011, 01:54 AMmoonnightingaleRe: kindly explain the step
- July 9th 2011, 01:55 AMGokuRe: kindly explain the step
http://upload.wikimedia.org/math/5/e...6beec9d791.png

x = d^2

y = (ht+hr)^2 / d^2 + 1