Transposition of Formula

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January 31st 2009, 11:13 AM
berniecole

Transposition of Formula

I want to find the height of a pole H

From a distance, I measure the angle (Theta 1) to the top of the pole.
I then move a distance D towards the pole and measure the angle to the top of the pole again (Theta 2). I do not know my distance from the pole so call it X

I know that H/X=Tan(Theta 2)
I also know that H/(X+D)=Tan(Theta 1)

I know by transposition that H=X(tan(Theta 2)) & H=(X+D)(Tan(Theta 1))

To find H I need to find X

This is where I get lost in my Transposition!!

Can anybody please help me, I'm loosing a lot of sleep over this
January 31st 2009, 11:26 AM
earboth

Quote:

Originally Posted by berniecole

I want to find the height of a pole H

From a distance, I measure the angle (Theta 1) to the top of the pole.
I then move a distance D towards the pole and measure the angle to the top of the pole again (Theta 2). I do not know my distance from the pole so call it X

I know that H/X=Tan(Theta 2) (1)
I also know that H/(X+D)=Tan(Theta 1) (2)

I know by transposition that H=X(tan(Theta 2)) & H=(X+D)(Tan(Theta 1))

To find H I need to find X

This is where I get lost in my Transposition!!

Can anybody please help me, I'm loosing a lot of sleep over this

Solve (1) for x. Plug in this term into (2) and solve for H:

$\dfrac Hx = \tan(\theta_2)~\implies~ x = \dfrac H{\tan(\theta_2)}$ Therefore the second equation becomes:

$H=(x+D)\tan(\theta_1)~\implies~ H=\left(\dfrac H{\tan(\theta_2)} + D\right) \tan(\theta_1)$

Solve this equation for H.
January 31st 2009, 11:42 AM
berniecole

How do I combine the 2 Hs in the final equation?

Regards
January 31st 2009, 12:02 PM
earboth

Quote:

Originally Posted by berniecole

How do I combine the 2 Hs in the final equation?

Regards

$<br /> H=(x+D)\tan(\theta_1)~\implies~ H=\left(\dfrac H{\tan(\theta_2)} + D\right) \tan(\theta_1)<br />$

1. Expand the bracket:

$H=\dfrac {H \cdot \tan(\theta_1)}{\tan(\theta_2)} + D \cdot \tan(\theta_1)$

2. Take the term containing H from the RHS to the LHS:

$H-\dfrac {H \cdot \tan(\theta_1)}{\tan(\theta_2)} = D \cdot \tan(\theta_1)$

3. Factor out H:

$H\left(1-\dfrac {\tan(\theta_1)}{\tan(\theta_2)}\right) = D \cdot \tan(\theta_1)$

4. Divide by the bracket.
January 31st 2009, 02:11 PM
berniecole

Transposition of Formula

Many thanks for your explanation, it was greatly appreciated.

Clearly I now need to develop my factoring skills, this is where I fell down.

Regards

Berniecole