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**Vedanti****Member**- Registered: 2017-12-19
- Posts: 7

The solution to the following problem confused me a little:

Problem: Points P and Q are on sides CB and CD, respectively, of square ABCD such that CP=CQ=AB/4. Find tan(angle PAQ).

Solution: We have tan(angle PAQ) = (tan(angle QAB - angle PAB) = tan(angle QAB) - tan(angle PAB))/(1 + (tan(angle QAB))(tan(angle PAB)).

The solution continues from here but I don't understand the part where it says (tan(angle QAB - angle PAB) = tan(angle QAB) - tan(angle PAB))/(1 + (tan(angle QAB))(tan(angle PAB)). How do you get this equation?

Thanks in advance for the help!

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,930

hi Vedanti

Welcome to the forum.

This is one of six compound angle formulas. You'll find them at the bottom of this page:

https://www.mathsisfun.com/algebra/trig … ities.html

tan (A-B) is the one used in your question.

Those formulas are the last topics covered by the MIF trigonometry pages and no proof is given.

You can probably find proofs elsewhere or I'll post one here if you ask. There are two ways I know of (1) using geometry, (2) using matrix multiplication. Say which youthink you'll find easier to follow.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Vedanti****Member**- Registered: 2017-12-19
- Posts: 7

Hi Bob!

If you could post the one using geometry that would be great.

Thanks for the help!

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 8,930

OK. I'll have a go. It's easier to work out the sine compound angle formula; then all the others come from that. I've labelled my diagram with letters A to F so I'll use alpha and beta for the angles. The geometric proof assumes that alpha + beta is less than 90 degrees. To prove the general case for any angles is somewhat harder. First my diagram:

alpha is made by the triangle ABC, with a right angle at C.

beta is made by the triangle ABF with a right angle at B.

Construct FE parallel to BC, and BD parallel to CE.

Note that DE = BC and DBA = alpha, therefore BFD = alpha as well.

If you then replace beta by -beta you get

and replace alpha with 90 - alpha

simplifies to

and replace alpha with - alpha

Now get the tan results with

divide top and bottom by cos(alpha).cos(beta)

and finally replacing beta with - beta

which is the one you needed.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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