Re: Complementary Angles?

Quote:

Originally Posted by

**aldrincabrera** theorem on geometry,.,it says "If two angles are complementary, then both are acute" so i have this question "is zero degree the complement of 90 degrees???if

In a course in *axiomatic geometry*, most authors use this definition:

*An *__angle__ is the union of two non-collinear rays with a common endpoint.

So that rules out both a degenerate angle and a so-called ‘straight angle’.

In fact, Jay Greenberg has the following theorem in his text:

$\displaystyle \left( {m\angle A} \right)^ \circ \text{ is a real number and }0<\left( {m\angle A} \right)^ \circ<180. $

So strictly speaking we would not have an angle of measure zero.

Re: Complementary Angles?

,.,sir,.,i was proving the theorem and i ended up with this,.plss tell me if im doing it correctly,.

here is my proof.,

let alpha and beta be complementary angles,

then alpha + beta = 90.

we show that alpha < 90 and beta<90.

Suppose alpha >equal to 90 or beta >equal to 90,

then alpha+beta >equal to 90 in either cases.

thus alpha<90 and beta <90,.

am i wrong sir??thnx

Re: Complementary Angles?

Quote:

Originally Posted by

**aldrincabrera** ,.,sir,.,i was proving the theorem and i ended up with this,.plss tell me if im doing it correctly,.

here is my proof.,

let alpha and beta be complementary angles,

then alpha + beta = 90.

we show that alpha < 90 and beta<90.

Suppose alpha >equal to 90 or beta >equal to 90,

then alpha+beta >equal to 90 in either cases.

thus alpha<90 and beta <90,.

If the **sum of two positive numbers** is 90 then each must be less than 90.

Re: Complementary Angles?

,.but sir,.,am i on the right track??

Re: Complementary Angles?

Let $\displaystyle \alpha$ and $\displaystyle \beta$ be complementary angles so indeed:

$\displaystyle \alpha+\beta=90$

If you want to show: $\displaystyle \alpha<90$ and $\displaystyle \beta<90$.

You have to do that for positive numbers!

Because if you take $\displaystyle \alpha=-240°$ and $\displaystyle \beta=330$

Then $\displaystyle \alpha+\beta=90$ but $\displaystyle \beta$ is not $\displaystyle <90$.

So you have to assure you work with positive numbers.