Finding complementary angles  TOPIC_SOLVED

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

Finding complementary angles

Postby futurebrewer on Sat Oct 01, 2011 8:28 pm


Find the angles that are complimentary and supplementary to A= 34º 19' 42"

So, if angle B is complementary to A, then B = 90º - A

ok I get that part.

here is where my book loses me:

so they re-write it like this: B = 90º - 34º 19' 42"

ok, so far so good.

then they rewrite it again: 89º 59' 60" - 34º 19' 42"


they go onto say 90º = 89º 59' 60"

I dont get it. how did they get the 89º 59' 60" ?

Any help would be greatly appreciated. Thanks for your time.
Posts: 2
Joined: Sat Oct 01, 2011 8:15 pm



Re: Finding complementary angles  TOPIC_SOLVED

Postby nona.m.nona on Sat Oct 01, 2011 10:28 pm

futurebrewer wrote:...A= 34º 19' 42"... book loses me...they re-write it like this: B = 90º - 34º 19' 42"...

then they rewrite it again: 89º 59' 60" - 34º 19' 42"

This is "borrowing", similar to what you learned back in grammar school when you were doing vertical subtraction. If you were subtracting 34 from 90, you would "borrow" from the "tens column" to convert the subtraction from 90 - 34 to 810 - 34. Then you'd subtract the 3 from the 8 and the 4 from the 10.

In this case, you are subtracting 34º 19' 42" from 90º 0' 0". There are sixty minutes in each degree, and sixty seconds in each minute. When one "borrows" in this context, one is borrowing sixties, rather than tens. But the process and reasoning are exactly the same.
Posts: 260
Joined: Sun Dec 14, 2008 11:07 pm

Return to Trigonometry