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 December 12th, 2007, 09:27 AM #1 Newbie   Joined: Dec 2007 Posts: 5 Thanks: 0 Help on a trigo question I got a question in my exams which says -: "In a triangle ABC, tan A = 1, 2 cos B = 1. Find the value of angle C." I figure that if they're using trigonometric ratios on A and B, shouldn't angle C be 90° ?
 December 12th, 2007, 09:35 AM #2 Senior Member   Joined: Oct 2007 From: France Posts: 121 Thanks: 1 tan(A)=1 ===> A=? cos(B)=1/2 ===>B=? A+B+C=? ===> C=?
December 12th, 2007, 09:45 AM   #3
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 Originally Posted by Richard André-Jeannin tan(A)=1 ===> A=? cos(B)=1/2 ===>B=? A+B+C=? ===> C=?
Yeah, I tried that way. Angle C comes 75° then. But how can that be, considering one of the angles has to be 90°? You can't use trigonometric ratios on a usual triangle..

December 12th, 2007, 01:46 PM   #4
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 You can't use trigonometric ratios on a usual triangle..
Why not?????????????????? The trig functions are defined using a right triangle, but once they are defined they apply to any angle. The angle could be stand alone or in any polygon, nor even a triangle.

December 12th, 2007, 06:41 PM   #5
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Originally Posted by mathman
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 You can't use trigonometric ratios on a usual triangle..
Why not?????????????????? The trig functions are defined using a right triangle, but once they are defined they apply to any angle. The angle could be stand alone or in any polygon, nor even a triangle.
Well, we don't have "defining" in our course yet. And in the book it says trigonometric ratios are applied on right angle triangles.

December 12th, 2007, 09:05 PM   #6
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 Originally Posted by SVXX Well, we don't have "defining" in our course yet. And in the book it says trigonometric ratios are applied on right angle triangles.
You can draw CD perpendicular to AB, with D lying on AB, so that ADC and BDC are right-angled triangles.

December 13th, 2007, 01:50 AM   #7
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Originally Posted by skipjack
Quote:
 Originally Posted by SVXX Well, we don't have "defining" in our course yet. And in the book it says trigonometric ratios are applied on right angle triangles.
You can draw CD perpendicular to AB, with D lying on AB, so that ADC and BDC are right-angled triangles.
Hmm I seem to get your point now..thanks! That amounts to 3 errors in my examination

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