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August 15th, 2013, 05:48 AM   #1
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find the sides and angles of triangle

My question relating to trigonometry. A triangle having altitude magnitude is 40 and its base 100 and base opposite angle is 110 degree, so find the rest of all sides and angles. I attach picture that I draw rough sketch on paint so sorry about sketch. It will clear you the question... please answer my question as quickly as possible.
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 August 15th, 2013, 07:24 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 When the altitude bisects the base, the angle at B is maximized, and would then be 2arctan(50/40), which is only about 102.68°, so the problem has no solution.
 August 15th, 2013, 11:42 AM #3 Newbie   Joined: Aug 2013 Posts: 3 Thanks: 0 Re: find the sides and angles of triangle @ skipjack why you take 50 for which method and principle?
 August 15th, 2013, 02:26 PM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: find the sides and angles of triangle Is angle A = angle C? Your diagram seems to indicate so.
 August 15th, 2013, 02:39 PM #5 Member   Joined: Aug 2013 Posts: 59 Thanks: 0 If we call the base point D, then is ?ADB = ?CDB = 90° true?
 August 16th, 2013, 02:46 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 I used 100/2 = 50 because I was considering the case where the altitude bisects the base. Using 50 - x and 50 + x, one would get arctan((50 - x)/40) + arctan((50 + x)/40) = 110°, which has no solution. The LHS is an even function and its maximum value occurs at x = 0.
 August 16th, 2013, 03:30 AM #7 Senior Member     Joined: Feb 2010 Posts: 706 Thanks: 141 Re: find the sides and angles of triangle Even if you don't assume that the altitude hits at the midpoint, there is no solution. Suppose that the altitude divides the side of 100 into two parts ... x on the left and 100-x on the right. You then get $\tan\ 110= \dfrac{\frac{x}{40}+\frac{100-x}{40}}{1-\frac{x(100-x)}{1600}}$ This simplifies to $x^2 - 100x + (1600-\frac{4000}{\tan\ 110})=0$ which has a negative discriminant and thus has no real solutions.
 August 17th, 2013, 04:26 AM #8 Newbie   Joined: Aug 2013 Posts: 3 Thanks: 0 Re: find the sides and angles of triangle It's not a midpoint.
August 17th, 2013, 06:17 AM   #9
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Re: find the sides and angles of triangle

Quote:
 Originally Posted by adilgrt007 It's not a midpoint.
Well then, with height 40 and base 100, angle B = 110 degrees is impossible; did you guess?

Highest possible is ~103 degrees; that's with height line dividing AC in 49:51 ratio.

 August 22nd, 2013, 10:25 AM #10 Newbie   Joined: Jun 2013 Posts: 8 Thanks: 0 Re: find the sides and angles of triangle You can apply the Generalized Pythagorean Theorem, let’s say that the foot of altitude is H so we have, note BH with x and HC with 100 – x so we have a equation in x with two solutions, both are valid.

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