My Math Forum Shortest distance triangle

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 May 10th, 2012, 09:07 AM #1 Newbie   Joined: May 2012 Posts: 17 Thanks: 0 Shortest distance triangle Hi I have another question that I need to get confirmed that I've done it correct. "In triangle ABC, BC = 6.8 cm, angle B = 54° and angle C = 43°. Find the shortest distance from point A (angel A) to side BC." This is how I solved it. $\frac{AB}{sin(43)}=\frac{6.8}{sin(83)}\hspace{6}\R ightarrow$ $AB=\frac{6.8\cdot sin(43)}{sin(83)}\Rightarrow$ $AB\approx4.67$ The shortest distance from point A to side BC, is the height of the triangle, isn't it? $sin(54)\cdot 4.67\approx3.8 cm$ The key says the answer is 3.8 cm. I was just wondering if I'm thinking correct, that the height from point A to side BC is the shortest distance.
 May 10th, 2012, 09:35 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 461 Math Focus: Calculus/ODEs Re: Shortest distance triangle Yes, the shortest distance from a point to a line is the perpendicular distance.
 May 10th, 2012, 09:38 AM #3 Member   Joined: Mar 2012 Posts: 60 Thanks: 0 Re: Shortest distance triangle The height is indeed the shortest distance between the point A and the opposit side.
 May 10th, 2012, 09:44 AM #4 Newbie   Joined: May 2012 Posts: 17 Thanks: 0 Re: Shortest distance triangle Great! Thanks!

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