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April 10th, 2011, 03:01 AM   #1
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geometry

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 April 10th, 2011, 07:32 AM #2 Newbie   Joined: Apr 2011 Posts: 6 Thanks: 0 Re: geometry 1) angle CAB = angle CBD therefore the two triangles are similar, triangle CAB & triangle CBD the ration of the sides must be equal 2) Lets AD = x, then AC = (x+3) 3) ( BC / AC ) = ( DC / BC) ( 5 / (x+3)) = ( 3 / 5 ) cross multiple: 3x + 9 = 25 x = (25 - 9) / 3 x= 5.333 answer is C) 5 1/3
April 10th, 2011, 07:41 AM   #3
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Re: geometry

Hello, khalidmustafa06!

Quote:
 $\text{In the diagram: }\:\angle ABC \,=\,\angle BDC \,=\,90^o,\;BC \.=\.5\text{ cm},\;DC \,=\,3\text{ cm}$ $\text{Find the length of }AD.$ [color=beige]. . [/color]$(A)\;8\frac{1}{3}\;\;\;\;(B)\;5\;\;\;\;(C)\;5\frac {1}{3} \;\;\;\; (D)\;2 \;\;\;\;(E)\;12$ Code:  B * * @|* * | * * 4| * 5 * | * * | @ * A * - - - - - - - - * - - * C D 3

$\text{We see that }\Delta BDC\text{ is a 3-4-5 right triangle.\; Hence: }\,BD \,=\,4.$

$\text{All three right triangles are similar. \;Let }\theta=\angle BCD =\angle ABD.$

$\text{In }\Delta ADB:\;\tan\,\theta \,=\,\frac{AD}{4}\;\;[1]$

$\text{In }\Delta BDC: \;\tan\,\theta \,=\,\frac{4}{3}\;\;[2]$

$\text{Equate [1] and [2]: }\:\frac{AD}{4} \:=\:\frac{4}{3} \;\;\;\Rightarrow\;\;\;AD \:=\:\frac{16}{3} \:=\:5\frac{1}{3}\;\;\text{ . . . answer (C)}$

 April 13th, 2011, 07:30 PM #4 Newbie   Joined: Apr 2011 Posts: 3 Thanks: 0 Re: geometry Measure of a tangent is opposite over adjacent angles. You will need a scientific or graphing calculator. Here's an example: I \ I \ I_ \ I_I__\ The angle to the right of the right angle is the example I will use. I will call it 'A'. TAN A=opposite/adjacent. The number of the opposite and adjacent sides should be given. This is a ratio. TAN A = ratio. Divide by TAN (hit TAN -1: you will have to use the 2nd key) on both sides. You will then have: A = TAN-1 (multiplied by) ratio. YOU MUST HAVE A CALCULATOR TO DO THIS PROBLEM. Repeat this for the other angle you need. Then add your results, giving you the final measure of the large angle.

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