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by sampath » Sat Aug 20, 2011 9:36 am
In the figure, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that �ADC = 60º and �ABD = 45º, what is the measure of angle x in degrees? (Note:
Figure is not drawn to scale.)

(A) 55 (B) 60 (C) 70 (D) 75 (E) 90

My question is that isn't the length of the height drawn from C to the line between A and D have value sqrt(3) (as it is a 30-60-90 triangle). If yes, then shouldn't the final value of [spoiler]x = 60 [/spoiler]instead of 75?
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by arashyazdiha » Sat Aug 20, 2011 11:42 am
Image
I uploaded the image maybe now it is better...
Let's say the way you went ahead,
draw a line from C,perpendicular to the line AD,and name the point that cut the AD line, point H. so <HCB will be 30.
Then draw from point H to point B.
As you know in the CHD triangle, HD will be 1/2 of the CD side. so HD is 1. and as BD is 1 so we can conclude that <DHB=30 because Triangle DHB is isosceles and <ADB=120. (<ADB = 180 -ADC)
Also <DAB=15 because <DAB = 180 - (<ADB+<ABC). So if <HBC is 30 <ABH =15, thus AH = HB = CH = sqrt(3) as you properly calculated this one. So it reveals that <CAD=45 and also <ACH=45 and as previously calculated <HCB=30 so X will be 45+30=75
I hope it helped
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