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maheshsrini
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Triangle Q 16 Apr 2011, 23:27
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In the attached figure, ABC is a right angled triangle and AD is the external bisector of angle A. Find AD, if AC=17 and BC=8



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Triangle Q 19 Apr 2011, 00:16
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Would external bisector of angle A here mean that angle DAB and BAC are equal?
If so, AD=AC
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Marcab
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Triangle Q 19 Apr 2011, 01:14
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sin A=8/17,
cos A=15/17
since its mentioned that ad is the external bisector of angle A, then angle DAB would be {180-A}/2 or 90-A/2.

also cos{90-A/2}=AB/AD=sin A/2
therefore AD=AB/(sin A/2)

now we know that cos A= 1-2sin^2 (A/2)
therefore sin^2 A/2=(1-cos A)/2
(1-15/17)/2
sin A/2= 1/(17^0.5)
therefore AD=15*(17^0.5)

ans :wink:

btw gud question but not of GMAT type
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Triangle Q 19 Apr 2011, 01:35
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siddharthasingh
sin A=8/17,
cos A=15/17
since its mentioned that ad is the external bisector of angle A, then angle DAB would be {180-A}/2 or 90-A/2.

also cos{90-A/2}=AB/AD=sin A/2
therefore AD=AB/(sin A/2)

now we know that cos A= 1-2sin^2 (A/2)
therefore sin^2 A/2=(1-cos A)/2
(1-15/17)/2
sin A/2= 1/(17^0.5)
therefore AD=15*(17^0.5)

ans :wink:

btw gud question but not of GMAT type
Show more

How did you get this?
now we know that cos A= 1-2sin^2 (A/2)
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Marcab
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Triangle Q 19 Apr 2011, 06:56
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dude..
if have ever studied trignometry, then u must have played around with the formula

cos A=1-2sin^2(A/2)
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Marcab
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Triangle Q 24 Apr 2011, 18:28
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:wink: :wink: :wink: :wink: :wink: :wink: :wink: :wink: :wink: :wink: :wink: :wink:



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