Posted by suzy on Tuesday, November 30, 2010 at 6:12pm.
obtuse triangle ABC with obtuse angle B. Point D on side AC such that angle BDC is also obtuse. Angle DAB is 2/3 of angle ABD. Angle BCD is 1/5 of angle CBD is 4 more than angle BAD. Find the measure of angle ABC.
- geometry - Reiny, Tuesday, November 30, 2010 at 6:47pm
The wording of " Angle BCD is 1/5 of angle CBD is 4 more than angle BAD." is rather ambiguous.
Did you mean
" Angle BCD is 1/5 of angle CBD, AND IT is 4 more than angle BAD." ?
- geometry - suzy, Tuesday, November 30, 2010 at 7:16pm
- geometry - Reiny, Tuesday, November 30, 2010 at 8:00pm
Ok, let me use < for angle
Let <ABD = 3x (I can avoid fractions this way)
then <DAB = 2x
by exterior angle theorem <BDC = 5x
<BCD = 2x + 4 (it said so)
<CBD is 5 times <BCD (it said so)
so <CBD = 5(2x+4) = 10x + 20
so by exterior angle theorem :
< BDC = 12x + 24
But CDA is a straight line, so
5x + 12x+24 = 180
x = 216/7 or 9.1765
then <ABC = 10x+20 + 3x
= 13x + 20 = 139.294°
let's check the rest
<ABD = 27.529
<BAD = 18.353
and 18.535/27.529 = .6666678 or 2/3
< BDA = 134.118
for a total sum of 134.118+27.529+18.353 = 180
In triangle BDC
<C = 4 more than <BAD = 22.353
which is 1/5 of <CBD making
<CBD = 111.765
Also we knew that <BDC = 5x = 45.882
check: is CDA a straight line?
is 17x+24 = 180 ?
17x + 24 = 17(9.1765) + 24 = 180.0005 , OK
what about the sum of angles in that triangle?
what is 111.765+45.882+22.353 = 180
<ABC = 10x+20 + 3x
= 13x + 20 = 139.294°
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