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
yes

- 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

So again

** <ABC = 10x+20 + 3x
**

= 13x + 20 = 139.294°

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