In the figure, point D divides side BC of triangle ABC into segments

Complete solution for all the angles is in the image below:



x=45+30=75.

Notes:
Sides with one blue segment crossing them are equal and sides with two blues segments crossing them are equal too.

CO is perpendicular to AD --> OD=1 (from 30-60-90 right triangle property as sides are in ratio \(1:\sqrt{3}:2\)) --> as OD=BD=1 then ODB is an isosceles triangle.

<CDO and <BDO are supplementary to each other (supplementary angles are two angles that add up to 180°), so <BDO=120 --> <DAB=180-(120+45)=15.

As ODB is an isosceles triangle --> <DOB=<DBO=30. <OBA=45-30=15 --> AOB is an isosceles triangle, so OA=OB. Also COB is an isosceles triangle, so CO=OB --> OA=CO=OB. So, AOC is an isosceles triangle --> <CAO=<OCA=45 (as <COA=90) --> x=45+30=75.

Answer: D.

P.S. You can attach image files directly to the post.


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step by step using this diagram can help to understand:

Bunuel - Can we not use exterior angle property

for triangle ACD
Angle at C = x
<CDA = 60
Since sum of all the angle of triangle is 180 then we

<CAD + <CDA + x = 180
<CAD = 180 -60 -x
<CAD = 120 -x

For the other triangle we know
<DAC = 15
<ADB = 120
<DBA =45

Exterior angle is sum of opp interior angle and not adjacent to it
So <CAD = <ADB + <DBA
120 -x = 120 + 45
x = 45.

Since this is not the ans could you please let me know what I'm doing wrong here?

Many thanks.

Cheers

<CAD is not exterior angle of <DAB. Exterior angle for <DAB is given below in green:Hope it's clear.
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say angle CAB=y
since sum of angles in a triangle is 180
x+y+45=180
x+y=135 equation 1

line AD is dividing BC in 2:1 ratio
hence
X+2/3Y+60=180
X+2/3y=120 equation 2
solving equation 1 &2 we get x=75
answer is D

I lost you from here. How do you know that angle cab is 45+15?

<CDO and <BDO are supplementary to each other (supplementary angles are two angles that add up to 180°), so <BDO=120 --> <DAB=180-(120+45)=15. As ODB is an isosceles triangle --> <DOB=<DBO=30. <OBA=45-30=15 --> AOB is an isosceles triangle, so OA=OB also COB is an isosceles triangle, so CO=OB --> OA=CO=OB. So, AOC is an isosceles triangle --> <CAO=<OCA=45 (as <COA=90) --> x=45+30=75.

<CAB = <DAB + <CAO = 15 + 45.
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hi Bunnel,

How can we be sure that AO is equals to CO?

ODB is an isosceles triangle --> <DOB=<DBO=30. <OBA=45-30=15 --> AOB is an isosceles triangle, so OA=OB. Also COB is an isosceles triangle, so CO=OB --> OA=CO=OB. So, AOC is an isosceles triangle --> <CAO=<OCA=45 (as <COA=90) --> x=45+30=75.
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@Bunnel

Can you please explain how this is derived?

"As ODB is an isosceles triangle"

May be the attached image can help explain how.

Nice explanation. What confuses me more is, how do we know that AOB is isosceles?
Thank you

Hi reto,

In triangle ADB, ∠ADB = 120 ( as∠ADB = 180 - ∠CDO = 180 - 60 =120). We now know two angles of the triangle ADB. So the third angle ∠DAB = 180 -120 -45 = 15.

Similarly in triangle ODB since ∠DBO =30 (as ODB is an isosceles triangle), we have ∠OBA = 45 - 30 = 15.

So we see that ∠OBA = ∠DAB = 15. These are two angles of triangle AOB. Hence AOB is an isosceles triangle with sides OA = OB

Hope it's clear

Regards
Harsh
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This looks much more simple!

Just so I am clear because line AD bisects line CB in a two to 1 ratio we can calculate the breakdown of angles at vertex A?

Thanks for your help

This is a very dangerous inference. The only formula that links sides to angles in by sine law from trigonometry that says:

In a triangle ABC with sides a,b,c opposite \(\angle (A), \angle (B) and \angle (C)\) respectively, then

a/sin(A) = b/sin(B) = c/sin(C). You do not have to remember this 'sine law' for GMAT.

It was a lucky coincidence that the user was able to arrive at the correct answer by assuming that the angles themselves got divided into 1:2 ratio if the sides were divided in the ratio 1:2.

The best way would be to follow what Bunuel did in his approach and for obtaining \(\angle (BAD) and \angle (CAD)\), after you find OD = DB = 1 ---> \(\angle {DBO} = \angle {DOB}\) = 30 degrees.

Also, as we know \(\angle{ABD}\) = 45 degrees ---->\(\angle {ABO}\) = 15 degrees. Additionally, \(\angle{ADB} = \angle {ADB} + \angle {DBO}\) ... (by external angle theorem of triangles).

Once you do this, you will see \(\angle{BAO}\) = 15 degrees. Then , you proceed to finding other angles as stated by Bunuel.
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We are solving it with the misconception that if we make a perpendicular from the point C to the line AD will perfectly connect with the point O(M). In this case fortunately matched, but if you try to use the same logic just modifying the value of line CD to 3, this method won't work. We need to find a real answer.

-UPDATE-

before start working with the original triangle, we need to understand the equilateral triangle properties. I'll explain later.

here are my way to solve it:

A GMAT way to answer the problem, the best way I can explain it. (click on the image and then in the "magnifying glass" at the top left to enlarge to full size)
Don't forget +1KUDOS !!!

This DOES NOT give us the correct solution. There is a calculation mistake here. Using the calculation mentioned, we get x=90 which is incorrect. This basically implies that the statement
"line AD is dividing BC in 2:1 ratio - > therefore corresponding angles also get divided in the same proportion" is invalid. The only way to go forward is to follow the approach that Bunuel mentioned.

Hope this helps clear the confusion around this.

Thanks.
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