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

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.

First let's add a line from point C so that it is PERPENDICULAR to line AD
Also, note that, ∠EDB is 120° since it lies on a line with a 60° angle.



Since we already know 2 angles in ∆CED, we can see that the remaining angle, ∠ECD, is 30°, which means ∆CED is a 30-60-90 right triangle.
So, ∆CED is a 30-60-90 right triangle AND we know that the hypotenuse CD has length 2, we can conclude that side ED has length 1.



Now draw a line from E to B.
Since ED = EB (both have length 1), we can see that ∆EDB is an ISOSCELES TRIANGLE, which means the two remaining angles are each 30°



Next, ∠EAB is 150° since it shares a line with a 30° angle
Also, since we were originally told that ∠CBA is 45°, we can conclude that ∠ABE is 15°


Now focus on ∆CBE
Notice that this triangle is an ISOSCELES triangle, because ∠ECB = ∠EBC = 30°
This means that side EC = side EB



Now focus on ∆EAB
Since we already know two of the angles in this triangle (150° and 15°), we can conclude that = ∠EAB = 15°



Stay focused on ∆EAB
This triangle is an ISOSCELES triangle, because ∠EAB = ∠EBA = 15°
This means that side AE = side EB



Now focus on ∆ACE
Since CE = EA, this is an ISOSCELES triangle
Since one angle = 90°, the other two angles are each 45°


At this point, we can see that x = 45° + 30° = 75°
Answer: D

Cheers,
Brent

This answer is wrong. Solving this system does not yield x=75, but x=90, and y=45.
This answer should not have so many kudos, it is very misleading.

Bunnel/Experts,

I have a doubt here.I understand angles and sides ratio will be proportional.

So If I see angle opposite to side with lenght 1 is 15 degree .So why not we
directly comprehend angle opposite to CB(3 units) is 45 degrees.

from this I will get angle y as 45 degrees

The solution you are quoting is not correct. Please check the discussion on previous pages.

Posted from my mobile device

This was indeed a difficult question. After looking at Bunuel's solution, it became obvious. However, solving a question like this on exam day might be too much.

Bunuel - is there a rationale (and is it obvious?) that you chose to drop a perpendicular from C to AD?

Probably one of the hardest geometry questions I have come across. Stared at the solutions for 10 mins trying to figure it out.....

Hey, but this violates one of the cardinal rules of Triangles. The sum of any two angles in a Triangle should always be treated then the third angle. In this if <ADB is 120, then Angle <ABD + <BAD should be greater than <ADB, but it is not the case. it is less than 120 degrees. Please correct me if I'm wrong

It seems you are mixing up two distinct properties of triangles:

1. The length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides.
2. The sum of the angles of a triangle always equals 180 degrees.

1. Draw the perpendicular from point C to AD. Name this point as E. YOu have all the angles of the ∆ CED with the right angle at E.
2. From the right angle properties, length ED =1 and CE =√3
3. Join B and E now. ∆ BED will be an isosceles triangle as ED and DB are both of equal length.
4. ∠EDB = 120°, ∠BED = ∠DBE = 30°
5. Now focus on ∆CEB
Notice that this triangle is an ISOSCELES triangle because ∠ECB = ∠EBC = 30°
This means that side EC = side EB
6. Now in ∆AEB
It's again an Isosceles triangle. Side AE = Side EB
7. From 5 and 6, Side AE = Side EC. Thus ∆AEC is also an isosceles triangle.
This ∠ACE = ∠EAC = 45°

so, ANSWER ∠ACD = 30° + 45° = 75°, i.e. D.