In the figure to the right, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that Angle ADC = 60 and Angle ABD = 45, what is the measure of angle x? (Note: Figure is not drawn to scale.) (A) 55 (B) 60 (C) 70 (D) 75 (E) 90

I think it should be E, Angle ADC is the exterior angle of Traingle ADC, Angle BAD is 15, as the segment AD divides the line CB in the ratio of 1:2, Angle BAD, must be half of Angle CAD, by this relation, Angle DAC is 30 , we have two angles of a triangle so angle C must be 90,

goyalsau wrote:I think it should be E, Angle ADC is the exterior angle of Traingle ADC, Angle BAD is 15, as the segment AD divides the line CB in the ratio of 1:2, Angle BAD, must be half of Angle CAD, by this relation, Angle DAC is 30 , we have two angles of a triangle so angle C must be 90, E is incorrect.

beat_gmat_09 wrote:goyalsau wrote:I think it should be E, Angle ADC is the exterior angle of Traingle ADC, Angle BAD is 15, as the segment AD divides the line CB in the ratio of 1:2, Angle BAD, must be half of Angle CAD, by this relation, Angle DAC is 30 , we have two angles of a triangle so angle C must be 90, E is incorrect. What is the OA, Do you have the explanation... IF you have please post it then,

goyalsau wrote:beat_gmat_09 wrote:goyalsau wrote:I think it should be E, Angle ADC is the exterior angle of Traingle ADC, Angle BAD is 15, as the segment AD divides the line CB in the ratio of 1:2, Angle BAD, must be half of Angle CAD, by this relation, Angle DAC is 30 , we have two angles of a triangle so angle C must be 90, E is incorrect. What is the OA, Do you have the explanation... IF you have please post it then, OA is D. I dont have explanation.

beat_gmat_09 wrote:In the figure to the right, point D divides side BC of triangle ABC into segments BD and DC of lengths 1 and 2 units respectively. Given that Angle ADC = 60 and Angle ABD = 45, what is the measure of angle x? (Note: Figure is not drawn to scale.) (A) 55 (B) 60 (C) 70 (D) 75 (E) 90 This is a very hard problem to solve without using trigonometry. The trick is to draw a 30:60:90 triangle and notice all the isosceles triangles that are created. For one possible solution, please see the attachment.