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18 Jun 2020, 01:51
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
40% (02:05) correct
60%
(02:44)
wrong
based on 42
sessions
History
Date
Time
Result
Not Attempted Yet
In a triangle ABC shown above, AC = DB. What is the measure of a blue angle?
A. 15°
B. 20°
C. 25°
D. 30°
E. 35°
Are You Up For the Challenge: 700 Level Questions
Attachment:
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18 Jun 2020, 05:24
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Attachment:
Untitled.png [ 6.42 KiB | Viewed 3787 times ]
In triangle ACD
\(\frac{a}{y} = \frac{sinx}{sin80}\).........(1)
In triangle ABD
\(\frac{a}{y} = \frac{sin(x-20)}{sin 20}\).........(2)
\(\frac{sinx}{sin80 }= \frac{sin(x-20)}{sin 20}\)
\(\frac{sinx}{4*sin20*cos20*cos40} = \frac{sin(x-20)}{sin 20}\).........................[sin2A = 2sinA*cosA]
sinx = 4cos40*cos20 * sin(x-20)
sinx = 2(cos20 + cos60) * sin(x-20)..............................[2cosAcosB = cos(A-B)+cos(A+B)]
sinx = (2cos20 +1)*sin(x-20)
sin x = 2 cos 20*sin (x-20) + sin(x-20)
sinx = sinx + sin(x-40) + sin (x-20)
sin(x-40) = - sin (x-20)
sin(x-40) = sin (20-x)
x-40 = 20-x
x=30
18 Jun 2020, 07:43
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IMO:- D
Solving data already Present
The third angle will be 80 degrees (Sum of angle is 180 degrees), Angle CAB will be 80 degrees
Hence Triangle ACB will be isosceles with AB = CB
Lets Suppose AC= z; AB=CB =x
This question can be done by some constructions (Refer Diagram for details)
Let us Construct Equilateral Triangle AEB
So ; AE= EB = AE = x
As each angle of Equilateral Triangle is 60 degrees
Angle CBE = 40 degrees ( Angle ABC =60 degree and Angle ABD = 20 degree)
Angle CAZ = 20 degrees ( Angle EAB = 60 degrees and Angle CAB = 80 degree)
Triangle CBE is an isosceles triangle as CB = EB =x (As we have constructed equilateral triangle AEB)
So Angle BCE = Angle BEC = (180-40)/2 = 140/2 = 70 degree
Hence Angle ACE = 150 degree
Lets Compare Triangle ACE & Triangle ADB
AC=BD=z
AE=AB=x
Angle CAB=Angle ABD
By Side - Angle - Side Theorem both triangles are congruent
Hence Angle ACE=Angle ADB = 150 degree
So required Angle ADC = 180-150= 30 degree
Solving data already Present
The third angle will be 80 degrees (Sum of angle is 180 degrees), Angle CAB will be 80 degrees
Hence Triangle ACB will be isosceles with AB = CB
Lets Suppose AC= z; AB=CB =x
This question can be done by some constructions (Refer Diagram for details)
Let us Construct Equilateral Triangle AEB
So ; AE= EB = AE = x
As each angle of Equilateral Triangle is 60 degrees
Angle CBE = 40 degrees ( Angle ABC =60 degree and Angle ABD = 20 degree)
Angle CAZ = 20 degrees ( Angle EAB = 60 degrees and Angle CAB = 80 degree)
Triangle CBE is an isosceles triangle as CB = EB =x (As we have constructed equilateral triangle AEB)
So Angle BCE = Angle BEC = (180-40)/2 = 140/2 = 70 degree
Hence Angle ACE = 150 degree
Lets Compare Triangle ACE & Triangle ADB
AC=BD=z
AE=AB=x
Angle CAB=Angle ABD
By Side - Angle - Side Theorem both triangles are congruent
Hence Angle ACE=Angle ADB = 150 degree
So required Angle ADC = 180-150= 30 degree
Attachments
Solution_Triangle.jpg [ 404.74 KiB | Viewed 3709 times ]
