The figure above shows a circle whose diameter AB lies on the x-axis

Question

Competition Mode Question https://gmatclub.com/forum/download/file.php?id=55087 The figure above shows a circle whose diameter AB lies on the x-axis as shown. Triangle ACB is a right-angled triangle whose side AC makes an angle of 30? with side AB. If the coordinates of points A and B are (0,0) and (4,0) respectively, what is the y-coordinate of point C? A. \frac{1}{\sqrt{3}} B. \frac{\sqrt{3}}{2} C. \frac{2}{\sqrt{3}} D. \sqrt{3} E. 2\sqrt{3} 1.png

nick1816 · Answer

ABC is a (30-60-90) triangle; hence  BC : AC : AB  =  1: √3 : 2  =  2 : 2√3 : 4

Suppose y- coordinate of C is y.

Area of triangle ABC =  \frac{1}{2} * BC * AC  =  \frac{1}{2} * y * AB

y = \frac{BC * AC}{AB}

y= \frac{2 * 2√3 }{ 4} = √3

yashikaaggarwal · Answer

ABC is a 30-60-90 right angle triangle.
That means AB:BC:AC are in ratio of 2:1:√3
Therefore the sides lengths will be 4:2:2√3

Distance between A and C is 2√3
2√3 = √(y2-y1)^2+(x2-x1)^2
Let Coordinate of C be (x,y) and A(0,0)
2√3 = √(y-0)^2+(x-0)^2
12 = y^2+x^2 ----------(1)

Distance between B and C is 2
2 = √(y2-y1)^2+(x2-x1)^2
Let Coordinate of C be (x,y) and B(4,0)
4 = (y-0)^2+(x-4)^2
4 = y^2 + X^2 + 16 -8x
Putting Value from equation 1
4 = 12 + 16 -8x
8x = 24 => X=3 ---------(2)
Y value => 12 = y^2+x^2
Putting Value of X
12= y^2+9
3=y^2
Y=√3

IMO D

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NitishJain · Answer

IMO  D

Triangle ACB is 30-60-90 triangle
30-60-90
1: √3 : 2

Side opposite to 90* is AB= 4

Hence: BC =2
& AC = 2√3

(Length of AC)^2  =  (x-0)^2  +  (y-0)^2  
==>  x^2  +  y^2  =12 -----------eq 1

Also:  (Length of BC)^2  =  (x-4)^2  +  (y-0)^2  
 (x-4)^2  +  y^2  =4

Substituting  y^2  =12 -  x^2  from eq1

x^2  + 16 -8x + (12 -  x^2 ) = 4

24 = 8x
or x =3

From eq1  3^2  +  y^2  =12 
 y^2  = 3
y = √3 or -√3

But we know y is in +ve x quadrant and hence -√3 can be rejected

y = √3

MayankSingh · Answer

IMO D

in triangle ABC
Sin 30 = BC/AB
=> BC = AB Sin 30 = 4 X 1/2 = 2

Cos 30 = AC/AB
=> AC = √3/2 x 4 = 2√3

Let Coordinate of C (a,b)

AC^2 = a^2 + b^2 = (2√3)^2 = 12 ...I
BC^2 = (a-4)^2 + b^2 = 4...............II

From I-II
a^2 - (a-4)^2 = 8
=> a= 3
So from I , b= √ (12-9) = √3

GMATinsight · Answer

ABC is a right triangle with angles 30º-60º-90º
i.e. ratio of Sides must be x:√3x:2x  
AB = 2x = 4
i.e. AC = √3x = 2√3
and BC = x = 2

Join Point C with Center of circle O
∆OCB is an equilateral Triangle with side = BC = 2

i.e. Y-coordinate of Point C = Height of Equilateral OBC  = (\frac{√3}{2})*2 = √3  
and, X-coordinate of Point C = Midpoint of O and C  = \frac{(2+4)}{2} = 3

i.e. Point C = (3, √3)

Answer: Option D

Superman249 · Answer

Ans is D

Two ways of doing it:-

AC = 2*sqrt(3) and BC = 2

By using Distance Formula x^2 +y^2 = Distance between point^2

Let's assume coordinate of C be x and y

We will have two-equation 
x^2 + y ^2 =12
and (x-4)^2 + y^2 = 4

Solving we get y= sqrt(3)

2nd Way  :- Better and easier
equation of line AC > y=x/sqrt(3) , as slope is 1/sqrt(3)
equation of BC > y=-sqrt(3)x + 4 sqrt (3)

Solving both equations to get the meeting point 
we get y= sqrt(3)

Eq

firas92 · Answer

Note that the y coordinate of C is simply the length of perpendicular from C to AB (which is the height of the triangle with base AB)

Since  \angleCAB  is  30 , the side lengths of ACB are in the ratio  BC:AC:AB=1:\sqrt3:2

So since length of  AB  is  4 ,  BC=2 ,  AC=2\sqrt3

Area of triangle ABC with base BC and height AC = Area of triangle ABC with base AB and height h (h=y coordinate of C)

\frac{1}{2}*AC*BC=\frac{1}{2}*AB*h 
 \frac{1}{2}*2\sqrt3*2=\frac{1}{2}*4*h

h=\sqrt3

Answer is  (D)

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exc4libur · Answer

The figure above shows a circle whose diameter AB lies on the x-axis as shown. Triangle ACB is a right-angled triangle whose side AC makes an angle of 30 with side AB. If the coordinates of points A and B are (0,0) and (4,0) respectively, what is the y-coordinate of point C?

Sides 30-60-90 triangle
x, xV3, 2x = x, xV3, 4 = 2, 2V3, 4

Distance between points
d^2=(x1-x2)^2+(y1-y2)^2
(2V3)^2=(a-0)^2+(b-0)^2
a^2+b^2=12

(2)^2=(a-4)^2+(b-0)^2
a^2+b^2+16-8a=4
 +16-8a=4
a=3

(3)^2+b^2=12
b^2=3
b=V3

Ans (D)

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vipulshahi · Answer

Given, Angle BAC = 30; AB = 4 ; Angle ACB = 90 ; Co-ordinate A=(0,0) B= (4,0)
Required = Co-ordinate C= (x,y) ; y= ??
Sol: AB = 4 ; In Tri. ABC, 2*a = 4 ; a = 2
AC = Sqt(3) * a = 2*Sqt(3) ; BC = 2
Lets calculate height , h = perpendicular drawn from point "C" to Line "AB" 
h=sqt(3)
Thus, y co-ordinate of point "C" is Sqrt(3)

IMO(D)

unraveled · Answer

Image
The figure above shows a circle whose diameter AB lies on the x-axis as shown. Triangle ACB is a right-angled triangle whose side AC makes an angle of 30? with side AB. If the coordinates of points A and B are (0,0) and (4,0) respectively, what is the y-coordinate of point C?

A. 1/√3

B. √3/2

C. 2/√3

D. √3

E. 2√3

The maximum y-coordinate of point C is 2 since radius of the circle is 2 i.e. the highest point from centre of the circle to the arc ACB.
So, E is out.

Triangle ACB is right angled at C. Thus, Angle ABC is 60. 
From 30-60-90 triangle, sides of the triangle are in ratio as BC:AC:AB
1:√3:2 OR 2:2√3:4 in this case

Area of triangle ACB = 1/2* AC * BC = 1/2 * AB * CD (where CD is perpendicular from C to AB)
 \frac{1}{2} * 2√3 * 2 = \frac{1}{2} * 4 * CD 
CD = √3

Answer D.

HoneyLemon · Answer

D ,  imo 
AB = 4 . 
Since ABC is an 30-60-90 triangle 
 BC = 2 and AC = 2 Root (3) .

Now in triangle ABC if we draw a perpendicular CD on AB . 
0.5 * AB * CD= 0.5 * BC * AC 
4 * CD = 4 root (3)
 CD = root (3)..

So Y coordinate of C is √3 . So D is my ans .

monikakumar · Answer

30 60 90 triangle
so, 1  \sqrt{3}  2
so, 2  2\sqrt{3}  4
dist between A and C=2 = \sqrt{3}  =  \sqrt{x^2+y^2} 
dist between B and C =2 = \sqrt{(x-4)^2+y^2} 
using distance formula, and solving
we get (3, \sqrt{3} )

Ans D

Kritisood · Answer

drop a perpendicular from C to point D on AB. we need to find CD to get the coordinate for y. 
CD is the height of the triangle.
This is 30 60 90 triangle AB=4 hence CB = 2 and CA =  2 \sqrt{3}

\frac{1}{2} *AC*AD= \frac{1}{2} *CA*CB

4*AD= 2*2*\sqrt{3}

AD= \sqrt{3}

bumpbot · Answer

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The figure above shows a circle whose diameter AB lies on the x-axis

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