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

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

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

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

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

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)

Posted from my mobile device

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
[12]+16-8a=4
a=3

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

Ans (D)

Posted from my mobile device

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)

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.

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 .

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

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=\(\sqrt{3}\)

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.