When I read this question, the first thing that struck me as odd was that they mentioned "CD is a positive integer" (which presumably means that the length of CD is a positive integer)
Length would obviously be positive but needn't be an integer. If it is specified and seeing that OCD is a right triangle (tangent makes 90 degree angle with radius), this might have something to do with pythagorean triplets.
The ones with 15 or factors of 15 (since OC = 15) are
3, 4, 5 (the sides can be 15, 20, 25)
5, 12, 13 (the sides can be 15, 36, 39)
8, 15, 17 (the sides can be 8, 15, 17)
(OC is one leg and CD is the other)

The only length given in the options is 20 (we have neither 36 nor 8)
Answer (C)
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Since AC || OD, < CAO = < COD.
—> if you draw a segment CB, then triangle ACB will be right-angled triangle.
—> Also, triangle ACB and triangle OCD are similar triangles.
—> \(\frac{AB}{AC} = \frac{OD}{OC}\)

\(\frac{30}{AC} = \frac{OD}{15}\)

AC*OD = 450
AC+ OD = 43
—> AC= 18, OD = 25
(Using Pythagorean theorem)
\(CD^{2} + OC^{2} = OD^{2}\)

—> \(CD^{2} = 625 —225 = 400\)
CD = 20

The answer is C

Posted from my mobile device

- Radius OA=OC=15=3*5
- CD is a positive integer and OCD is a right triangle at C, so:
OC^2+CD^2=OD^2
(3.5)^2+CD^2=OD^2

A quick and sensible value of positive integer CD to satisfy the above equation is CD=20=4*5, and thus, we easily satisfy
OC^2+CD^2=OD^2
(3*5)^2+(4*5)^2=(5*5)^2

Check AC=43-OD then AC=43-25=18.
AC^2+CB^2=AB^2
(3*6)^2+(4*6)^2=(5*6)^2 (OK!!)
We can roughly infer that angle B = angle D.

FINAL ANSWER IS (C)

Posted from my mobile device

From the given figure, radius, AO = OC = AC = OB = 15
AC+OD=43
OD = 43 – 15 = 28

So AOC is a equilateral triangle. ∠ACO = ∠AOC = ∠OAC = 60
As AC is parallel to OD , ∠CAO = ∠DOB = 60.
Therefore, ∠COD = 60.

As DC is tangent, ∠DOC = 90

Now, 90:60:30 = 2x : √2x :1x
2x = 28
or, x = 14
Opposite side of 60 = √2x = 1.4 *14 = 20(Appoximate)

Answer: 20(C)

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It is assumed here that AC is a radius too and hence of length 15. That is not true.
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we can find DE ; AC+OD=43
∆ACB ~ ∆ OCB
AB/AC= OD/OC
30/AC= OD/15
AC+OD=43
AC= 13, OD= 25
CD^2+OC^2=OD^2
CD^2 = 625-225 ; 400
CD= 20
IMO C


In the figure given below, O is the center and AB is the diameter of the circle of radius 15. From point D, two tangents DC and DB are drawn to the circle. If AC is parallel to OD and AC+OD=43, find the length of CD given CD is positive integer

A. 16
B. 18
C. 20
D. 22
E. 24

whats wrong in this, can someone explain this