inscribed triangles, very tough DS

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inscribed triangles, very tough DS

by tj123 » Sat Aug 08, 2009 4:31 pm
attached.

can someone explain why both angle ADB and angle ACB are equal. i dont understand just based on the understanding that they span the same arc?

OA A
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by cramya » Sat Aug 08, 2009 5:00 pm
The rule is angles (inscribed) in the same segment of a circle are equal.
The chord AB divides the circle in to a major segment and a minor segment.The angles above are major segment angles.Makes no difference if these were present in the minor segment

Hope this helps!

Regards,
CR

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by ghacker » Mon Aug 10, 2009 10:59 pm
Very simple question

But the thing with circles is that there are certain thing that you have to know about them ..................Then the questions become extremely easy


One thing to remember is that angles (inscribed )in the same segment are equal

So DAC ) and DBC ) are equal so too are ADC) and ACB )

The other Thing to remember is that the angle in a half circle is 90 , we are told that DAB triangle is a right triangle so DB will be the diameter

Hence we know all the angles and from A ;we know the value of a side of a right triangle( with all the angles known )

Hence A is sufficient

B is insufficient because there is no length given for the calculations

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by woo » Sun Aug 16, 2009 5:12 am
We know that angle ADB and angle ACB are equal

because the two triangle share a base.

We also know that angle ACB is 60 thus, ADB is also 60.

Triangle ABD is therefore a special triangle with angles 30, 60, 90.

From condition 1 we are told that AD is 4.

We know for the special triangle the ratio 3:4:5 holds.

Therefore we know length of a side of the equilateral.

Therefore we can find the radius of the circle thus, the area.