# Circle Theorems and Tangents help? Picture included.?

Hi! I need some help with a question on Circle Theorems and tangents. I REEEALLY don't understand it.

Here's a picture of the question: http://twitpic.com/3n0w2d

I've been asked to find Angles “ADC”, “OCB” and “TCD” and I'm not sure how to do it.

~Kit~

Update:

SHOOT!!!! Sorry...wrong angles...we've been given them!!!

Ummm, we have to find "RCB" "ABC" and "OAD"

Relevance

Seriously? You have been asked to find the three angles that are already labeled on the sketch?

Followup:

That's better now.

ABC = 180° - 131° = 49° (opposite angles, cyclic quadrilateral)

AOC = 2·49° = 98° (central angle is twice the inscribed angle)

OCA = (180° - 98°)/2 = 41° (isosceles triangle)

ACT = ABC = 49° (inscribed in same segment as ABC)

TCB = 49° + 41° + 38° = 128° (sum of its parts)

RCB = 180° - TCB = 52° (linear pair of angles)

DCA = TCA - TCD = 49° - 21° = 28° (sum of its parts)

DCO = 28° + 41° = 69° (sum of its parts)

OAD = 360° - 131° - 69° - 98° = 62° (angle sum of quadrilateral DCOA)

• Anonymous
4 years ago

The graph of a circle you place up looks to me to be precisely that of Thales's Theorem,¹ that's additionally commonplace because of the fact the prolonged evidence of the Pythagorean Theorem. there are various "circle theorems," yet this one is complication-loose to beginning geometry. right here is what i stumbled on approximately this actual one-- Quote: evidence provided AC is a diameter, attitude at B is persevering with suggestions-blowing parent for the evidence. We use right here information: * the sum of the angles in a triangle is comparable to 2 suggestions-blowing angles (a hundred and eighty°), * the backside angles of an isosceles triangle are equivalent. enable O be the middle of the circle. considering that OA = OB = OC, OBA and OBC are isosceles triangles, and via the equality of the backside angles of an isosceles triangle, OBC = OCB and BAO = ABO. enable ? = BAO and ? = OBC. the three inner angles of the ABC triangle are ?, ? + ? and ?. because of the fact the sum of the angles of a triangle is comparable to 2 suggestions-blowing angles, we've alpha+ left (alpha + beta suggestions-blowing) + beta = a hundred and eighty^circ then {2}alpha + {2}beta =a hundred and eighty^circ or in simple terms alpha + beta =ninety^circ end quote. I desire you reliable success in this actual evidence. My potential isn't geometry, although, extending algebraic concepts to the evidence works out rather nicely for me. i'm hoping it does the comparable for you. And that's *not* a dumb question, nor does it make you look that way. The *dumb* way is to repeat without understand-how what that's you are trying interior the learn itself. Your approach is the potential via which one fairly learns.