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Need Hellp

shook  Nov 29, 2018
edited by Guest  Nov 29, 2018
edited by shook  Nov 29, 2018
 #1
avatar+92814 
+1

DCO....since CO = DO.....the angles opposite those sides are equal

So DCO = x

 

CBO....angle OBQ = 90°  [tangent meeting a radius forms a right angle ]

So CBO = 90 - 2x

 

OCB   .......since OB = OC, the angles opposite those sides are equal

So OCB = CBO =  90 - 2x

 

BCD = DCO + OCB =  (x) + ( 90 - 2x) =  ( 90 - x)   [angle addition postulate ]

 

DOB is a central angle that is twice the measure of BCD = 2(90 - x) = 180 - 2x  = measure of minor arc   DB   [ a central angle measures twice  an inscribed angle intercepting the same arc ]

 

Note that arc DCB = 360° - minor arc DB  =  360 - (180 - 2x) = 180 + 2x

[ the sum of arcs in acircle = 360° ]

 

 

Angle  y  =  (1/2) [ difference of  arc DCB - arc DOB] 

 

[ The angle formed by the intersection of 2 tangents, 2 secants or 1 tangent and 1 secant outside the circle equals half the difference of the intercepted arcs ]

Read more: https://www.mathwarehouse.com/geometry/circle/tangents-secants-arcs-angles.php#ixzz5YI50pVkl

 

(1/2) [ (180 + 2x) - (180 - 2x) ]  =

 

(1/2) [  2x + 2x ]  =

 

(1/2) [ 4x]  =   

 

2x

 

 

 

 

cool cool cool

CPhill  Nov 29, 2018
 #2
avatar+55 
0

I only got 2/6 i need proof that links to circle theorums

shook  Dec 5, 2018

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