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https://vle.mathswatch.co.uk/images/questions/question15392.png

 

 

winkwinkwink

lynx7  Feb 13, 2018
edited by lynx7  Feb 13, 2018

Best Answer 

 #1
avatar+7339 
+4

Let's find  m∠DCO .

Notice that  DO  and  CO  are radii of circle O, so they have the same length. So triangle DCO is isosceles, and the base angles have the same measure.

m∠DCO  =  x

 

 

Let's find  m∠CBO .

Since  AQ  is tangent to circle O at point  B ,  OB  is perpendicular to  AQ  .

m∠QBO  =  90°

2x  +  m∠CBO   =   90°

m∠CBO   =   90° - 2x

 

 

Let's find  m∠OCB .

Notice that  BO  and  CO  are radii of circle O, so they have the same length. So triangle OCB is isosceles, and the base angles have the same measure.

m∠OCB   =   m∠CBO

m∠OCB   =   90° - 2x

 

 

Let's find  m∠BCD .

m∠BCD   =   m∠DCO  +  m∠OCB

m∠BCD   =   ( x )  +  ( 90° - 2x )

m∠BCD   =   x + 90° - 2x

m∠BCD   =   90° - x

 

 

Let's find  m∠DOB .

∠BCD  is an inscribed angle, so its measure is half of that of its intercepted arc, ∠DOB .

m∠BCD   =   (1/2) * m∠DOB

2 * m∠BCD   =   m∠DOB

2 * (90° - x)   =   m∠DOB

m∠DOB   =   180° - 2x

 

 

Let's find  m∠DAB .

The sum of the interior angles of quadrilateral DOBA  =  360°

And   m∠ABO  =  90°   and   m∠ADO  =  90°  ,  so...

  m∠DOB   +  90°  +  90°  +  m∠DAB   =   360°

(180° - 2x)  +  90°  +  90°  +  m∠DAB  =  360°

m∠DAB  =  360°  -  180° + 2x  -  90°  -  90°

m∠DAB  =  2x

y  =  2x

hectictar  Feb 13, 2018
 #1
avatar+7339 
+4
Best Answer

Let's find  m∠DCO .

Notice that  DO  and  CO  are radii of circle O, so they have the same length. So triangle DCO is isosceles, and the base angles have the same measure.

m∠DCO  =  x

 

 

Let's find  m∠CBO .

Since  AQ  is tangent to circle O at point  B ,  OB  is perpendicular to  AQ  .

m∠QBO  =  90°

2x  +  m∠CBO   =   90°

m∠CBO   =   90° - 2x

 

 

Let's find  m∠OCB .

Notice that  BO  and  CO  are radii of circle O, so they have the same length. So triangle OCB is isosceles, and the base angles have the same measure.

m∠OCB   =   m∠CBO

m∠OCB   =   90° - 2x

 

 

Let's find  m∠BCD .

m∠BCD   =   m∠DCO  +  m∠OCB

m∠BCD   =   ( x )  +  ( 90° - 2x )

m∠BCD   =   x + 90° - 2x

m∠BCD   =   90° - x

 

 

Let's find  m∠DOB .

∠BCD  is an inscribed angle, so its measure is half of that of its intercepted arc, ∠DOB .

m∠BCD   =   (1/2) * m∠DOB

2 * m∠BCD   =   m∠DOB

2 * (90° - x)   =   m∠DOB

m∠DOB   =   180° - 2x

 

 

Let's find  m∠DAB .

The sum of the interior angles of quadrilateral DOBA  =  360°

And   m∠ABO  =  90°   and   m∠ADO  =  90°  ,  so...

  m∠DOB   +  90°  +  90°  +  m∠DAB   =   360°

(180° - 2x)  +  90°  +  90°  +  m∠DAB  =  360°

m∠DAB  =  360°  -  180° + 2x  -  90°  -  90°

m∠DAB  =  2x

y  =  2x

hectictar  Feb 13, 2018

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