I’m really stuck on this circle theorems question, can someone help please

Step-by-step explanation:

O is the center of the circle.  So OD, OC, and OB are all radii of the circle, and therefore congruent.

That means that ΔDCO and ΔCBO are isosceles triangles.  The base angles of isosceles triangles are congruent.  So ∠DCO = x.

ADP and ABQ are tangent to the circle.  Therefore, they are perpendicular to OD and OB.

Since ∠ABO = 90°, ∠CBO = 90° − 2x.  Which means ∠OCB = 90° − 2x.

∠BCD is the sum of ∠DCO and ∠OCB:

∠BCD = x + 90° − 2x

∠BCD = 90° − x

Angles of a triangle add up to 180°.  So:

∠DOC = 180° − x − x = 180° − 2x

∠BOC = 180° − (90° − 2x) − (90° − 2x) = 4x

Central angles add up to 360°.  So ∠DOB is:

∠DOB = 360° − (180° − 2x) − (4x)

∠DOB = 180° − 2x

The interior angles of kite ADOB add up to 360°.

360° = 90° + 90° + (180° − 2x) + y

360° = 360° − 2x + y

y = 2x

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2