How do I solve for x using the Circle Theorems?
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got as far as stating that OBP=90˚ (as angle between tangent and radius is always 90˚), and thus CBO=90˚- 2x. CBO=OCB as they are bases in a isosceles. COB=180-90-2x-90-2x. But after this, i am clueless.
I am stuck with this Question. It is from a GCSE Further Maths past paper. Despite seeing online tutorials, and checking the answer scheme, I still don't understand how you solve this question. Could you please show me a step by step explanation of how you solve this question. Thank you.
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enter image description here
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Apr 14 '15 at 10:57
arjunn
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Jun 12 '20 at 10:38
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Just figured it out, check below for the answer! – arjunn Apr 15 '15 at 18:03
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At first forget about the condition that ∠CBP=2x. (The condition ∠CBP=2∠ODC determines the exact position of C.)
You can express ∠BCD with y. Using ∠OCD=∠ODC=x you can express ∠OCB=OBC with x and y and hence also ∠CBP.
Now use the condition that ∠CBP equals 2x. This gives an equation in x and y.
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Apr 15 '15 at 13:10
coproc
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Try expressing the same angle both in terms of x and in terms of y. Then you can set the expressions equal to each other and rearrange. For example, you can do this with either ∠BOD (either the reflex or obtuse angle) or ∠BCD. Does this help?
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Apr 15 '15 at 9:27
bnosnehpets
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Apr 15 '15 at 15:30
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enter image description hereAngle BOD = 180 − y
Angle OCD = x
Angle OBC = 90 − 2x
Angle BCO= 90 − 2x
Angle BOD reflex = 360 - (90 − 2x) − (90 − 2x) − x − x = 180 + 2x.
180 − y + 180 + 2x = 360,
thus y = 2x
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Apr 15 '15 at 18:02
arjunn
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Apr 15 '15 at 18:27
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I am also doing Further Maths GCSE and what i would do is draw a line from B to D (this is a lot shorter) this would create the alternate segment theorem for angles therefore BDC would be 2x and BDO and DBO=x which means BOD=180−2x which means y=360−90−90−(180−2x)
y=2x
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Feb 7 '16 at 19:40
ahm
Answer:
Y=150/2=5
x+y=180°
x+75=180°
x=105°
Step-by-step explanation:
hope it is helpful for you and please mark me brainiest please