From the given figure, find the value of $x$ and $y$.
My Attempt,
$$\angle QPB=\angle QAB=50$$ $$2\angle x=\angle y$$
Now, how to calculate further..?
(Challenging) From the given figure, find the value of $x$ and $y$.
1
$\begingroup$
geometry
circles
angle
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0May be $PB||AQ$? – 2017-02-23
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0@ Roman83, why? Any reason!? – 2017-02-23
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1$x$ can be anything between $0$ and $90$ degree. – 2017-02-23
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0@ Seyed, $x=55^\circ $!! – 2017-02-23
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0As I said it can be anything between $0$ and $90$ – 2017-02-23
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1@Ramanujan You should listen to the smarter people who already told you and showed you (below, by giving you counterexamples!) that your problem does not have a unique solution. You are missing a piece of information that makes the problem uniquely solvable. You copied the problem wrong from the textbook or the chalkboard... – 2017-02-23
1 Answers
2
Hint:
Look at your figure and note that $\angle PBA=\angle PQA = x$.
So, for fixed $\angle QPB=50°$, changing the position of $P$ the value of $x$ is such that:
If $P\to B$ then $x \to 90°$,
if $P\to A$ then $x \to 0°$
In the figures you can see two situations.
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0@ Emilio Novati, I couldn't understand. Could you please elaborate? – 2017-02-23
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0Do you see that $\angle PBA= \angle PQA =x$ ? So if $P$ is near $B$ the angle become near $90°$, and if $P$ is near $A$ the angle become near $0°$.(For $P=B$ the line $PB$ become the tangent to the circle in $B$). – 2017-02-23
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0Yeah. Then after?? – 2017-02-23
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0@ Emilio Novati, but the answer in my book is $x=55^\circ $ and $y=110^\circ $? – 2017-02-23
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1Maybe that there is another condition that fix the unique solution? – 2017-02-23
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2I don't understand the downvote. Where I'm wrong? – 2017-02-23
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0@ Emilio Novati, but why aren't we getting the answer I have stated?? – 2017-02-23
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0Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/54133/discussion-between-emilio-novati-and-ramanujan). – 2017-02-23