Question
In the given figure, $AB$ is a diameter of the circle such that $\angle\text{A}=35^\circ$ and $\angle\text{Q}=25^\circ,$ find $\angle\text{PBR.}$
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Answer
Let us first consider the triangle $\triangle\text{ABQ}.$
It is known that in a triangle the sum of all the interior angles add up to $180^\circ .$
So here in our triangle $\triangle\text{ABQ}$
we have, $\angle\text{BAQ}+\angle\text{AQB}+\angle\text{ABQ}=180^\circ$
$\angle\text{ABQ}=180^\circ-\angle\text{BAQ}-\angle\text{AQB}$
$=180^\circ-35^\circ-25^\circ$
$\angle\text{ABQ}=120^\circ$ By a property of the circle we know that an angle formed in a semi-circle will be $90^\circ .$
In the given circle since $‘AB’$ is the diameter of the circle the angle $\angle\text{APB}$ which is formed in a semi-circle will have to be $90^\circ .$
So, we have $\angle\text{APB}=90^\circ$
Now considering the triangle $\triangle\text{APB}$
we have, $\angle\text{APB}+\angle\text{BAP}+\angle\text{ABP}=180^\circ$
$\angle\text{APB}=180^\circ-\angle\text{APB}-\angle\text{BAP}$
$=180^\circ-90^\circ-35^\circ$
$\angle\text{ABP}=55^\circ$ From the given figure it can be seen that,
$\angle\text{ABP}+\angle\text{PBQ}=\angle\text{ABQ}$
$\angle\text{PBQ}=\angle\text{ABQ}-\angle\text{ABP}$
$=120^\circ-55^\circ$
$\angle\text{PBQ}=65^\circ$
Now, we can also say that, $\angle\text{PBQ}+\angle\text{PBR}=180^\circ$
$\angle\text{PBR}=180^\circ-\angle\text{PBQ}$
$=180^\circ-65^\circ$
$\angle\text{PBR}=115^\circ$
Hence the measure of the angle $\angle\text{PBR}$ is $115^\circ .$
It is known that in a triangle the sum of all the interior angles add up to $180^\circ .$
So here in our triangle $\triangle\text{ABQ}$
we have, $\angle\text{BAQ}+\angle\text{AQB}+\angle\text{ABQ}=180^\circ$
$\angle\text{ABQ}=180^\circ-\angle\text{BAQ}-\angle\text{AQB}$
$=180^\circ-35^\circ-25^\circ$
$\angle\text{ABQ}=120^\circ$ By a property of the circle we know that an angle formed in a semi-circle will be $90^\circ .$
In the given circle since $‘AB’$ is the diameter of the circle the angle $\angle\text{APB}$ which is formed in a semi-circle will have to be $90^\circ .$
So, we have $\angle\text{APB}=90^\circ$
Now considering the triangle $\triangle\text{APB}$
we have, $\angle\text{APB}+\angle\text{BAP}+\angle\text{ABP}=180^\circ$
$\angle\text{APB}=180^\circ-\angle\text{APB}-\angle\text{BAP}$
$=180^\circ-90^\circ-35^\circ$
$\angle\text{ABP}=55^\circ$ From the given figure it can be seen that,
$\angle\text{ABP}+\angle\text{PBQ}=\angle\text{ABQ}$
$\angle\text{PBQ}=\angle\text{ABQ}-\angle\text{ABP}$
$=120^\circ-55^\circ$
$\angle\text{PBQ}=65^\circ$
Now, we can also say that, $\angle\text{PBQ}+\angle\text{PBR}=180^\circ$
$\angle\text{PBR}=180^\circ-\angle\text{PBQ}$
$=180^\circ-65^\circ$
$\angle\text{PBR}=115^\circ$
Hence the measure of the angle $\angle\text{PBR}$ is $115^\circ .$
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