PQRS is a cyclic quadrilateral. Given $$\angle QPS = {73^ \circ }$$, $$\angle PQS = {55^ \circ }$$ and $$\angle PSR = {82^ \circ }$$, calculate $$\angle RQS$$.
Answer
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Hint: Here in this question, we need to find the value of angle $$\angle RQS$$, first we need to analyze the figure given in question, then use the properties of angles also keep in mind to use the corresponding angles and opposite angles properties of angles, using this information to approach towards the solution of the given problem.
Complete step by step answer:
Consider the given question:
Given, PQRS is a cyclic quadrilateral, then $$\angle QPS = {73^ \circ }$$, $$\angle PQS = {55^ \circ }$$ and $$\angle PSR = {82^ \circ }$$.
We need to calculate the angle $$\angle RQS$$.
As we know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e., $${180^ \circ }$$.
In the given figure, $$\angle S = \angle PSR = {82^ \circ }$$ and $$\angle Q = \angle PQR$$ are opposite angles.
Then,
$$ \Rightarrow \,\,\angle S + \angle Q = {180^ \circ }$$
$$ \Rightarrow \,\,{82^ \circ } + \angle Q = {180^ \circ }$$
Subtract $${82^ \circ }$$ on both the side, then we have
$$ \Rightarrow \,\,\angle Q = {180^ \circ } - {82^ \circ }$$
$$ \Rightarrow \,\,\,\,\angle Q = {98^ \circ }$$
Therefore, $$\angle Q = \angle PQR = {98^ \circ }$$.
By the figure,
$$ \Rightarrow \,\,\angle PQS + \angle RQS = \angle PQR$$
$$ \Rightarrow \,\,{55^ \circ } + \angle RQS = {98^ \circ }$$
Subtract $${55^ \circ }$$ on both the sides, then we get
$$ \Rightarrow \,\,\angle RQS = {98^ \circ } - {55^ \circ }$$
$$\therefore \,\,\,\,\,\angle RQS = {43^ \circ }$$
Hence, the value of required angle $$\angle RQS = {43^ \circ }$$
Note:
For this type of problem, drawing a diagram is more important and it makes proof easy. To solve this kind of problem we need to know properties of cyclic quadrilaterals and it’s angle sum properties.
Complete step by step answer:
Consider the given question:
Given, PQRS is a cyclic quadrilateral, then $$\angle QPS = {73^ \circ }$$, $$\angle PQS = {55^ \circ }$$ and $$\angle PSR = {82^ \circ }$$.
We need to calculate the angle $$\angle RQS$$.
As we know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e., $${180^ \circ }$$.
In the given figure, $$\angle S = \angle PSR = {82^ \circ }$$ and $$\angle Q = \angle PQR$$ are opposite angles.
Then,
$$ \Rightarrow \,\,\angle S + \angle Q = {180^ \circ }$$
$$ \Rightarrow \,\,{82^ \circ } + \angle Q = {180^ \circ }$$
Subtract $${82^ \circ }$$ on both the side, then we have
$$ \Rightarrow \,\,\angle Q = {180^ \circ } - {82^ \circ }$$
$$ \Rightarrow \,\,\,\,\angle Q = {98^ \circ }$$
Therefore, $$\angle Q = \angle PQR = {98^ \circ }$$.
By the figure,
$$ \Rightarrow \,\,\angle PQS + \angle RQS = \angle PQR$$
$$ \Rightarrow \,\,{55^ \circ } + \angle RQS = {98^ \circ }$$
Subtract $${55^ \circ }$$ on both the sides, then we get
$$ \Rightarrow \,\,\angle RQS = {98^ \circ } - {55^ \circ }$$
$$\therefore \,\,\,\,\,\angle RQS = {43^ \circ }$$
Hence, the value of required angle $$\angle RQS = {43^ \circ }$$
Note:
For this type of problem, drawing a diagram is more important and it makes proof easy. To solve this kind of problem we need to know properties of cyclic quadrilaterals and it’s angle sum properties.
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