Appropriate Question :-
In the figure, PQRS is cyclic quadrilateral. If angle SPR=25° and angle PRS= 60°, find the of angle RQP. Give reasons also.
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
In [tex] \triangle[/tex] PRS,
We know, sum of all interior angles of a triangle is supplementary.
So, using this
[tex]\rm \: \angle PRS + \angle SPR + \angle RSP = 180\degree \\ [/tex]
[tex]\rm \: 60\degree + 25\degree + \angle RSP = 180\degree \\ [/tex]
[tex]\rm \: 85\degree + \angle RSP = 180\degree \\ [/tex]
[tex]\rm \: \angle RSP = 180\degree - 85\degree \\ [/tex]
[tex]\rm \: \bf\implies \:\angle RSP = 95\degree \\ [/tex]
Now, PQRS is a cyclic quadrilateral.
We know, sum of the opposite angles of a cyclic quadrilateral is supplementary.
So, using this, we get
[tex]\rm \: \angle RSP + \angle RQP = 180\degree \\ [/tex]
[tex]\rm \: 95\degree + \angle RQP = 180\degree \\ [/tex]
[tex]\rm \: \angle RQP = 180\degree - 95\degree \\ [/tex]
[tex]\rm\implies \:\boxed{ \rm{ \:\bf \: \angle RQP = 85\degree \: \: }} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information :-
1. Angle in same segments are equal.
2. Angle in semi-circle is right angle.
3. Angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference of a circle by the same arc.
4. Exterior angle of a cyclic quadrilateral is equals to its interior opposite angle.
5. Equals chords of a circle are equidistant from center.
6. Equal chords subtends equal angles at the center.
[tex]\large\underline{\sf{Solution-}} [/tex]
[tex]\rm \:\angle SPR = 25\degree∠SPR=25°[/tex]
[tex]\rm \:\angle PRS = 60\degree∠PRS=60°[/tex]
[tex]In \ \triangle \: PRS,[/tex]
[tex]\begin{gathered}\rm \: \angle PRS + \angle SPR + \angle RSP = 180\degree \\ \end{gathered} [/tex]
∠PRS+∠SPR+∠RSP=180°
[tex]\begin{gathered}\rm \: 60\degree + 25\degree + \angle RSP = 180\degree \\ \end{gathered} [/tex]
60°+25°+∠RSP=180°
[tex]\begin{gathered}\rm \: 85\degree + \angle RSP = 180\degree \\ \end{gathered} [/tex]
85°+∠RSP=180°
[tex]\begin{gathered}\rm \: \angle RSP = 180\degree - 85\degree \\ \end{gathered} [/tex]
∠RSP=180°−85°
[tex]\begin{gathered}\rm \: \bf\implies \:\angle RSP = 95\degree \\ \end{gathered} [/tex]
⟹∠RSP=95°
[tex]\begin{gathered}\rm \: \angle RSP + \angle RQP = 180\degree \\ \end{gathered} [/tex]
∠RSP+∠RQP=180°
[tex]\begin{gathered}\rm \: 95\degree + \angle RQP = 180\degree \\ \end{gathered} [/tex]
95°+∠RQP=180°
[tex]\begin{gathered}\rm \: \angle RQP = 180\degree - 95\degree \\ \end{gathered} [/tex]
∠RQP=180°−95°
[tex]\begin{gathered}\rm\implies \:\boxed{ \rm{ \:\bf \: \angle RQP = 85\degree \: \: }} \\ \end{gathered} [/tex]
⟹ ∠RQP=85°
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Verified answer
Appropriate Question :-
In the figure, PQRS is cyclic quadrilateral. If angle SPR=25° and angle PRS= 60°, find the of angle RQP. Give reasons also.
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
In [tex] \triangle[/tex] PRS,
We know, sum of all interior angles of a triangle is supplementary.
So, using this
[tex]\rm \: \angle PRS + \angle SPR + \angle RSP = 180\degree \\ [/tex]
[tex]\rm \: 60\degree + 25\degree + \angle RSP = 180\degree \\ [/tex]
[tex]\rm \: 85\degree + \angle RSP = 180\degree \\ [/tex]
[tex]\rm \: \angle RSP = 180\degree - 85\degree \\ [/tex]
[tex]\rm \: \bf\implies \:\angle RSP = 95\degree \\ [/tex]
Now, PQRS is a cyclic quadrilateral.
We know, sum of the opposite angles of a cyclic quadrilateral is supplementary.
So, using this, we get
[tex]\rm \: \angle RSP + \angle RQP = 180\degree \\ [/tex]
[tex]\rm \: 95\degree + \angle RQP = 180\degree \\ [/tex]
[tex]\rm \: \angle RQP = 180\degree - 95\degree \\ [/tex]
[tex]\rm\implies \:\boxed{ \rm{ \:\bf \: \angle RQP = 85\degree \: \: }} \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information :-
1. Angle in same segments are equal.
2. Angle in semi-circle is right angle.
3. Angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference of a circle by the same arc.
4. Exterior angle of a cyclic quadrilateral is equals to its interior opposite angle.
5. Equals chords of a circle are equidistant from center.
6. Equal chords subtends equal angles at the center.
Appropriate Question :-
In the figure, PQRS is cyclic quadrilateral. If angle SPR=25° and angle PRS= 60°, find the of angle RQP. Give reasons also.
[tex]\large\underline{\sf{Solution-}} [/tex]
Given that,
[tex]\rm \:\angle SPR = 25\degree∠SPR=25°[/tex]
[tex]\rm \:\angle PRS = 60\degree∠PRS=60°[/tex]
[tex]In \ \triangle \: PRS,[/tex]
We know, sum of all interior angles of a triangle is supplementary.
So, using this
[tex]\begin{gathered}\rm \: \angle PRS + \angle SPR + \angle RSP = 180\degree \\ \end{gathered} [/tex]
∠PRS+∠SPR+∠RSP=180°
[tex]\begin{gathered}\rm \: 60\degree + 25\degree + \angle RSP = 180\degree \\ \end{gathered} [/tex]
60°+25°+∠RSP=180°
[tex]\begin{gathered}\rm \: 85\degree + \angle RSP = 180\degree \\ \end{gathered} [/tex]
85°+∠RSP=180°
[tex]\begin{gathered}\rm \: \angle RSP = 180\degree - 85\degree \\ \end{gathered} [/tex]
∠RSP=180°−85°
[tex]\begin{gathered}\rm \: \bf\implies \:\angle RSP = 95\degree \\ \end{gathered} [/tex]
⟹∠RSP=95°
Now, PQRS is a cyclic quadrilateral.
We know, sum of the opposite angles of a cyclic quadrilateral is supplementary.
So, using this, we get
[tex]\begin{gathered}\rm \: \angle RSP + \angle RQP = 180\degree \\ \end{gathered} [/tex]
∠RSP+∠RQP=180°
[tex]\begin{gathered}\rm \: 95\degree + \angle RQP = 180\degree \\ \end{gathered} [/tex]
95°+∠RQP=180°
[tex]\begin{gathered}\rm \: \angle RQP = 180\degree - 95\degree \\ \end{gathered} [/tex]
∠RQP=180°−95°
[tex]\begin{gathered}\rm\implies \:\boxed{ \rm{ \:\bf \: \angle RQP = 85\degree \: \: }} \\ \end{gathered} [/tex]
⟹ ∠RQP=85°
[tex]\rule{190pt}{2pt}[/tex]
Additional Information :-
1. Angle in same segments are equal.
2. Angle in semi-circle is right angle.
3. Angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference of a circle by the same arc.
4. Exterior angle of a cyclic quadrilateral is equals to its interior opposite angle.
5. Equals chords of a circle are equidistant from center.
6. Equal chords subtends equal angles at the center.