Question

In the figure below, PQRS is cyclic quadrilateral. If angle SPR=25° and PRS= 60°, find the of angle ROP. Give reasons also.​

Answer 1

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.

$\large\underline{\sf{Solution-}}$

Given that,

• $\rm \:\angle SPR = 25\degree$

• $\rm \:\angle PRS = 60\degree$

In $\triangle$ PRS,

We know, sum of all interior angles of a triangle is supplementary.

So, using this

$\rm \: \angle PRS + \angle SPR + \angle RSP = 180\degree$

$\rm \: 60\degree + 25\degree + \angle RSP = 180\degree$

$\rm \: 85\degree + \angle RSP = 180\degree$

$\rm \: \angle RSP = 180\degree - 85\degree$

$\rm \: \bf\implies \:\angle RSP = 95\degree$

Now, PQRS is a cyclic quadrilateral.

We know, sum of the opposite angles of a cyclic quadrilateral is supplementary.

So, using this, we get

$\rm \: \angle RSP + \angle RQP = 180\degree$

$\rm \: 95\degree + \angle RQP = 180\degree$

$\rm \: \angle RQP = 180\degree - 95\degree$

$\rm\implies \:\boxed{ \rm{ \:\bf \: \angle RQP = 85\degree \: \: }}$

$\rule{190pt}{2pt}$

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.

Answer 2

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.

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \:\angle SPR = 25\degree∠SPR=25°$

$\rm \:\angle PRS = 60\degree∠PRS=60°$

$In \ \triangle \: PRS,$

We know, sum of all interior angles of a triangle is supplementary.

So, using this

$\begin{gathered}\rm \: \angle PRS + \angle SPR + \angle RSP = 180\degree \\ \end{gathered}$

∠PRS+∠SPR+∠RSP=180°

$\begin{gathered}\rm \: 60\degree + 25\degree + \angle RSP = 180\degree \\ \end{gathered}$

60°+25°+∠RSP=180°

$\begin{gathered}\rm \: 85\degree + \angle RSP = 180\degree \\ \end{gathered}$

85°+∠RSP=180°

$\begin{gathered}\rm \: \angle RSP = 180\degree - 85\degree \\ \end{gathered}$

∠RSP=180°−85°

$\begin{gathered}\rm \: \bf\implies \:\angle RSP = 95\degree \\ \end{gathered}$

⟹∠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

$\begin{gathered}\rm \: \angle RSP + \angle RQP = 180\degree \\ \end{gathered}$

∠RSP+∠RQP=180°

$\begin{gathered}\rm \: 95\degree + \angle RQP = 180\degree \\ \end{gathered}$

95°+∠RQP=180°

$\begin{gathered}\rm \: \angle RQP = 180\degree - 95\degree \\ \end{gathered}$

∠RQP=180°−95°

$\begin{gathered}\rm\implies \:\boxed{ \rm{ \:\bf \: \angle RQP = 85\degree \: \: }} \\ \end{gathered}$

⟹ ∠RQP=85°

$\rule{190pt}{2pt}$

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.