In the given figure, ÐPQR = …

Here,  ∠PQR=100^∘

Take a point S in the major arc. Join PS and RS.

{tex}\because{/tex} PQRS is a cyclic quadrilateral.
{tex}\therefore \angle \mathrm { PQR } + \angle \mathrm { PSQ } = 180 ^ { \circ }{/tex}
|The sum of either pair of opposite angles of a cyclic quadrilateral is 180^o
{tex}\Rightarrow 100 ^ { \circ } + \angle P S R = 180 ^ { \circ }{/tex}
{tex}\Rightarrow \angle P S R = 180 ^ { \circ } - 100 ^ { \circ }{/tex}
{tex}\Rightarrow \angle P S R = 80 ^ { \circ }{/tex} ......... (1)
Now, {tex}\angle \mathrm { POR } = 2 \angle \mathrm { PSR }{/tex}
|The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle
{tex}= 2 \times 80 ^ { \circ } = 160 ^ { \circ }{/tex} ........ (2) |Using (1)
In {tex}\triangle O P R{/tex}
{tex}\because O P = O R{/tex} |Radii of a circle
{tex}\therefore \angle \mathrm { OPR } = \angle \mathrm { ORP }{/tex} ....... (3)
|Angles opposite to equal sides of a triangle are equal
In {tex}\triangle O P R{/tex}
{tex}\angle O P R + \angle O R P + \angle P O R = 180 ^ { \circ }{/tex} | Sum of all the angles of a triangle is 180^o
{tex}\Rightarrow \angle \mathrm { OPR } + \angle \mathrm { OPR } + 160 ^ { \circ } = 180 ^ { \circ }{/tex} |Using (2) and (1)
{tex}\Rightarrow 2 \angle O P R + 160 ^ { \circ } = 180 ^ { \circ }{/tex}
{tex}\Rightarrow 2 \angle O P R = 180 ^ { \circ } - 160 ^ { \circ } = 20 ^ { \circ }{/tex}
{tex}\Rightarrow \angle \mathrm { OPR } = 10 ^ { \circ }{/tex}