Advertisements
Advertisements
Question
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
Options
41°
23°
67°
18°
Advertisements
Solution
Here we have a cyclic quadrilateral PQRS with PR being a diameter of the circle. Let the centre of this circle be ‘O’.
We are given that `angleQPR` and `angleSPR = 72°` . This is shown in fig (2).

So we see that,
\[\angle QPS = \angle QPR + \angle RPS\]
\[ = 67°+ 72° \]
\[ = 139°\]
In a cyclic quadrilateral it is known that the opposite angles are supplementary.
`angleQPS + angleQRS = 180°`
`angleQRS = 180° - angleQPS`
`= 180° - 139°`
= 41°
APPEARS IN
RELATED QUESTIONS
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

In the figure, `square`ABCD is a cyclic quadrilateral. Seg AB is a diameter. If ∠ ADC = 120˚, complete the following activity to find measure of ∠ BAC.
`square` ABCD is a cyclic quadrilateral.
∴ ∠ ADC + ∠ ABC = 180°
∴ 120˚ + ∠ ABC = 180°
∴ ∠ ABC = ______
But ∠ ACB = ______ .......(angle in semicircle)
In Δ ABC,
∠ BAC + ∠ ACB + ∠ ABC = 180°
∴ ∠BAC + ______ = 180°
∴ ∠ BAC = ______
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.

ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
