Advertisements
Advertisements
प्रश्न
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.
उत्तर
Join PR.
i. ∠PRQ = 90°
(Angle in a semicircle)
∴ In right triangle PQR,
∠RPQ = 90° – ∠PQR
= 90° – 58°
= 32°
ii. Also, SR || PQ
∠PRS = ∠RPQ = 32° (Alternate angles)
In cyclic quadrilateral PRST,
∠STP = 180° – ∠PRS
= 180° – 32°
= 148°
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
APPEARS IN
संबंधित प्रश्न
In the figure, m∠DBC = 58°. BD is the diameter of the circle. Calculate:
1) m∠BDC
2) m∠BEC
3) m∠BAC
Calculate the area of the shaded region, if the diameter of the semicircle is equal to 14 cm. Take `pi = 22/7`
ABC is a right angles triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Calculate the value of x, the radius of the inscribed circle.
Prove that the parallelogram, inscribed in a circle, is a rectangle.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate:
- ∠DAB,
- ∠DBA,
- ∠DBC,
- ∠ADC.
Also, show that the ΔAOD is an equilateral triangle.
In the given figure, AB is the diameter of a circle with centre O.
If chord AC = chord AD, prove that:
- arc BC = arc DB
- AB is bisector of ∠CAD.
Further, if the length of arc AC is twice the length of arc BC, find:
- ∠BAC
- ∠ABC
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate : ∠DBA
Also, show that the ΔAOD is an equilateral triangle.
In the given figure, O is the centre of the circle and ∠PBA = 45°. Calculate the value of ∠PQB.
In the given figure AC is the diameter of the circle with centre O. CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°.
Calculate
- ∠BEC
- ∠BCD
- ∠CED
