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प्रश्न
In the figure given below 'O' is the centre of the circle. If QR = OP and ∠ORP = 20°. Find the value of 'x' giving reasons
उत्तर
Now,
OP = QR....given
So,OP = OT = OQ = QR
In ΔRQP
RQ = QO
So `angleQRO = angleQOR = 20^@`
So by sum of angles in ΔRQP
`angleRQO = 140^@`
Now
`angleRQO + angleOQP = 180^@` .....linear pair
`angle OQP = 40^@`
in ΔPOQ
OQ = PO...radii
So `angleQPO - angle OQP = 40^@`
So by sumof angle in ΔOQP
`anglePOQ - 100^@`
Now
`anglePOT + anglePOQ +angle QOR = 180^@` ......angles in straight line
`x= 60^@`
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संबंधित प्रश्न
Use ruler and compasses only for the following questions. All constructions lines and arcs must be clearly shown
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cm and ∠ABC = 120°
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Using a ruler and compasses only:
1) Construct a triangle ABC with the following data: AB = 3.5 cm, BC = 6 cm and ABC = 120°
2) In the same diagram, draw a circle with BC as diameter. Find a point P on the circumference of the circle which is equidistant from AB and BC.
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Use ruler and compasses only for this question:
(i) Construct A ABC, where AB = 3.5 cm, BC = 6 cm and ∠ ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
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