Advertisements
Advertisements
प्रश्न
In the figure, given below, AD = BC, ∠BAC = 30° and ∠CBD = 70°.
Find:
- ∠BCD
- ∠BCA
- ∠ABC
- ∠ADB
उत्तर
In the figure, ABCD is a cyclic quadrilateral
AC and BD are its diagonals
∠BAC = 30° and ∠CBD = 70°
Now we have to find the measure of
∠BCD, ∠BCA, ∠ABC and ∠ADB
We have ∠CAD = ∠CBD = 70° ...[Angles in the same segment]
Similarly, ∠BAD = ∠BDC = 30°
∴ ∠BAD = ∠BAC + ∠CAD
= 30° + 70°
= 100°
i. Now ∠BCD + ∠BAD = 180° ...[Opposite angles of cyclic quadrilateral]
`=>` ∠BCD + ∠BAD = 180°
`=>` ∠BCD + 100° = 180°
`=>` ∠BCD = 180° – 100°
`=>` ∠BCD = 80°
ii. Since AD = BC,
ABCD is an isosceles trapezium and AB || DC
∠BAC = ∠DCA ...[Alternate angles]
`=>` ∠DCA = 30°
∴ ∠ABD = ∠DAC = 30° ...[Angles in the same segment]
∴ ∠BCA = ∠BCD – ∠DAC
= 80° – 30°
= 50°
iii. ∠ABC = ∠ABD + ∠CBD
= 30° + 70°
= 100°
iv. ∠ADB = ∠BCA = 50° ...[Angles in the same segment]
APPEARS IN
संबंधित प्रश्न
In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC . Show your steps of working.
AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find : ∠PRB
In the given figure, AOC is a diameter and AC is parallel to ED. If ∠CBE = 64°, calculate ∠DEC.
In the given circle with diameter AB, find the value of x.
If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66° and ∠ABC = 80°.
Calculate:
- ∠DBC,
- ∠IBC,
- ∠BIC.
In the given figure, AB = AD = DC = PB and ∠DBC = x°. Determine, in terms of x :
- ∠ABD,
- ∠APB.
Hence or otherwise, prove that AP is parallel to DB.
In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°, find the number of degrees in ∠ABC.
In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find: ∠ ACB.
Hence, show that AC is a diameter.
In the given Figure. P is any point on the chord BC of a circle such that AB = AP. Prove that CP = CQ.
In the following figure, O is the centre of the circle, ∠ PBA = 42°.
Calculate:
(i) ∠ APB
(ii) ∠PQB
(iii) ∠ AQB