the tangent at a point c of a circle and a diameter ab when extended

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The tangent at a point C of a circle and a diameter AB when extended intersect at P. If PCA = 110^0 , find CBA .Hint : Join C with centre O.

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The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110

Question 0

, find ∠CBA .Hint : Join C with centre O.

A70

o

B90

o

C50

o

D30

o Medium Open in App Solution Verified by Toppr

Correct option is A)

Let O be the center of the circle.

A,O,B,P all are on the same line and P and C are points on the tangent.

AB is a diameter of a circle.

∴  ∠BCA=90 o

[ Angle inscribe in a semi-circle. ]

C is the point on the circle where the tangent touches the circle.

⇒  So, ∠OCP=90 o . ⇒  ∠PCA=∠PCO+∠OCA ⇒  110 o =90 o +∠OCA ⇒  ∠OCA=20 o In △AOC,

⇒ AO=OC             [ Radius of a circle. ]

⇒  ∠OCA=∠CAO=20 o In △ABC,

⇒  ∠CAB+∠CBA+∠BCA=180

o ⇒  20 o +∠CBA+90 o =180 o ⇒  110 o +∠CBA=180 o ∴  ∠CBA=70 o .

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In the figure, tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠ PCA = 110∘, find ∠ CBA

In the figure, tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠ PCA = 110∘, find ∠ CBA

Byju's Answer Standard X Mathematics

Tangent Perpendicular to Radius at Point of Contact

In the figure... Question

In the figure, tangent at a point C of a circle and a diameter AB when extended intersect at P. If

∠ P C A = 110 ∘ , find ∠ C B A

Open in App Solution ∠PCA = 110°

PC is the tangent to the circle whose centre is O.

Construction

Join points C and O.

∠BCA = 90° [Since angle in a semi circle is 90°]

Also ∠PCO = 90° [Since radius ⊥ tangent]

From the figure we have,

∠PCA =∠PCO + ∠OCA

i.e. 110° = 90° + ∠OCA

∴ ∠OCA =20°

Now in ΔAOC, AO = OC [Radii]

So, ∠OCA = ∠OAC =20°

In ΔABC, we have

∠BCA = 90° & ∠CAB = 20°

∴ ∠CBA = 70° Suggest Corrections 62

SIMILAR QUESTIONS

Q. The tangent at a point C of a circle and a diameter AB when extended intersect at P. If

∠ P C A = 110 0

, find CBA [see Fig. 9.21].

Q. In Fig . 10.69, the tangent at a point C of a circle and a diameter AB when extended intersect at P . If

∠ PCA =1100, find ∠

CBA. [Hint: Join CO.]

figure

Q. Question 12

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If

∠ P C A = 110 ∘ , find ∠ CBA.

Q. Question 12

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If

∠ P C A = 110 ∘ , find ∠ CBA.

Q. The tangent at a point C of a circle and a diameter ab when extended intersect it p. If angle PCA = 110, find angle CBA

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Theorems MATHEMATICS Watch in App EXPLORE MORE

Tangent Perpendicular to Radius at Point of Contact

Standard X Mathematics

स्रोत : byjus.com

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110º , find ∠CBA [see Fig. 9.21]

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110º , find ∠CBA [see Fig. 9.21] - The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110º, then ∠CBA = 70°

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110º , find ∠CBA [see Fig. 9.21]

Solution:

Given, O is the centre of a circle.

AB is the diameter of the circle

AB is extended to P and PC is a tangent to the circle at point C.

Given, ∠PCA = 110º

We have to find ∠CBA

We know that angle in a semicircle is always equal to 90°

So, ∠BCA = 90°

We know that the radius of a circle is perpendicular to the tangent at the point of contact.

i.e., OC ⟂ PC So, ∠OCP = 90° From the figure, ∠PCA = ∠BCA + ∠PCB 110° = 90° + ∠PCB ∠PCB = 110° - 90° ∠PCB = 20°

By alternate segment theorem,

We know that the angle between the tangent and the chord of a circle is equal to the angle made by the chord in the alternate segment.

∠PCB = ∠CAB So, ∠CAB = 20°

Considering triangle ABC,

We know that the sum of all three interior angles of a triangle is equal to 180°

∠BCA + ∠CBA + ∠CAB = 180°

90° + ∠CBA + 20° = 180°

110° + ∠CBA = 180° ∠CBA = 180° - 110°

Therefore, ∠CBA = 70°

✦ Try This: In the figure, PQ is a tangent at a point C to a circle with centre O. If AB is a diameter and ∠CAB = 30°, find ∠PCA.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10NCERT Exemplar Class 10 Maths Exercise 9.4 Problem 12

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110º , find ∠CBA [see Fig. 9.21]

Summary:

The tangent at a point C of a circle and a diameter AB when extended intersect at P. If ∠PCA=110º, then ∠CBA = 70°

☛ Related Questions:

In Fig. 9.20. O is the centre of a circle of radius 5 cm, T is a point such that OT = 13 cm and OT i . . . .

If an isosceles triangle ABC, in which AB = AC = 6 cm, is inscribed in a circle of radius 9 cm, find . . . .

A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tan . . . .

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