Share
Notifications

View all notifications
Advertisement

In the Adjoining Figure, O is the Centre of the Circle and Ab is a Tangent to It at Point B. ∠Bdc = 65° Find ∠Bao. - Mathematics

Login
Create free account


      Forgot password?

Question

In the adjoining figure, O is the centre of the circle and AB is a tangent to it at point B. ∠BDC = 65° Find ∠BAO.

Solution 1

AB is a straight line.
∴ `∠`ADE +  `∠`BDE = 180°
⇒ `∠`ADE  + 65° = 180°
 ⇒ `∠`ADE = 115°          ………..(i)
AB i.e. DB is tangent to the circle at point B and BC is the diameter.

∴`∠`DB = 90°
In ΔBDC,
`∠`DBC +  `∠`BDC + `∠`DCB = 180° 
⇒ 90° +  65° +  `∠`DCB  = 180°
⇒ `∠` DCB = 25°
Now, OE = OC (radii of the same circle)
∴ `∠`DCB or `∠`OCE =  `∠`OEC = 25°
Also,
`∠`OEC = `∠`DEC =  25°

(vertically opposite angles)
In  ΔADE,
`∠`ADE + `∠`DEA + `∠`DAE = 180°
From (i) and (ii)
115° + 25°  + `∠` DAE  = 180°
⇒ `∠`DAE or `∠`BAO  = 180° - 140° =40°
∴`∠`BAO = 40°

Solution 2

As AB is a tangent to the circle at B and OB is radius, OB + AB ⇒ ∠ CBD = 90°.

In ΔBCD,
∠ BCD + ∠ CBD + ∠ BDC = 180°
∠ BCD + 90° + 65° = 180°
∠ BCD + 155° = 180°
∠ BCD = 180° - 155°
∠ BCD = 25°
∠ BOE = 2∠ BCE      ....(angle at centre = double the angle at the remaining part of circle)
∠ BOE = 2 x 25°
∠ BOE = 50°
∠ BOA = 50°

In ΔBOA,
∠ BAO + ∠ ABO + ∠ BOA = 180°
∠ BAO + 90° + 50° = 180°
∠ BAO + 140° = 180°
∠ BAO = 180° - 140°
∠ BAO = 40°

  Is there an error in this question or solution?
Advertisement

APPEARS IN

 Selina Solution for Concise Mathematics for Class 10 ICSE (2020 (Latest))
Chapter 18: Tangents and Intersecting Chords
Exercise 18 (B) | Q: 16 | Page no. 284
Advertisement
In the Adjoining Figure, O is the Centre of the Circle and Ab is a Tangent to It at Point B. ∠Bdc = 65° Find ∠Bao. Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.
Advertisement
View in app×