In the given figure, ∆ABC and ∆DBC are inscribed in a circle such that ∠BAC = 60° and ∠DBC = 50°.
(a) 50°
(b) 60°
(c) 70°
(d) 80°

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Solution

(c) 70°
∠BDC = ∠BAC = 60° (Angles in the same segment of a circle)
In Δ BDC, we have:
∠DBC + ∠BDC + ∠BCD = 180° (Angle sum property of a triangle)
∴ 50° + 60° + ∠BCD = 180°
⇒ ∠BCD = 180° - (50° + 60°) = (180° - 110°) = 70°

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Similar questions

Q. If the given figure,

Δ A B C

and

Δ D B C

are inscribed in a circle such that

∠ B A C = 60^{\circ}

and

∠ D B C = 50^{\circ}

. Find

∠ B C D

$I n t h e g i v e n f i g u r e, △ A B C i s i n s c r i b e d i n a c i r c l e . T h e b i s e c t o r o f ∠ B A C m e e t s B C a t D a n d t h e c i r c l e a t E, i f E C i s j o i n e d t h e n ∠ E C D = 30^{0} . F i n d ∠ B A C .$

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In the given figure, side BC of $Δ A B C$ is produced to D such that $∠ A B C = 70^{\circ}$ and $∠ A C D = 120^{\circ}$ . Then, $∠ B A C =$ ?

(a) $60^{\circ}$
(b) $50^{\circ}$
(c) $70^{\circ}$
(d) $35^{\circ}$

Q. In the given figure, ∠DAB = 60° and ∠ABD = 50°, then ∠ACB = ?
(a) 50°
(b) 60°
(c) 70°
(d) 80°

In the figure ABCD is a ||^gm in which ∠DAB = 80^o and ∠DBC = 60^o, then ∠CDB and ∠ADB are equal to

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In the given figure, ∆ABC and ∆DBC are inscribed in a circle such that ∠BAC = 60° and ∠DBC = 50°. (a) 50° (b) 60° (c) 70° (d) 80°

(c) 70° ∠BDC = ∠BAC = 60° (Angles in the same segment of a circle) In Δ BDC, we have: ∠DBC + ∠BDC + ∠BCD = 180° (Angle sum property of a triangle) ∴ 50° + 60° + ∠BCD = 180° ⇒ ∠BCD = 180° - (50° + 60°) = (180° - 110°) = 70°

In the given figure, ∆ABC and ∆DBC are inscribed in a circle such that ∠BAC = 60° and ∠DBC = 50°.
(a) 50°
(b) 60°
(c) 70°
(d) 80°

(c) 70°
∠BDC = ∠BAC = 60° (Angles in the same segment of a circle)
In Δ BDC, we have:
∠DBC + ∠BDC + ∠BCD = 180° (Angle sum property of a triangle)
∴ 50° + 60° + ∠BCD = 180°
⇒ ∠BCD = 180° - (50° + 60°) = (180° - 110°) = 70°