Math, asked by SilverButterfly, 1 month ago

In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate :

(i) ∠BDC (ii) ∠BEC (iii) ∠BAC​

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Answers

Answered by ItzAshi
68

Step-by-step explanation:

Question :

In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate :

(i) ∠BDC (ii) ∠BEC (iii) ∠BAC

Solution :

In ∆ BCD; ∠DBC = 58°

(i) ∠BCD = 90° (Angle in the semicircle as BD is diameter)

{\bold{\sf{∴  \:  \:  \:  \:  \: ∠DBC \:  +  \: ∠BCD  \: + \:  ∠BDC  \: = \:  180°}}} \\

{\bold{\sf{➠  \:  \:  \:  \:  \: 58° \: +  \: 90° \:  + \: ∠BDC  \: = \:  180°}}} \\

{\bold{\sf{➠  \:  \:  \:  \:  \: ∠BDC  \: =  \: 180° \:  – \:  (90°  \: + \:  58°)}}} \\

{\bold{\sf{➠ \:  \:  \:  \:  \:  180° \: - \:  148°}}} \\

{\bold{\sf{➠  \:  \:  \:  \:  \: 32°}}} \\

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(ii) ∠BEC + ∠BDC = 180° ( ∵ BECD is a cyclic quadrilateral)

{\bold{\sf{➠ \:  \:  \:  \:  \: ∠BEC  \: =  \: 180° \:  - \:  ∠BDC}}} \\

{\bold{\sf{➠  \:  \:  \:  \:  \: 180° \:  -  \: 32°}}} \\

{\bold{\sf{➠ \:  \:  \:  \:  \: ∠BEC  \: = \:  148°}}} \\

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(iii) ∠BAC = ∠BDC (Angle of same segment are equal)

{\bold{\sf{➠ \:  \:  \:  \:  \: ∠BAC \:  =  \: 32°}}} \\

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