Mathematics
In the given figure, AB is a diameter of a circle with centre O. If ADF and CBF are straight lines, meeting at F such that ∠BAD = 35° and ∠BFD = 25°, find :
(i) ∠DCB
(ii) ∠DBC
(iii) ∠BDC
Circles
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Answer
(i) We know that,
Angles in same segment are equal.
∠DCB = ∠BAD = 35°
Hence, the value of ∠DCB = 35°.
(ii) From figure,
∠ADB = 90° (∵ Angle in semicircle is 90°.)
∴ ∠BDF = 90°
In △DBF,
⇒ ∠DBF + ∠DFB + ∠BDF = 180° (∵ By angle sum triangle property).
⇒ ∠DBF = 180° - (∠DFB + ∠BDF)
⇒ ∠DBF = 180° - (90° + 25°)
⇒ ∠DBF = 65°.
⇒ ∠DBF + ∠DBC = 180° (Linear pairs)
⇒ ∠DBC = 180° - 65°
⇒ ∠DBC = 115°.
Hence, the value of ∠DBC = 115°.
(iii) In △DCB,
⇒ ∠DCB + ∠CBD + ∠BDC = 180° (∵ By angle sum triangle property).
⇒ ∠BDC = 180° - (∠DCB + ∠CBD)
⇒ ∠BDC = 180° - (35° + 115°)
⇒ ∠BDC = 180° - (35° + 115°)
⇒ ∠BDC = 30°.
Hence, the value of ∠BDC = 30°.
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