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In a DABC, let D be the mid-point of BC. If ∠ADB = 45º and ∠ACD = 30º. then ∠BAD equals.

(a) 45º 

(b) 60º 

(c) 30º 

(d) 15º

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(c) 30º.

Draw BP ⊥ AC and join P to D. 

In ΔBPC, 

∠PBC = 180º – (∠BPC + ∠BCP) 

= 180º – (90º + 30º) = 180º – 120º = 60º

In ΔBPC, 

sin 30º = \(\frac{BP}{BC}\) ⇒ \(\frac{BP}{BC}\) = \(\frac{1}{2}\) ⇒ BP = \(\frac{1}{2}\) BC = BD

Now in ΔBPD, BP = BD ⇒ ∠BDP = ∠PBD = 60º 

⇒ ∠BPD = 60º ⇒ ΔBPD is equilateral 

∴ PB = PD and ∠ADP = 60º – 45º = 15º 

In ΔADC, ext. ∠ADB = ∠ACD + ∠DAC 

45º = 30º + ∠DAC ⇒ ∠DAC = 15º 

∴ In ΔAPD, ∠ADP = ∠PAD ⇒ PD = PA 

We have PD = PA = PB 

⇒ P is the circumcentre of ΔADB as circumcentre is equidistant from the vertices of a Δ.

⇒ BAD = \(\frac{1}{2}\) ∠BPD

( Angle subtended by an are at the centre of a circle is half the angle subtended by the same are at any other point on the remaining part of the circumference.)

⇒ BAD = \(\frac{1}{2}\)x 60º = 30º.

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