Question

In triangle abc d is the mid point of bc if angle adb =45 and angle acd =30 then what is the measure of angle bad in degree

Answer 1

$\huge\boxed{Answer\:=\:15\:°}$

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$\huge\underline\mathfrak\green{Explanation}$

★Given : In ∆ABC, D is the mid point of BC.

Angle ADB = 45°

Angle ACD = 30°

★To find : Measure of angle BAD.

★Proof : In ∆ADC,

Angle ADB is the exterior angle.

Therefore, by using exterior angle property,

Angle DAC + Angle ACD = Angle ADB

Angle DAC + 30° = 45°

So, Angle DAC = 45° - 30° = 15°

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Now,

Angle BAD = Angle DAC ( ∆s are congruent )

Therefore, Angle BAD = 15° ( required answer )

Answer 2

SOLUTION

[In Attachment]

Draw BL perpendicular to AC & join L to D.

=)∠BLC= 90°

In ∆BLC,

∠BLC+ ∠BCL+ ∠CBL= 180°

=) 90° + 30°+ ∠CBL= 180°

=) ∠CBL= 180°-120

=) ∠CBL= 60°

In ∆BLC,

$sin30 \degree = \frac{BL}{BC} \\ \\ = > \frac{1}{2} = \frac{BL}{BC} \\ \\ = > BL= \frac{BC}{2} \\ \\ = > BL = BD \: \: \: \: \: [D\: is \: the \:midpoint ]\\ \\ Hence \: in \: \triangle \: BLD \: we \: have \: BL= BD \\ = > \angle \:BDL= \angle \: BLD \\ and \: DBL= 60 \degree$

Hence,

∠BDL + ∠BDL+∠DBL= 180°

=) 2∠BDL+60°= 180°

=) 2∠BDL= 180°-60°

=) ∠BDL= 120/2

=) ∠BDL= 60°

∠BLD= ∠DBL

Hence,

∆BLD is an equilateral ∆.

=) LB= LD........(1)

∠BDL= ∠ADL+ ∠ADB

=) 60°= ∠ADL+ 45°

=) ∠ADL= 60°-45°

=) ∠ADL= 15°

In ∆ADC,

Exterior angle should be equal to sum of opposite interior Angles.

=) 45°= 30° + ∠CAD

=) ∠CAD= 15°

Hence,

In ∆DAL,

=)∠ADL= ∠DAL= 15°

=)LD= LA..........(2)

By (1) & (2), we get

LD= LA=LB

Hence L is circumcentre of ∆BDA & angels subtended by arc of circle at the centre is twice the angle subtended at circumference.

=)∠BAD= 1/2× ∠BLD

=) ∠BAD= 1/2× 60°

=) ∠BAD= 30° (Answer)

HOPE it helps ☺️