Question

please brainly answer this

Answer 1

Q : 70 # In triangle ABC, let D be the mid-point of BC . If $\angle ADB = 45 \degree$ and $\angle ACD = 30\degree$ ,determine $\angle BAD$ ?

Final Answer : 30°
( Option D)

Construction: Draw $AO \perp OC$ where CB produced to O with
BD = DC = a (say)
BO = b (say)

Steps:
$In\: \Delta AOD, \\ tan(45\degree ) = \frac{AO}{DO} \\ => 1 = \frac{AO}{DO}\\ => AO = DO \\ => AO = a + b$

$\Delta AOC \\ tan(30 \degree) = \frac{AO}{OC} \\ => \frac{1}{\sqrt{3}} = \frac{a+b}{2a+b} \\ => \frac{a}{b} = \frac{\sqrt{3}-1}{2-\sqrt{3}}$

$\Delta AOB, \\ tan(\theta) = \frac{AO}{OB} \\ => tan(\theta) = \frac{a+b}{b} \\ => tan(\theta) = \frac{a}{b} + 1 \\ => tan(\theta) = 2+\sqrt{3} \\ => \theta = 75\degree$

4) Now, By Exterior Angle Property,
$\angle ABO = \angle BAD + \angle ADB \\ => 75\degree= \angle BAD + 45\degree \\ => \angle BAD = 30\degree$

Answer 2

Go through the steps attached above ,

I hope this helps you.

: )