Find out the angle of <ABC

Question

Help me to solve it please.how could it be done.I tried but nothing comes out.Help me please

Answer 1

Hint: Let $\angle BAD = \alpha$, so by sine rule, $$\frac{\sin \alpha}{BD}=\frac{\sin 45^{\circ}}{AB}$$ and $$\frac{\sin (\alpha+ 15^{\circ})}{2BD}=\frac{\sin 30^{\circ}}{AB}$$ so $$\frac{2\sin \alpha}{\sin (\alpha+ 15^{\circ})}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}$$

After finding $\alpha$, you can then find $\angle ABC$

Answer 2

we draw BO$\perp$AC and we join OD

• see in $\bigtriangleup$BOC $\angle$BOC=90 and OD is midian. so OD=BD=DC
• NOW $\angle$OCD=30 so $\angle$OBD=60=$\angle$NOD[ where BO meets AD at N]
• now $\angle$NBD=60 and $\angle$NDB=45 so $\angle$BND=75
• $\angle$BND=75 and $\angle$NOD=60 so $\angle$ODN=15=$\angle$OAD. so AO=OD
• now $\angle$ODB=45+15=60. so $\bigtriangleup$BOD is equilateral. so OD=BO =BD AND AO=OD
• so we have AO=BO and we see $\angle$AOB=90 so $\angle$OAB=45
• we name $\angle$DAB=x so x+15 =45 so x=30
• in $\bigtriangleup$ABD $\angle$DAB=30 and $\angle$ADB=45 so $\angle$ABC=105

Answer 3

Do construction as shown WLOG, we can let BD = DC = 1.

Applying special angle relations to $\triangle AHC$, we have $h : h + 1 = 1 : \sqrt{3} \leftrightarrow h = \frac{\sqrt(3) +1} {2}$

Then calculate $h - 1$ for later use. $h – 1 = \frac {\sqrt(3) – 1} {2}$

In $\triangle AHB, AB^2 = h^2 + (h – 1)^2 \leftrightarrow AB = \sqrt 2$

Then, $\dfrac {AB}{BD} = \dfrac {\sqrt 2}{1}$ and $\dfrac {BC}{AB} = \dfrac {2}{\sqrt 2} = \dfrac {\sqrt 2}{1}$.

By ratio of 2 sides and the included angle, $\triangle ABC\sim\triangle DBA$.

Therefore, $\angle BAD = \angle BCA = 30^{\circ}$

Hence $\angle ABC = 105^{\circ}$

Note:- This figure is commonly used in calculating sin (15).