# We have a triangle $ABC$ and we must find the angle $ABO$

We have a triangle $$ABC$$ and some information related to:

So I don't know use which rule to find the angle $$ABO$$. Options are : $$30$$ or $$35$$ or $$40$$ or $$45$$ degrees.

• I am not too sure but is seems the answer should be $20^{\circ}$ – Mohammad Zuhair Khan Oct 25 '18 at 7:44
• @Raptor but there is not any options like what you say! – user602338 Oct 25 '18 at 7:45
• It must be 30 35 40 45 – user602338 Oct 25 '18 at 7:45
• $BH$ Isn't a staright line @Raptor – user602338 Oct 25 '18 at 7:48
• As per my calculator answer should be 40 – geeky me Oct 25 '18 at 8:05

First off, I would like to apologise for the low quality image.
As $$\angle OBC=\angle OCB,$$ we conclude that $$\triangle OBC$$ is isosceles, thus $$OB=OC=2$$.

Then we use the fact that $$\sin \theta= \frac {\text{opp}}{\text{hypo}}.$$ Thus $$\angle ACO= 30^\circ.$$

As $$AH=CH,$$ we conclude that $$\triangle OAC$$ is isosceles, thus $$\angle ACO=\angle CAO=30^\circ$$.

Using the fact that the angles in a triangle sum to $$180^\circ,$$ we find that $$\angle BOC=140^\circ$$ and $$\angle AOC=120^\circ.$$

$$\therefore \angle ABO=100^\circ$$

As $$\angle OAC=30^\circ$$ and $$OH=1, OA=2.$$

As both $$OB$$ and $$OA$$ are equal to $$2,$$ $$\triangle OAB$$ is isosceles and $$\angle ABO=\angle BAO=x^\circ$$

$$x+x+100=180 \implies x=40^\circ$$

$$\therefore\angle ABO=40^\circ$$

• Why did you assume that the BH is not a straight line? – NoChance Oct 25 '18 at 9:20
• OP's comments. If $BH$ was straight, then the answer would have been $20^\circ$ which is clearly not an option. – Mohammad Zuhair Khan Oct 25 '18 at 9:21
• Good work. Thanks. – NoChance Oct 25 '18 at 9:25
• Thanks :) $\$ $\$ – Mohammad Zuhair Khan Oct 25 '18 at 9:25
• @Raptor thanks a lot. You helped me with this! – user602338 Oct 25 '18 at 9:26