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

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

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

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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 $\angle ABO=\angle BAO=x^\circ$

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

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

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

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