10
$\begingroup$

A Circle

$$\angle(ABC) = 30°\\ \angle(BCO) = 20°\\ \angle(OCD) = 20°$$

How do i find $\angle(ODC)$? so i wanted to show my teacher this but he is not available yet. Can someone help me to solve? geometry problems on circle seems hard to me. Thanks!

$\endgroup$
6
  • $\begingroup$ Just a hint: $OC=OB$. Not a solution but a good point to start. $\endgroup$
    – N74
    May 27 '17 at 11:25
  • $\begingroup$ I can't solve this problem, needs help from someone. $\endgroup$
    – Physicer
    May 27 '17 at 11:36
  • $\begingroup$ What methods are you expected to use. The sine rule, perhaps? $\endgroup$ May 27 '17 at 11:37
  • $\begingroup$ I think sine rule better so i use sine rule. $\endgroup$
    – Physicer
    May 27 '17 at 11:42
  • $\begingroup$ Anyone helps me? $\endgroup$
    – Physicer
    May 27 '17 at 11:44
4
$\begingroup$

Without loss of generality, we can consider $R=OC=OB=1$.

From $\Delta OBC: \frac{OB}{\sin20}=\frac{BC}{\sin140}\Rightarrow BC=\frac{\sin140}{\sin20}=2\cos20.$

From $\Delta BCD: \frac{BC}{\sin110}=\frac{CD}{\sin30}\Rightarrow CD=\frac{\cos20}{\sin110}=1.$

Finally from $\Delta OCD: \frac{OC}{\sin{x}}=\frac{CD}{\sin{(x+20)}}\Rightarrow \sin x=\sin{(x+20)}\Rightarrow x=80.$

$\endgroup$
4
  • $\begingroup$ where does the 140 comes from ? $\endgroup$
    – zwim
    May 27 '17 at 12:16
  • $\begingroup$ it is the angle opposite to the side BC and $\Delta OBC$ is isosceles... $\endgroup$
    – farruhota
    May 27 '17 at 12:20
  • 1
    $\begingroup$ OK. From $CD=1$ you can conclude the same way then, $x=(180-20)/2$ in isocele CDO, simpler than last line. $\endgroup$
    – zwim
    May 27 '17 at 12:26
  • $\begingroup$ @zwim (+1) Thank you. I just wanted to keep the rhyme. $\endgroup$
    – farruhota
    May 27 '17 at 12:34
3
$\begingroup$

Join OA and AC

angle AOC = 2xangle ABC=60 deg (center angle and circumference angle)

OC=OA (radii)

Triangle OAC is equilateral

AC=OA=OC

angle CAB =180-30-80=70 deg

angle CDA=40+30=70 deg

Therefore CA=CD=CO

ANGLE ODC=angle BAC+angle ABC

Angle ODC=(180-100)=80°
● ● ● ANGLE ODC IS 80°

$\endgroup$
0
$\begingroup$

Hint...with a bit of angle-chasing you should be able to establish $\angle ADC=70$.

You can then use the sine rule in triangles $ODB$ and $ODC$ (assume the radius is $1$)

The final answer seems to be $x=80$ which would suggest there must be a better way.

$\endgroup$
2
  • $\begingroup$ Ah, I didn't understand it. can you use some numerical? because i don't un derstand knowns $\endgroup$
    – Physicer
    May 27 '17 at 11:47
  • $\begingroup$ Can you get $\angle ADC=70$? You also need to employ the sine rule and also the compound angle formula for $\sin(A+B)$. On the pther hand, as I said, there may be a better way because the answer is exactly $80$... $\endgroup$ May 27 '17 at 12:03
0
$\begingroup$

$$/angle(ABC)=30°// /angle(AOC)=2*angle(ABC)=60°//$$ So AOC is equilateral triangle (AO=AC=CO=R) $$/angle(CAB)=/angle(COB)/2=70° //angle(ACD)=70°$$ So CD=CB=R and CDO is isosceles triangle so $$/Angle(CDO)=angle(DOC) Angle(CDO)=90°-Angle(DCO)/2=160°/2=80°$$

$\endgroup$
0
$\begingroup$

Produce CD and meet the circle at E Join OA and AC angle APC = 60 deg (centre and circumference angles) OA = AC = OC Join BE angle CAB = angle CEB = 70 deg (CE = CB and angle BCE = 40deg) Therefore AD = CD = CD x = 80 deg

$\endgroup$
2
  • $\begingroup$ @JoséCarlosSantos: Let us not exaggerate: whay would $AB$ be better than AB? MathJax is meant for complicated formulae that would be ambiguous or difficult to read as plain text. Let us not transform it into an obligation. $\endgroup$
    – Alex M.
    Nov 8 '17 at 14:38
  • 1
    $\begingroup$ @AlexM. I think that $\angle CAB=\angle CEB=70^\circ$ is more pleasent to read than angle CAB = angle CEB = 70 deg. $\endgroup$ Nov 8 '17 at 14:42
-1
$\begingroup$

BCD is equal to 40 (OCD plus BCO). So the unknown corner of the triangle is 180 - (40 + 30) which is 110. ADC, CDB and ODB are supplementary, so ADC is 70. COB is 180 - (20 + 15) which is 155. I'm pretty sure DOB is equal to COB, so 155 + 155 equals 310 therefore COD should be 50? 180 - (20 + 50) is 110 so x is 110. I'm pretty sure that's the answer, however I'm only a Year Seven, that happens to love Geometry and I've just started using this. Basically I'm saying, don't think my word is final. Hope it helps anyways. :) please tell me if I'm not right, I would like to know to correct answer if I've gotten it wrong.

$\endgroup$
4
  • $\begingroup$ It's wrong, it seems $x = 80$ on my answerkey $\endgroup$
    – Physicer
    May 27 '17 at 11:42
  • $\begingroup$ As I said, I'm only in Year Seven. Thanks for letting me know though! $\endgroup$
    – Violet
    May 27 '17 at 11:46
  • $\begingroup$ its dw! thank you again. $\endgroup$
    – Physicer
    May 27 '17 at 11:46
  • 1
    $\begingroup$ The segment $OB$ doesn't bisect $\hat B$ so $COB$ is not $180-(20+15)$. Anyway the triangle $BCO$ is isosceles, so $OBC=OCB$ $\endgroup$
    – N74
    May 27 '17 at 11:47
-2
$\begingroup$

First join $AC$ and $AO$. Now, $\angle ABC$ is $30^\circ$,so AOC is 60*.AO=OC.so,OAC=OCA=(180*-60*)÷2=60*. Triangle AOC is equilateral Triangle.so,AC=OC=OA.angleACD=(ACO-DCO)=(60*-20*)=40*.In triangle ACB ,CAB=70*,ACD=40*.So,ADC=70*.So,AC=CD. Now AC=CO=CD,So,OC=CD.in triangle CDO,DCO=20*,and OC=CD.So,CDO=COD=(180*-20*)÷2=80*.HENCE,CDO=80*.(ANS.).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.