In the figure, [tex] \angle[/tex] AOB = [tex] \frac{1}{2} [/tex][tex] \angle[/tex] BOC and [tex] \angle[/tex] BOC = [tex] \frac{2}{3} [/tex][tex] \angle[/tex] COD . Show that [tex] \angle[/tex] COD = [tex] \frac{1}{2} \angle[/tex] AOD mentioning necessary axioms.

~Thanks in advance !

Question

13-02-2021
Mathematics

contestada

In the figure, [tex] \angle[/tex] AOB = [tex] \frac{1}{2} [/tex][tex] \angle[/tex] BOC and [tex] \angle[/tex] BOC = [tex] \frac{2}{3} [/tex][tex] \angle[/tex] COD . Show that [tex] \angle[/tex] COD = [tex] \frac{1}{2} \angle[/tex] AOD mentioning necessary axioms.

~Thanks in advance !

$In the figure tex angletex AOB tex frac12 textex angletex BOC and tex angletex BOC tex frac23 textex angletex COD Show that tex angletex COD tex frac12 angletex class=$

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xKelvin xKelvin · Answer 1 · 2021-02-13T20:00:49+01:00

Answer:

See Below.

Step-by-step explanation:

Statements: Reasons:

[tex]1)\text{ }\angle AOD=180[/tex] Straight Angle Theorem

[tex]2)\text{ } \angle AOB+\angle BOC+\angle COD=\angle AOD[/tex] Angle Addition

[tex]3)\text{ } \angle AOB+\angle BOC+\angle COD=180[/tex] Substitution

[tex]\displaystyle 4)\text{ } \angle AOB=\frac{1}{2}\angle BOC[/tex] Given

[tex]\displaystyle 5)\text{ } \angle BOC=\frac{2}{3}\angle COD[/tex] Given

[tex]\displaystyle 6)\text{ } \frac{1}{2}\angle BOC+\frac{2}{3}\angle COD+\angle COD=180[/tex] Substitution

[tex]\displaystyle 7)\text{ } \frac{1}{2}\Big(\frac{2}{3}\angle COD\Big)+\frac{2}{3}\angle COD+\angle COD=180[/tex] Substitution

[tex]\displaystyle 8)\text{ } \frac{1}{3}\angle COD+\frac{2}{3}\angle COD+\angle COD=180[/tex] Multiplication

[tex]\displaystyle 9)\text{ } 2\angle COD=180[/tex] Addition

[tex]10)\text{ } \angle COD=90[/tex] Division Property of Equality

[tex]\displaystyle 11)\text{ } \frac{1}{2}\angle AOD=90[/tex] Division Property of Equality

[tex]\displaystyle 12)\text{ } \angle COD=\frac{1}{2}\angle AOD[/tex] Substitution

Аноним Аноним · Answer 2 · 2021-02-20T16:31:59+01:00

Аноним Аноним

20-02-2021

Answer:

this is your answer. good luck.

Answer Link

In the figure, [tex] \angle[/tex] AOB = [tex] \frac{1}{2} [/tex][tex] \angle[/tex] BOC and [tex] \angle[/tex] BOC = [tex] \frac{2}{3} [/tex][tex] \angle[/tex] COD . Show that [tex] \angle[/tex] COD = [tex] \frac{1}{2} \angle[/tex] AOD mentioning necessary axioms.~Thanks in advance !

Respuesta :

Otras preguntas

In the figure, [tex] \angle[/tex] AOB = [tex] \frac{1}{2} [/tex][tex] \angle[/tex] BOC and [tex] \angle[/tex] BOC = [tex] \frac{2}{3} [/tex][tex] \angle[/tex] COD . Show that [tex] \angle[/tex] COD = [tex] \frac{1}{2} \angle[/tex] AOD mentioning necessary axioms.

~Thanks in advance !