MHB State the angles in terms of x , Circle theorems

[mathlearn]

Problem


View attachment 6048

O is the center of the circle and AB is the diameter of this circle , C & D are points on this circle , If $\angle CDB=x^\circ$ ,State the following angles in terms of $x$

$\angle CAB$

$\angle CBA$

Workings & what is known

$OC=OB=OA$ radii of the same circle

$\therefore \angle {CBO} = \angle {OCB} = \angle {OAC} = \angle {OCA}$

$ACB=90^\circ$ Thales theorem

As $\angle BCO = \angle ACO $ it can be said that $\angle ACB$ is bisected

Many Thanks :)

 

[MarkFL]

We are given that the quadrilateral $ABDC$ is cyclic, therefore:

$$x^{\circ}+\angle{CAB}=180^{\circ}$$

You are correct that $\angle{ACB}=90^{\circ}$

And so use:

$$\angle{CBA}+\angle{CAB}+\angle{ACB}=180^{\circ}$$

 

[mathlearn]

We are given that the quadrilateral $ABDC$ is cyclic, therefore:

$$x^{\circ}+\angle{CAB}=180^{\circ}$$

You are correct that $\angle{ACB}=90^{\circ}$

And so use:

$$\angle{CBA}+\angle{CAB}+\angle{ACB}=180^{\circ}$$

Thanks :)

$\angle CAB+x=180$
$\angle CAB =180-x $

That's for $\angle CAB$ , Now For $\angle CBA$ as I think it is equal to $\angle CAO$ as in the figure $\angle CBA= 180-x$

Correct?

Many Thanks :)

 

[MarkFL]

Using the sum of the interior angles of a triangle, I wrote:

$$\angle{CBA}+\angle{CAB}+\angle{ACB}=180^{\circ}$$

Now substitute:

$$\angle{CBA}+\left(180-x\right)^{\circ}+90^{\circ}=180^{\circ}$$

What do you get?

 

[mathlearn]

Using the sum of the interior angles of a triangle, I wrote:

$$\angle{CBA}+\angle{CAB}+\angle{ACB}=180^{\circ}$$

Now substitute:

$$\angle{CBA}+\left(180-x\right)^{\circ}+90^{\circ}=180^{\circ}$$

What do you get?

Thanks :D , I get by substituting,

$\displaystyle \angle{CBA}+\left(180-x\right)^{\circ}+90^{\circ}=180^{\circ}$

$\displaystyle \angle{CBA}=x^{\circ} -90^{\circ}$

Correct ? :)

Many Thanks :)

 

[MarkFL]

Yes, that's what I got as well. (Yes)

 

[mathlearn]

Yes, that's what I got as well. (Yes)

Thank you very much MarkFL (Happy) (Smile) (Party)