Calculate angle AEC and ADC

[chwala]

Homework Statement

See attached

Relevant Equations

Deductive Geometry- Alternate segment theorem

Question;

Text Solution;

My reasoning;
$∠ABC= 180^0 -(32^0+60^0)$
=$88^0$

$∠ABC=∠ACF=88^0$ (Alternate segment theorem).

$∠EBC=92^0$ i.e angles lying on a straight line and $∠BCE=180^0 -(88^0+60^0)=32^0$ therefore;
$∠BEC=180^0 -(32^0+92^0)=56^0$
$∠ECA=∠ADC=32^0+60^0=92^0$ (by Alternate segment theorem).

Cheers...there may be another way of looking at this...

 

[Lnewqban]

Regarding $∠EBC=92^0$:
Doesn't the problem figure contradicts the fact that an angle inscribed across a circle's diameter is always a right angle?

 

[chwala]

Regarding $∠EBC=92^0$:
Doesn't the problem figure contradicts the fact that an angle inscribed across a circle's diameter is always a right angle?

...but we are not told if $AC$ is the diameter of the circle. We have only been given an indication of the tangent line, that is line $ECF$.

 

[Heisenberg x Shrek]

Regarding $∠EBC=92^0$:
Doesn't the problem figure contradicts the fact that an angle inscribed across a circle's diameter is always a right angle?

AC isn't a diameter, that was never mentioned

 

[SammyS]

Homework Statement:: See attached
Relevant Equations:: Deductive Geometry- Alternate segment theorem

Question;

View attachment 315298

Text Solution;

View attachment 315299

My reasoning;
$∠ABC= 180^0 -(32^0+60^0)$
=$88^0$

$∠ABC=∠ACF=88^0$ (Alternate segment theorem).

$∠EBC=92^0$ i.e angles lying on a straight line and $∠BCE=180^0 -(88^0+60^0)=32^0$ therefore;
$∠BEC=180^0 -(32^0+92^0)=56^0$
$∠ECA=∠ADC=32^0+60^0=92^0$ (by Alternate segment theorem).

Cheers...there may be another way of looking at this...

You are correct that angles $∠ACF=88^\circ$ and $∠EBC=92^\circ$ .
However, your reasoning leading to determining angle $∠BCE$ is faulty, because you do not know a value for angle $∠ACF$ either.

An angle you can easily determine is $∠ADC$. (There's a cyclic quadrilateral involved.)

 

[chwala]

You are correct that angles $∠ACF=88^\circ$ and $∠EBC=92^\circ$ .
However, your reasoning leading to determining angle $∠BCE$ is faulty, because you do not know a value for angle $∠ACF$ either.

An angle you can easily determine is $∠ADC$. (There's a cyclic quadrilateral involved.)

@sammy but we know that $∠ECA=∠ADC$ using the alternate segment theorem. Angle $BCE=32^0.$ I do not need $∠ACF$ to determine this.

 

[chwala]

...Just thinking is it possible to determine all the angles in the given diagram? my thinking is as shown on the diagram below;

My reasoning being $∠EBC$ is similar to $∠ADC$ they have a common angle i.e $92^0$.

$∠CAD=DCF=BEC=56^0$

$∠ACD=88^0 - ∠DCF=32^0$

 

[SammyS]

@sammy but we know that $∠ECA=∠ADC$ using the alternate segment theorem. Angle $BCE=32^0.$ I do not need $∠ACF$ to determine this.

Apologies !

I overlooked your reasoning establishing that $∠ABC=∠ACF$ .

 

[SammyS]

...Just thinking is it possible to determine all the angles in the given diagram? my thinking is as shown on the diagram below;

My reasoning being $∠EBC$ is similar to $∠ADC$ they have a common angle i.e $92^0$.

$∠CAD=DCF=BEC=56^0$

$∠ACD=88^0 - ∠DCF=32^0$

Yes, $∠CAD=∠DCF$ .

But no, you can not determine $∠ACD$ , $∠CAD$ , nor $∠DCF$ .