# 3 Questions on Circle

• Dec 31st 2005, 08:49 AM
ling_c_0202
3 Questions on Circle
In circle1.jpg,

DB is a diameter of the circle, AC = BC and ∠DAC = 23∘ .
Find ∠ABD.

In circle2.jpg,

AD is the diameter of the circle
If AB = BC = CD, find x + y + z.

In circle3.jpg,

In the figure XAB and XDC are straight lines. If DX = 5, AX = 6 and AB = A, Find CD

Can anyone give me some idea please? :confused:

HAPPY NEW YEAR! ;)
• Dec 31st 2005, 09:47 AM
CaptainBlack
Quote:

Originally Posted by ling_c_0202
In circle1.jpg,

DB is a diameter of the circle, AC = BC and ∠DAC = 23∘ .
Find ∠ABD.

$\triangle ABC$ is isosceles, so:

$\angle CAB=\angle CBA$.

As $AB$ is a diameter:

$\angle DAB=90^\circ$.

So:

$\angle CAB=90-23=67^\circ$.

Hence:

$\angle ACB=180-2 \times \angle CAB = 46^\circ$

Now:

$\angle ACB=\angle ADB$,

as they are angles subtended by a cord to a circle from points on the
circumference of the circle.

Now look at the angle sum of $\triangle ABD$:

$180^\circ = \angle ABC+\angle ADB + \angle DAB$,

or:

$\angle ABC=180 - 46-90=44^\circ$

RonL
• Dec 31st 2005, 09:14 PM
ticbol
Quote:

Originally Posted by ling_c_0202
In circle1.jpg,
DB is a diameter of the circle, AC = BC and ∠DAC = 23∘ .
Find ∠ABD.

In circle2.jpg,
AD is the diameter of the circle
If AB = BC = CD, find x + y + z.

In circle3.jpg,
In the figure XAB and XDC are straight lines. If DX = 5, AX = 6 and AB = A, Find CD

Can anyone give me some idea please? :confused:

HAPPY NEW YEAR! ;)

Happy New Year to you too!

Lots of time for me to answer your question re circles. However, maybe next time, try to ask only one problem per posting. Here, in this posting, you have 3 problems. Better is if you made 3 postings for the 3 problems. There are some of us, me, for example, who do not like to give partial answers, so if one of the many problems cannot be answered yet with the rest, we will not post the partial answer. If the many problems were separated into their own postings, then we can just answer those that we can answer and leave those that we cannot answer yet.

The 3 problems are exercises about angles made by two straight lines (secants, chords, diameters) piercing a circle. I don't know now the exact wordings, but in a way they mean these:
----An angle made by two secants, or lines from an external point, is measured by half the difference of the two intercepted arcs. ----(1)
----So if the external point is moved to a point on the circumference, the angle made is called an inscribed angle and it's measure is half of the intercepted arc. ---(2)
---inscribed angles subtended by the same arc are congruent. ------(3)
---a diameter divides the circumference into two equal parts or two semi-circles, so a semi-circle is 180deg in measure. An inscribed angle inside a semi-circle, or an inscribed angle that is subtended by a diameter is 90deg in measure. -----(4)

------------
Problem #1, In circle1.jpg:

>>>> DB is a diameter of the circle, AC = BC and ∠DAC = 23∘ .
Find ∠ABD. <<<<

angle DAB = 90deg ----see (4) above.
So, angle CAB = 90 -23 = 67deg -----------(i)

angle DAC = angle DBC = 23deg ---------see (3) above.

AC=BC, so triangle ACB is isosceles. Then, angle CAB = angle CBA = 67deg.

Therefore,
angle ABD = (angle CBA) minus (angle DBC) = 67 -23 = 44deg ----answer.

----------------
Problem #2, in circle2.jpg:

>>>> AD is the diameter of the circle
If AB = BC = CD, find x + y + z. <<<<<

arc ABCD is a semi-circle. Chords AB, BC, and CD are of same lengths. So, arcs AB, BC, and CD have the same lengths also. Hence, arc AB = arc BC = arc CD = 180/3 = 60deg each.
Call the center of the circle, point O. Draw the radii OB, OC, OD.
Central angles AOB, BOC, and COD are subtended by equal arcs of 60deg each, so, the 3 central angles are 60deg each also. [A central angle and its intercepted arc are equal in measures.]

Concentrate on triangle AOB:
AO=OB=radii, so triangle AOB is isosceles. Then, angle BAO = angle ABO = (180-60)/2 = 60deg each also. That means triangle AOB is equiangular.

That means also that:
--- triangles AOB, BOC, and COD are all equiangulars.
--- angle ABC = angle BCD = 60+60 = 120deg each.

Angle AED = 90deg --------inscribed angle subtended by a diameter.

In polygon ABCDE:
---it has 5 sides, so the sum of its interior angles is (5-2)(180) = 540deg.
---angle y = angle ABC = 120deg
So,
x +y +(angle BCD) +z +(angle DEA) = 540
Substitutions,
x +y +120 +z +90 = 540
x +y +z = 540 -120 -90 = 330deg --------------------answer.

--------------------
Problem #3, in circle3.jpg:

>>>> In the figure XAB and XDC are straight lines. If DX = 5, AX = 6 and AB = A, Find CD. <<<<

Draw chords AC and BD.
Let arc AD = 2u, in degrees
And arc BC = 2v, in degrees

angle ABD = angle ACD -----------both are subtended by the same arc AD.
So, angle ABD = angle ACD = (1/2)(2u) = u degrees.
Mark those on the figure.

angle BAC = angle BDC -------------subtended by arc BC.
So, angle BAC = angle BDC = v degrees.
Mark those on the figure.

angle BXC = angle CXB -----------Identity, or the same thing.
Let w = angle BXC = angle CXB, in degrees.
Mark that on the figure.

----I am trying to prove here that triangles BDX and CAX are similar----

angle XAB = 180deg
So, angle XAC = (180-v) deg
Hence, the interior angles of triangle CAX are: u, (180-v), and w.

angle XDC = 180deg
So, angle XDB = (180-v) deg
Hence, the interior angles of triangle BDX are: u, (180-v), and w also.

That means triangles CAX and BDX are similar---and hence, proportional.
(XD)/(XB) = (XA)/(XC)
Substitutions,
(5)/(6+4) = (6)/(5 +DC)
Cross multiply,
5*(5 +DC) = (10)*6
(5 +DC) = 60/5 = 12
DC = 12 -5 = 7 -----------------answer.

--------------------
If this answer is not long for you, then submit 4 problems per posting.
• Jan 1st 2006, 05:12 AM
ling_c_0202
thank you for your help. I understand how to solve it now, obviously I missed some conditions when I thought.

sorry, I will ask only a question next time.
Thank you very much for you reminder. :p