# calculate the angle

• Apr 18th 2010, 03:17 PM
slapmaxwell1
calculate the angle
ok well when you look at the attachment you can see the diagram i am supposed to be using to figure out the answer to the questions. i dont know where to start i havent had geometry since high school, and to be honest we dont use much of it in calculus. so anyways im stuck on #16 and #15 on the handout. if arc HI = 80 degrees and arc JI = 75 what is the measure of angle HKI? see my first question is why did they break up the arcs? if you add them together wouldnt arc HJ = 155 degrees?Attachment 16443 my next problem im having is that the center of the circle is Q but the center of the inscribed box is K, i found some identities but they are using the center of the circle as reference, which means i cant use them? any help that could be provided would be greatly appreciated..thanks in advance...
• Apr 18th 2010, 03:53 PM
masters
Quote:

Originally Posted by slapmaxwell1
ok well when you look at the attachment you can see the diagram i am supposed to be using to figure out the answer to the questions. i dont know where to start i havent had geometry since high school, and to be honest we dont use much of it in calculus. so anyways im stuck on #16 and #15 on the handout. if arc HI = 80 degrees and arc JI = 75 what is the measure of angle HKI? see my first question is why did they break up the arcs? if you add them together wouldnt arc HJ = 155 degrees?Attachment 16443 my next problem im having is that the center of the circle is Q but the center of the inscribed box is K, i found some identities but they are using the center of the circle as reference, which means i cant use them? any help that could be provided would be greatly appreciated..thanks in advance...

Hi slapmaxwell1,

[16]

\$\displaystyle \widehat{HG}=80\$

\$\displaystyle \widehat{JI}=75\$

As far as I can tell here, \$\displaystyle \widehat{HG}\$ is not needed.

Angle JHI is an inscribed angle = one-half of arc JI. \$\displaystyle \angle JHI=37.5^{\circ}\$

\$\displaystyle \widehat{HG}=75\$

Angle GIH is an inscribed angle = one-half of arc GH. \$\displaystyle \angle GIH =37.5^{\circ}\$

This means that triangle HIK is isosceles. The vertex angle HKI that your are looking for = 180 - 75 = 105 degrees.

[15]

\$\displaystyle \widehat{HL}=40^{\circ}\$

Calculate \$\displaystyle \widehat{IO}\$

A diameter perpendicular to a chord bisects the chord and its subtended arc.

That means \$\displaystyle \widehat{HL}=\widehat{LI}=40^{\circ}\$

Now find \$\displaystyle \widehat{IO}\$ by subtracting this from 180.

• Apr 18th 2010, 04:51 PM
slapmaxwell1
ok you wouldnt happen to have the formulas? or some info i can read to help me through these problems? i mean in case you get tied up or busy? :O) i wish someone could go through all the math books and jus write down or summarize the formulas and explain how they work, that would be a very valuable book. thanks for your help, im going to print this out and study it right now.
• Apr 25th 2010, 08:04 AM
slapmaxwell1
thx again for your help, i really appreciate you taking the time to answer my question. i did get ahold of a geometry book with a list of theorems? i wish it had pictures to better illustrate the properties...lol