1. ## 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?IMG.pdf 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...

2. 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?IMG.pdf 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]

$\widehat{HG}=80$

$\widehat{JI}=75$

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

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

$\widehat{HG}=75$

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

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

[15]

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

Calculate $\widehat{IO}$

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

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

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

Answer: $140^{\circ}$

3. 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.

4. 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