Answer:
Step-by-step explanation:
1)
The given angle is 57° at least I think it is . so we need to look at the triangle being formed inside the circle. the legs from the center to the edge are both the radius. so they are the same lengths. so that triangle is an isosceles. and the two angle at the circles perimeter are going to be both the same, b/c that's a rule of an isosceles triangle. so now we not both angles on the perimeter of the circle of that inner triangle are both 57° :) that's great, b/c now we can find the angle of the inner point of that triangle. it's going to be 180 = 57 + 57 + inner angle x
180 - 57 -57 = inner angle x
66 = x , so that inside angle on the isosceles is 66°
that's great to know, that angle is also part of the bigger triangle that goes outside the circle and I'm pretty sure the angle that is at the top of the triangle , at least in the picture, is a right angle or 90° sooo we now know two angle of the triangle and can figure our the "?" one. :)
180 = 66 + 90 + ?
180-66-90 = ?
24 = ?
so that angle at the point with the "?" is 24 ° :)
2)
The given angle is 24° :0 look above.. that last we we solved was also 24°
everything is the same in drawing 2, as in drawing 1. We can simply use all the things we know about drawing 1.. The inside angle at the center of the circle will be 66 ° and the two angle out by the perimeter of the circle will be 57° just like in drawing 1. They made that a little too easy :D Le'ts hope we get a few more like that :)
3)
I'm just sure why the drew the circle in this drawing. we don't need it to solve this triangle. Maybe you're not supposed to use this very common way to solve the triangle yet? but maybe it's good to introduce it anyway? since we are given the hypotenuse (long side) and the adjacent side (next to) the angle we can just use SOH CAH TOA to find any angle or side of this triangle. I'll put the useful remind of how trig functions ( sin, cos, ect.) fit on triangles and then we'll use them to find the parts we want. :)
Use SOH CAH TOA to recall how the trig functions fit on a triangle
SOH: Sin(Ф)= Opp / Hyp
CAH: Cos(Ф)= Adj / Hyp
TOA: Tan(Ф) = Opp / Adj
copy the above down. and put in in a handy place on your computer , for quick reference any time you have to solve a triangle. it's really super helpful :)
okay now look at the triangle in question 3) what do we know, and what do we want to find? we know the Hyp and Adj and one angle and want to find the Opposite (Opp) side from the known angle. sooo we have a choice of SOH or TOA.. let's use SOH,
SOH ⇒ Sin(Ф) = Opp / Hyp
Sin(53) = Opp/ 10
algebra skill engage :P
10* Sin(53) = Opp
7.9863551 = Opp
so that side with the question mark is 8.0 long. ( rounded to nearest 10th)
4)
On this triangle we are given the Opp and the Adj sides and an angle and are asked to find the Hyp
which trig function would you like to use?
Let's use CAH this time. We could also use SOH , but we just used that one. let's try out the "tools" of the trade :)
CAH ⇒ Cos(Ф) = Adj / Hyp
Cos(37) = 8/Hyp
use algebra skillz
Hyp = 8 / Cos(37)
use calculator to find Cos(37) :/
Hyp = 10.01708
? = 10.0 ( rounded to nearest 10th )
5)
oh for this one we do need the circle, and I don't know why they want people to know this , nobody uses this math :? Anyway,
there is a confusing formula for this type of problem
∠ = (bigger arc - smaller arc ) / 2
since we're given the 65° , we know that 2*65= 130 which is the angle of the bigger arc - smaller arc = 130 °
now we have to use our really keen algebra skillz to see that
if the total all the way around a circle is 360 and we have 130 of those left so
eq. 1) 360 = big arc + small arc
then
360-smal arc = big arc
and
eq. 2) 130 = big arc - small arc
plug in for "big arc" what we know from above :P
130 = 360-small arc - small arc
130 = 360 - 2small arc
130 + 2 small arc = 360
2 small arc = 230
small arc = 115
:0 wow.. we solved it.. that was a lot of fancy algebra huh :P
? = 115° ( the small arc)
6)
same as 5
given 72°
72*2=144°
eq. 1) 360 = big arc + small arc
eq. 2 ) 144 = big arc - small arc
solve for big arc this time
360-bigl arc = small arc
now plug into eq. 2
144 = big arc - (360-big arc)
144 = big arc - 360 + big arc
144 = 2big arc - 360
144+ 360 = 2 big arc
504 = 2 big arc
252° = big arc
yay solved !! phew.. that was a lot of algebra.. :P
5 and 6 are tough, ask the teacher how many others got it :)
