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is that a 9? or a 4?
here's the template again |dw:1362774782473:dw| I would add this to your notes if you don't have it already
notice how the number in front of the sqrt(3) on the longer leg is 9 so that must mean that x = 9
how did you get 30 and 2x?
well the given angle is 60, which adds to the right angle 90 to get 90+60 = 150 the remaining angle must be 30 in order for the three to add to 180
since 150 + 30 = 180
as for the 2x... the hypotenuse is always twice as long as the shorter leg in a 30-60-90 triangle
so that explains the x for the shorter leg and 2x for the hypotenuse
should I write that process down?
yes I would if you're not familiar with it
would it be a good idea to write it on my homework?
if you want to, there's nothing wrong with using this template
you can draw the template off to the side and show that x must be 9
when you match up the terms
why did you get rid of the 9?
what do you mean?
it was 9 sqrt 3? now, it's X sqrt 3?
no, I said x was 9 so x = 9
oh i see what you mean
you said in the picture above..
9 * sqrt(3) ... what we're given x * sqrt(3)... template
notice how the x and the 9 match up
so that's how I got that x = 9
the y and the 2x match up, so y = 2x y = 2*9 y = 18
you are so confusing me. Can you write it out like, step 1, step 2. And start from the beginning cause I'm lost
did you see how I got x = 9?
here is your triangle (on top) and the general template triangle (on bottom) |dw:1362776258556:dw|
notice how these two sides match up |dw:1362776369974:dw|
can we not use the template? it's confusing
have you learned trig yet?
can you just show me the process of how to solve this?
I'm just curious because if you have, then we can use that
I just wanna solve it..
I know, but I need to know if you know trig so it will make sense if I use it
have you learned things like sine, cosine, tangent, etc?
so you've seen things like sine = opposite/hypotenuse ?
I'll assume yes
tan(30) = opposite/adjacent tan(30) = x/(9*sqrt(3)) sqrt(3)/3 = x/(9*sqrt(3)) ... use a calculator or the unit circle here sqrt(3)*9*sqrt(3) = 3x 9*sqrt(3)*sqrt(3) = 3x 9*3 = 3x 27 = 3x 3x = 27 x = 27/3 x = 9 so you can see that we get the same answer as we did before
sure this method is a bit more complicated, but it works it all hinges on whether you know trig or not
do I write that down?
yeah, or you could skip a few steps I guess
wherever you see a 'sqrt', use the square root symbol of course
Should I write this down and skip all the other steps? :) sqrt(3)*9*sqrt(3) = 3x 9*sqrt(3)*sqrt(3) = 3x 9*3 = 3x 27 = 3x 3x = 27 x = 27/3 x = 9
I would have the previous steps in there, these ones tan(30) = opposite/adjacent tan(30) = x/(9*sqrt(3)) sqrt(3)/3 = x/(9*sqrt(3)) ------------------------------------------------------- I guess you could skip step 2 to get just these steps tan(30) = opposite/adjacent sqrt(3)/3 = x/(9*sqrt(3))
I'm confused. Is this 3 times 9? sqrt(3)*9*sqrt(3) = 3x
yeah it's just a complicated way of saying 9*3
so it's actually 9 times 3 sqrt 3= 3x
no the sqrt(3) terms multiply to 3 so that's why 9*sqrt(3)*sqrt(3) = 3x turns into 9*3 = 3x
I got the x. What's the Y?
well y = 2x so y = 2*9 = 18
so y = 18
you can use the pythagorean theorem to confirm, but you'd get the same answer (just in a more complicated way)
how did you find that Y= 2x?
well you would use the template...but just remembered you don't like that template very much
so you can use a^2 + b^2 = c^2 to find the value of y
a^2 + b^2 = c^2 9^2 + (9*sqrt(3))^2 = y^2 81 + 243 = y^2 keep going to solve for y
okay, 2nd problem! :p
Recreation. A skateboarding ramp is 12 inches high and rises at an angle of 17 degrees. How long is the base of the ramp? Round to the nearest inch.
tan(17) = 12/x x*tan(17) = 12 x = 12/tan(17) x = ???
do you not know X?
not until you solve for it
by typing 12/tan(17) into the calculator
yeah it's not too bad to find
so, that's the answer?
yes it is
Last question. I was wrong btw. There's only 3 questions left, not 4. :P
ok works for us lol
The lengths of the diagonals of a rhombus are in 2 inches and 5 inches. Find the measures of the angles of the rhombus to the nearest degree.
we get this triangle |dw:1362782979724:dw|
let x and y be the two unknown angles |dw:1362783041003:dw|
tan(x) = 1/2.5 tan(x) = 0.4 x = arctan(0.4) x = ??
do I draw all of these?
you don't have to, but it helps to see how to get the answer
so what do you get for x?
wait, what's arctan?
it's the inverse of tangent
if you don't have a calculator that can handle arctan, then look here http://www.google.com/search?hl=&q=arctan%280.4%29+in+degrees&sourceid=navclient-ff&rlz=1B3GGLL_enUS420US420&ie=UTF-8
how do I get it on my calculator?
at the top, you'll see that " arctan(0.4) = 21.8014095 degrees"
if you have a TI-83, 85, 86, etc you can hit 2nd, then tan to get arctan
where's arc on the calculator?
what kind of calculator do you have?
it said "Atan" when i did that, not arctan
But I still got the right answer
oh same thing
atan is just a shorter version of arctan
np what did you get for x?
btw, do I find Y?
y = 90 - x y = 90 - 21.8 y = 68.2
now that you know x and y, you would double them to find the angles of the rhombus
21.8*2 = 43.6 which is the measure of one set of angles of the rhombus 68.2*2 = 136.4 which is the measure of another set of angles of the rhombus
Add X and Y?
what do you mean
oh..double them seperately.
because we want the angles of the rhombus, not the triangle
okay, what do I do after?
you're done, you found the angles of the rhombus
oh, so x= 43.6 and y=136.4?
no those are the two angles of a rhombus
the other two are congruent to those two
we have something like this |dw:1362784125897:dw|
should I draw that?
the diagonals are 5 and 2, I just didn't have enough room to draw them in
no, they just want the angles
you can if you want
where are the diagonals?
lol like i said, i didn't have enough room
|dw:1362784257953:dw| see how things get cut off?
so, I'm done now?
yes pretty much