anonymous
  • anonymous
Can someone please help me solve this!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Nnesha
  • Nnesha
\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\] that's all you need to find all 6 ratios
Nnesha
  • Nnesha
|dw:1443972456164:dw| tan B = 2/9

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Nnesha
  • Nnesha
where you should put 2 and 9 ?
Nnesha
  • Nnesha
remember we are looking at angle B
anonymous
  • anonymous
|dw:1443972579055:dw|
Nnesha
  • Nnesha
good now apply the Pythagorean theorem to find 3rd side of right triangle
anonymous
  • anonymous
I got 9^2+2^2=c^2, 81+4=c^2, c^2=85,sqrt(85)=c
Nnesha
  • Nnesha
looks good |dw:1443972735567:dw| now use the definition iposted above to write sin cos and csc
anonymous
  • anonymous
for cos of B I got 9(sort)/85
anonymous
  • anonymous
sqrt*
Nnesha
  • Nnesha
looks good and we don't have to rationalize the denominator
anonymous
  • anonymous
and for cscB i'm not sure what that would be
Nnesha
  • Nnesha
what's the reciprocal of sin ??
Nnesha
  • Nnesha
like cos and sec are reciprocal cos theta = 1/sec theta
anonymous
  • anonymous
so csc(b)=85/2sqrt(85)?
Nnesha
  • Nnesha
what is sin B = ?
anonymous
  • anonymous
they are asking for sinA
AlexandervonHumboldt2
  • AlexandervonHumboldt2
@sana9 the original definitions were not those which nnesha gives you. please keep in mind that you might have a question like what is sin of 360? you will never find a 360 degrees in a triangle. so the standart school definitions will not work for such question. so keep in mind these universal things: Sinus: sin(a)=(yb)/R Cosine: cos(a)=(xb)/R Tangent: tg(a)=(yb)/(xb) picture : https://upload.wikimedia.org/wikipedia/ru/thumb/8/8f/Trig_functions.gif/250px-Trig_functions.gif sourse `wikipedia`
Nnesha
  • Nnesha
i know they are asking for sin but what is sin B = ?
anonymous
  • anonymous
sinb=2sqrt(85)/85
Nnesha
  • Nnesha
look at this definition \[\large\rm cos \theta =\frac{ 1 }{ \sec \theta}\]\[\sin \theta = \frac{ 1 }{ \csc \theta } \]\[\tan \theta =\frac{ 1 }{ \cot \theta }\] or tan theta = sin/ cos
Nnesha
  • Nnesha
so if you find sin B you can easily find csc B
anonymous
  • anonymous
for csc(b) I believe it's 85/2sqrt(85)
Nnesha
  • Nnesha
|dw:1443973415613:dw| don't rationalize the denominator
Nnesha
  • Nnesha
http://prntscr.com/8nn9ru
Nnesha
  • Nnesha
sin = opposite over hypotenuse so sin B =?
anonymous
  • anonymous
ok so let me try one more time :)
Nnesha
  • Nnesha
alright take ur time!
anonymous
  • anonymous
cos(b)=9/sqrt(85)
anonymous
  • anonymous
csc(b)=sqrt(85)/2
Nnesha
  • Nnesha
|dw:1443973622264:dw| right
Nnesha
  • Nnesha
looks good hm
anonymous
  • anonymous
sin(a)=9/sqrt(85)
Nnesha
  • Nnesha
right so what's sin B = ???
anonymous
  • anonymous
sinb=2/sqrt(85)
Nnesha
  • Nnesha
right what about tan a ??
anonymous
  • anonymous
tan=sin/cos so
Nnesha
  • Nnesha
right
Nnesha
  • Nnesha
so what is the exact value of tan A ?
anonymous
  • anonymous
1
anonymous
  • anonymous
so I have another question regarding find sin cos and tan
anonymous
  • anonymous
this involves side length so I was sure how to work this out!
anonymous
  • anonymous
wasn't*
Nnesha
  • Nnesha
no it's not 1 remember tan = sin over cos and sin = opposite over hyp cos = adjacent over hyp \[\huge\rm tan = \frac{ \frac{ o }{ h } }{ \frac{ a }{ h } }=\frac{o}{h} \times \frac{h}{a}\] multiply top fraction with the reciprocal of the bottom fraction \[\huge\rm tan =\frac{o}{\cancel{h}} \times \frac{\cancel{h}}{a}=\frac{o}{a}\] o=opposite a=adjacent so tan = opposite over adjacent
anonymous
  • anonymous
thank you!
Nnesha
  • Nnesha
np so what is tan A ?? :+)
Nnesha
  • Nnesha
^first question
anonymous
  • anonymous
2/9?
Nnesha
  • Nnesha
|dw:1443974262921:dw| remember opposite and adjacent sides depends on angle 9 is adjacent side of angle B but 9 is opposite side of angle A adjacent side that touches the angle we are looking at
anonymous
  • anonymous
9/2!
Nnesha
  • Nnesha
good ! cot =2/9 cot and tan are reciprocal of each other i think u got it just need more practice :=)
anonymous
  • anonymous
:) thanks so much. I'm sure how to solve angles and length when knowing only 2 side lengths in a right triangle
anonymous
  • anonymous
not sure*
Nnesha
  • Nnesha
|dw:1443974435122:dw| same question given is b and c
Nnesha
  • Nnesha
good then did you try to find 3rd side ? :=)
Nnesha
  • Nnesha
oh not sure
Nnesha
  • Nnesha
same question like last one one given are the 2 legs of `right triangle ` then always you can use Pythagorean theorem to find 3rd side
anonymous
  • anonymous
for a I got that it equal 8
anonymous
  • anonymous
it's just finding the angles which confuses me :/
Nnesha
  • Nnesha
8 is correct
Nnesha
  • Nnesha
ah i see |dw:1443974659273:dw| just focus one angle A for this question its easy to find angle bec u know all 3 sides you can use any ratio doesn't matter u will get the same answer!
Nnesha
  • Nnesha
|dw:1443974742902:dw|
Nnesha
  • Nnesha
let's pick sin sin A = opposite over adjacent what is opposite and adjacent side of angle A ?
anonymous
  • anonymous
I thought sin was opposite over hyp
Nnesha
  • Nnesha
right lol sorry mistake thinking about tan
anonymous
  • anonymous
sin(a) =8/17
Nnesha
  • Nnesha
right \[\huge\rm sin A = \frac{ 8 }{ 17}\] when we have to find anlde we use invs sin \[\large\rm A = \sin^{-1} \frac{8}{17}\]
Nnesha
  • Nnesha
angle **
anonymous
  • anonymous
so it's 28 degrees
Nnesha
  • Nnesha
let me check :D wait plz
anonymous
  • anonymous
ok thank you :)
Nnesha
  • Nnesha
right ! good job!
anonymous
  • anonymous
and to find angle B we would do sin(B)=15/17 which is sin-1(15/17)=61.9 and it's exactly 62?
Nnesha
  • Nnesha
right you can round it if you want or if the direction says so
Nnesha
  • Nnesha
ohh wait directions is NO DECIMAL so round it
anonymous
  • anonymous
the direction just mentions the exact measure
Nnesha
  • Nnesha
|dw:1443975183145:dw| so no decimal round it
anonymous
  • anonymous
62 :D
Nnesha
  • Nnesha
good :=)
anonymous
  • anonymous
i'm having trouble inputting the degrees sign into the answer
Nnesha
  • Nnesha
hmm
Nnesha
  • Nnesha
i guesss you don't have to put degree sign bec it says angle A so definitely the number u will put is an angle hmm i think ....
anonymous
  • anonymous
it's wanting the degrees symbol :/
Nnesha
  • Nnesha
hmmm does it work \[\huge\rm a^o\] i'm not sure how to put on that website
Nnesha
  • Nnesha
is it a practice website or online school documeent ?
anonymous
  • anonymous
anonymous
  • anonymous
this is what showed up
Nnesha
  • Nnesha
hmm well i think the answer we got is correct not sure what theyyy want us to do did they gave u the correct ??
Nnesha
  • Nnesha
answer >???*
anonymous
  • anonymous
no :/
anonymous
  • anonymous
It makes sense in terms of the math, but when I enter it in the system it's didn't recognize it
Nnesha
  • Nnesha
ye right let me think hmmm
Nnesha
  • Nnesha
how did you put degree sign ?
Nnesha
  • Nnesha
i mean degree how did you put little 0 ? :P
anonymous
  • anonymous
I did option shift and then 8
Nnesha
  • Nnesha
maybe they just want us to write the number like 28 and 62
anonymous
  • anonymous
I tried that but it was incorrect I think they want me to show that it's degrees
Nnesha
  • Nnesha
ye answer is right but `syntax incomplete ` :.........
anonymous
  • anonymous
-,-
Nnesha
  • Nnesha
sorry don't know i do it b hand so
anonymous
  • anonymous
I just emailed my instructor hopeful he responds:) thanks again so much for all your help
Nnesha
  • Nnesha
oh okay please let me know what was the mistake :=)
Nnesha
  • Nnesha
it was pleasure :=)
Nnesha
  • Nnesha
good work!
anonymous
  • anonymous
:) I think it's just something that has to do with the keys I don't think it's the actual math
Nnesha
  • Nnesha
ye right :=)
anonymous
  • anonymous
:D

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