Can someone please help me solve this!

- anonymous

Can someone please help me solve this!

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- schrodinger

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- anonymous

##### 1 Attachment

- 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

|dw:1443972456164:dw|
tan B = 2/9

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## More answers

- Nnesha

where you should put 2 and 9 ?

- Nnesha

remember we are looking at angle B

- anonymous

|dw:1443972579055:dw|

- Nnesha

good now apply the Pythagorean theorem to find 3rd side of right triangle

- anonymous

I got 9^2+2^2=c^2, 81+4=c^2, c^2=85,sqrt(85)=c

- Nnesha

looks good |dw:1443972735567:dw|
now use the definition iposted above
to write sin cos and csc

- anonymous

for cos of B I got 9(sort)/85

- anonymous

sqrt*

- Nnesha

looks good and we don't have to rationalize the denominator

- anonymous

and for cscB i'm not sure what that would be

- Nnesha

what's the reciprocal of sin ??

- Nnesha

like cos and sec are reciprocal
cos theta = 1/sec theta

- anonymous

so csc(b)=85/2sqrt(85)?

- Nnesha

what is sin B = ?

- anonymous

they are asking for sinA

- 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

i know they are asking for sin
but what is sin B = ?

- anonymous

sinb=2sqrt(85)/85

- 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

so if you find sin B you can easily find csc B

- anonymous

for csc(b) I believe it's 85/2sqrt(85)

- Nnesha

|dw:1443973415613:dw|
don't rationalize the denominator

- Nnesha

http://prntscr.com/8nn9ru

- Nnesha

sin = opposite over hypotenuse
so sin B =?

- anonymous

ok so let me try one more time :)

- Nnesha

alright take ur time!

- anonymous

cos(b)=9/sqrt(85)

- anonymous

csc(b)=sqrt(85)/2

- Nnesha

|dw:1443973622264:dw|
right

- Nnesha

looks good hm

- anonymous

sin(a)=9/sqrt(85)

- Nnesha

right so what's sin B = ???

- anonymous

sinb=2/sqrt(85)

- Nnesha

right what about tan a ??

- anonymous

tan=sin/cos so

- Nnesha

right

- Nnesha

so what is the exact value of tan A ?

- anonymous

1

- anonymous

so I have another question regarding find sin cos and tan

- anonymous

this involves side length so I was sure how to work this out!

##### 1 Attachment

- anonymous

wasn't*

- 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

thank you!

- Nnesha

np so what is tan A ?? :+)

- Nnesha

^first question

- anonymous

2/9?

- 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

9/2!

- Nnesha

good !
cot =2/9
cot and tan are reciprocal of each other
i think u got it just need more practice :=)

- anonymous

:) thanks so much. I'm sure how to solve angles and length when knowing only 2 side lengths in a right triangle

- anonymous

not sure*

- Nnesha

|dw:1443974435122:dw| same question
given is b and c

- Nnesha

good then did you try to find 3rd side ? :=)

- Nnesha

oh not sure

- 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

for a I got that it equal 8

- anonymous

it's just finding the angles which confuses me :/

- Nnesha

8 is correct

- 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

|dw:1443974742902:dw|

- Nnesha

let's pick sin
sin A = opposite over adjacent
what is opposite and adjacent side of angle A ?

- anonymous

I thought sin was opposite over hyp

- Nnesha

right lol sorry mistake thinking about tan

- anonymous

sin(a) =8/17

- 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

angle **

- anonymous

so it's 28 degrees

- Nnesha

let me check :D wait plz

- anonymous

ok thank you :)

- Nnesha

right ! good job!

- 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

right you can round it if you want or if the direction says so

- Nnesha

ohh wait directions is NO DECIMAL so round it

- anonymous

the direction just mentions the exact measure

- Nnesha

|dw:1443975183145:dw|
so no decimal
round it

- anonymous

62 :D

- Nnesha

good :=)

- anonymous

i'm having trouble inputting the degrees sign into the answer

- Nnesha

hmm

- 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

it's wanting the degrees symbol :/

- Nnesha

hmmm does it work \[\huge\rm a^o\] i'm not sure how to put on that website

- Nnesha

is it a practice website or online school documeent ?

- anonymous

##### 1 Attachment

- anonymous

this is what showed up

- 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

answer >???*

- anonymous

no :/

- anonymous

It makes sense in terms of the math, but when I enter it in the system it's didn't recognize it

- Nnesha

ye right let me think hmmm

- Nnesha

how did you put degree sign ?

- Nnesha

i mean degree how did you put little 0 ? :P

- anonymous

I did option shift and then 8

- Nnesha

maybe they just want us to write the number
like 28 and 62

- anonymous

I tried that but it was incorrect I think they want me to show that it's degrees

- Nnesha

ye answer is right but `syntax incomplete ` :.........

- anonymous

-,-

- Nnesha

sorry don't know i do it b hand so

- anonymous

I just emailed my instructor hopeful he responds:) thanks again so much for all your help

- Nnesha

oh okay please let me know what was the mistake :=)

- Nnesha

it was pleasure :=)

- Nnesha

good work!

- anonymous

:) I think it's just something that has to do with the keys I don't think it's the actual math

- Nnesha

ye right :=)

- anonymous

:D

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