Question

In triangle RST, angle S is a right angle and cscR=17/15. what is tan T?

Answer 1

Hello @berrypicking

Answer 2

Well @vikrantg4 , your method has a great mistake.
\[\large{\frac{RT}{TS} = \frac{17}{15}}\]
is right but saying that RT = 17 and ST = 15 is wrong. as 2/4 = 1/2 but 2 is not equal to 1 and 4 is not equal to 2.

Answer 3

I didn't say that but I do have the same diagram and am not sure what to do next

Answer 4

now. @berrypicking . first of all :

\[\large{\csc R = \frac{17}{15}}\]
Right?

Answer 5

and csc R = RT/TS , do you know why?

Answer 6

yep

Answer 7

no I know csc =1/sin

Answer 8

Ok good. So now we have :
\[\large{\frac{RT}{TS} = \frac{17}{15}}\]
Mark this as equation (1) .

Answer 9

see, as csc = 1/sin

and sin = opposite / hypotenuse
thus csc = hypotenuse / opposite

opposite to angle R is ST and hypotenuse of the triangle is RT
Thus csc R = RT/TS , got it?

Answer 10

I know the opposiste and hypo but tell me why does cscR=RT/TS

Answer 11

cosec = hypotenuse / opposite, agree?

Answer 12

ok

Answer 13

what is hypotenuse of the given triangle?

Answer 14

RT

Answer 15

right and opposite to angle R ?

Answer 16

angle T

Answer 17

Oh! We are talking about sides dude. So what is opposite "side" of angle R

Answer 18

gotcha ST

Answer 19

right so what is cosec R now?

Answer 20

RT/ST

Answer 21

Great work.

Answer 22

So RT/ST = 17/15 , agreed?

Answer 23

ok

Answer 24

wait!

Answer 25

so tan T = 15/RS. How do I find RS

Answer 26

why tan T = 15/RS ?

Answer 27

tan=opp/adj????

Answer 28

yes. so?

Answer 29

aren't I trying to find tan T what am I missing?

Answer 30

see tan T = opp. / adj. is right.

what is opposite side to angle T ?

Answer 31

right to angle T -it is RS/ST

Answer 32

opp. side to angle T is RS ... check that again

and adjacent side to angle T is ST

so tan T = RS/ST

Answer 33

now , as tan T = \(\large{\frac{RS}{ST}}\) and from equation (1)

\[\large{\frac{RT}{TS} = \frac{17}{15}}\]
\[\large{\frac{RT}{17} \times 15 = TS}\]
\[\large{\frac{15}{17} \times RT = TS}\]
thus : \[\large{\tan T = \frac{RS}{ST} = \frac{RS}{\frac{15}{17}\times RT}}\]
getting it?

Answer 34

Now, \[\large{\tan T = \frac{RS}{RT} \times \frac{17}{15}}\]
but RS/RT = cot R ( cot = adj. / opp. )

thus : \[\large{\tan T = \cot R \times \frac{17}{15}}\]

Now our aim is to calculate cot R ... since :
cot = 1 /tan

thus cot R = 1 / tan R

as tan R = sin R / cos R

since sin R = 1/ cosec R and cosec R = 17/15 thus sin R = 15/17

since \(\large{\sin R = \sqrt{1- \cos^2 R}}\)
Hence 15/17 = \(\sqrt{1- \cos^2 R}\)
SQuare both sides:
\[\large{\frac{225}{289} = 1 - cos^2 R}\]
\[\large{1- \frac{225}{289} = \cos ^2 R}\]
\[\large{ \frac{64}{289} = \cos^2 R}\]
Now we have cos R = 8/17

hence tan R = sin R / cos R = (15/17)/(8/17) = 15/8
got it?