Help... i have exams tomorrow. give 10 points. urgent

given: tana=5/12 and 0
Calculate:

A=(26sina+12seca)/(15cota)

I know that tana=sina/cosa but tana=5/12 and sina can't be 5, the maximum of sina is 1!!!

HELP!!!!!

.Q.Given: tana=5/12 and 0
Calculate:

A=(26sina+12seca)/(15cota)

Ans : Given : tana=5/12 => cota = 12/5
As Known : sec^2a = 1 + tan^a = 1 + (5/12)^2 = 169 / 144
=> seca = 13 / 12
Now : cosa = 12/13
Apply : sina = tana * cosa = 5 / 13
Therefore : cota=12/5 , seca = 13 / 12 , sina = 5 / 13

Now : A=(26sina+12seca)/(15cota)
=> A = {26*(5/13) + 12*(13/12) } / [15*(12/5)]
=> A = {10+13} / [3*12]
=> A=23 / 36

I wish you will appreciate my answer.

I know that tana=sina/cosa but tana=5/12 and sina can't be 5.
No. ratio 12/13 is also =24/26 right . that doesnot mean 12=24 and 13=26. It is just a ratio.
Hence sina/cosa=5/12 doesnot mean sina=5 and cosa=12. Get your basics right.

for the given triangle which is a right angled triangle ,hypotenuse=sqrt(5*5+12*12)=13

hence
sina=5/13
cosa=12/13
seca=1/cosa=13/12
tana=5/12
cota=12/5
Now calculate A=(26sina+12seca)/(15cota)
A=23/36

hahaha

your approach is very funny !

actually, tan is perpendicular divided by base. so perp=5 and base=12. This implies hypotenuse=13

This means sina = 5/13 and cosa = 12/13, obviously then sina/cosa = 5/12 = tana

therefore
A = (26*(5/13) + 12*(13/12)) / (15*(12/5)) = (10+13)/(36) = 23/36

This is how it is solved.

your approach.. LOL

tana=5/12
sec²a=1+tan²a =1+(5/12)² =169/144
seca=13/12 and cosa=1/seca=12/13
sina=tana×cosa =(5/12)×(12/13)= 5/13
cota=1/tana=12/5
Hence
(26sina+12seca)/(15cota)
=(26×5/13+12/(12/13)/(15×12/5)
=(10+13)/(36)=23/36

tanA =opp / adj = 5/12
hyp = sqrt ( opp^2 + adj^2)
hyp = 13
sinA = opp / hyp = 5/13
cos A = adj / hyp = 12 / 13
cotA = 12/5
(26sina+12seca)/(15cota) = ( 26*5/13 + 12 * 13/12 ) / (15 * 12/5 )
= 23 / 36