Question

if sin theta =5/13 ,find values of trigonometric ratio's​

Answer 1

sin theta = 5/13

Here, PBP

-------------------

HHB

where ,

sin theta=P/H

cos theta = B/H

tan theta = P/B

and p is perpendicular, B is base , H is hypotaneous

we have, sin theta = 5/13 = P/H

P = 5, H = 13

by pythagors theorm

H² = P² + B²

13²= 5²+B²

B²= 144

B= 12

so, other trignometric ratios are

cos theta = B/H

= 12/13

sec theta = 1/cos theta = H/B

= 13/12

cosec theta = 1/sin theta

= H/P = 13/5

tan theta = P/B = 5/12

cot theta = 1/tan theta

cot theta = B/P = 12/5

Answer 2

Step-by-step explanation:

As per the data given in the question,

We have,

sin∅ = 5/13

since, sin = p/h

(where p= perpendicular, B= base , H= hypotaneous)

so, p =5, h=13, b=?

By pythagors theorm

$p^{2} +b^{2} =h^{2}\\ 5^{2}+ b^{2} =13^{2} \\25+b^{2} =169\\b^{2} =144\\b=12$

So, the trigonometric ratio's​ are:

cos∅ = b/h = $\frac{12}{13}$

tan∅ = p/b= $\frac{5}{12}$

cosec∅ = h/p = $\frac{13}{5}$

sec∅ = h/b = $\frac{13}{12}$

cot∅ = b/p = $\frac{12}{5}$

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