Question

If cosec 0 = 41/40 find sin 0 and sec0

Answer 1

Answer:

if 0 is represented as theta

given :   cosec 0 = 40/41

sin 0 = 1/cosec 0

         = 1/(41/40)

         = 40/41

for sec 0 we can go with Pythagoras theorem:

sin0 = p/h

        :>p = 40

         :>h = 41

b = $\sqrt{h^{2} - p^{2} }$

   =$\sqrt{41^{2} - 40^{2} }$

    = 9

sec 0 = 41/9

i think this helps

Step-by-step explanation:

Answer 2

Answer:

Cosec 0= 41/40

Sin 0=1/cosec 0

Sin 0= 1/41/40

Sin 0= 40/41

Lets,take p=40k and h=41k

As we know,

H²=P²+B²

So, we have to find the value of B

B²=H²-P²

B²=(41K)²-(40K)²

B²=1681K²-1600K²

B²=81K²

B=

$\sqrt{ {81k}^{2} }$

B=9k

So, Sec 0=41k/9k

Sec 0=41/9

Sin 0=40/41