Question

Find sinA=3/5, A is acute find sin 3A​

Answer 1

Answer:

$\LARGE{ \underline{\underline{ \pink{ \bf{Required \: answer:}}}}}$

Given, Sin A = 3/5 and is an acute angle

So, In ΔABC we have ∠B = 90° ,

And AC = 5 and BC = 3

Note:-

By Pythagoras theorem:

$AB = \sqrt{(AC}² \: - \: BC²)$

$= \sqrt{5}² - 3²)$

$= \sqrt{25 - 9}$

$= \sqrt{16} = 4$

Now,

cos A = AB / AC = 4/5

tan A = BC / AB = 3/4

cot A = 1 / tanθ = 4/3.

sec A = 1 / cosθ = 5/4

cosecA = 1 / sinθ = 5/3.

Answer 2

$\huge\boxed{\fcolorbox{white}{pink}{Answer:-}}$

Given, Sin A = 3/5 and is an acute angle

So, In ΔABC we have ∠B = 90° ,

And AC = 5 and BC = 3

Note:-

By Pythagoras theorem:

AB = \sqrt{(AC}² \: - \: BC²)AB=

(AC

²−BC²)

= \sqrt{5}² - 3²)=

5

²−3²)

= \sqrt{25 - 9}=

25−9

$\huge☠\star\mathfrak\red{Nisha.here!!☠}$

= \sqrt{16} = 4=

16

=4

Now,

cos A = AB / AC = 4/5

tan A = BC / AB = 3/4

cot A = 1 / tanθ = 4/3.

sec A = 1 / cosθ = 5/4

cosecA = 1 / sinθ = 5/3.