Question

if 3 tan A= 5 then find sin A and cos A​

Answer 1

Answer:

sinA = 5/√34

cosA=3/√34

Step-by-step explanation:

Given that 3tanA = 5 . Then ,

=> 3tanA=5.

=> tanA = 5/3 .

Now , if we consider a right angled ∆ABC in which tanA = 5/3 and tan∅=perpendicular/base.

So , hypontenuse will be :

=> hypontenuse ²=base²+perp.²

=>hypontenuse ²=3²+5².

=>hypontenuse ²=9+25

=>hypontenuse ²=34.

=>hypontenuse=√34.

And we know sinA = perp./hypontenuse= 5/√34.

and cosA=base/hypontenuse= 3/√34.

Answer 2

Step-by-step explanation:

QUESTION.....

If 3 tan A=4, then find sin A and cos A

ANSWER......

LET ABC is a right angle triangle 

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC 

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC By using Pythagoras theorem

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC By using Pythagoras theoremAC2=AB2+BC2 

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC By using Pythagoras theoremAC2=AB2+BC2 =(3)2+(4)2 

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC By using Pythagoras theoremAC2=AB2+BC2 =(3)2+(4)2 =9+16=25 

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC By using Pythagoras theoremAC2=AB2+BC2 =(3)2+(4)2 =9+16=25 ∴AC=25=5 

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC By using Pythagoras theoremAC2=AB2+BC2 =(3)2+(4)2 =9+16=25 ∴AC=25=5 sinA=HypotenuseOpposite side of∠A 

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC By using Pythagoras theoremAC2=AB2+BC2 =(3)2+(4)2 =9+16=25 ∴AC=25=5 sinA=HypotenuseOpposite side of∠A        =ACBC=54 

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC By using Pythagoras theoremAC2=AB2+BC2 =(3)2+(4)2 =9+16=25 ∴AC=25=5 sinA=HypotenuseOpposite side of∠A        =ACBC=54 ∴cosA=HypotenuseAdjacent side of∠A 

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC By using Pythagoras theoremAC2=AB2+BC2 =(3)2+(4)2 =9+16=25 ∴AC=25=5 sinA=HypotenuseOpposite side of∠A        =ACBC=54 ∴cosA=HypotenuseAdjacent side of∠A          =ACAB=53 

C is a right angle triangle 3tanA=4  ⇒tanA=34=ABBC By using Pythagoras theoremAC2=AB2+BC2 =(3)2+(4)2 =9+16=25 ∴AC=25=5 sinA=HypotenuseOpposite side of∠A        =ACBC=54 ∴cosA=HypotenuseAdjacent side of∠A          =ACAB=53 ∴sinA= ⅘ and  cosA= ⅘.