Question

If tan A=3/4 , sin=5/13 , A and B are acute angles ,find sin (A+B) and cos(A-B)​

Answer 1

Answer:

sinA =3/5 and sinB=5/13 where A and B are acute angles.

SinA= 3/5 then, Cos A= 4/5. ( By Pythagoras theorem)

SinB = 5/13. Then ,Cos B= 12/13.(By Pythagoras theorem)

Cos(A+B)= CosA.CosB – SinA.SinB

Then put the value

Cos(A+B)= 4/5×12/13 – 3/5×5/13

Cos(A+B) = 48/65 – 15/65

Cos(A+B)=33/65. ( Answer).

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Answer 2

in above attachment is Answer......

I hope it will helpful......

in right ∆ABC, right angle at C...

Given:-

SinQ= P/h=AC/AB

SinB=5/13

To find side BC we have to apply Pythagoras theorem:-

i.e,

h²= p²+b²

b²=h²-p²

b²=(13)²-(5)²

b²=169-25

b²= 144

b=√144=12 units.

so, we get:-

Bc=12

then,CosB= 12/13

in other triangle..

∆ABC,right angle at C...

Given:-

tanA=3/4= BC/AC

similarly we have to find side AB

by applying Pythagoras theorem:-

h²= p²+b²

h²=3²+4²

h²= 9+16

AB=√25=5unit

So, sinA=3/5 and cosA=4/5

Now....

sin(A+B)= sinA cosB + cosA sinB

sin(A+B)= (3/5x12/13)+(4/5x5/13)

sin(A+B)=36/65+20/65

sin(A+B)=56/65

And..

Cos(A-B)=cosA cosB - sinA sinB

Cos(A-B)=(4/5•12/13)-(3/5•5/13)

Cos(A-B)=48/65-15/65

Cos(A-B)=33/65