if 3 COSA-4sinA=0,find sinA+2cosA/3cos-sinA
Answers
Answer:
Let △PQR be a right angled triangle where ∠Q=90
0
and ∠R=A as shown in the above figure:
Now it is given that 4sinA−3cosA=0 that is
4sinA=3cosA
⇒
cosA
sinA
=
4
3
⇒tanA=
4
3
We know that, in a right angled triangle, tanθ is equal to opposite side over adjacent side that is tanθ=
Adjacentside
Oppositeside
, therefore, opposite side PQ=3 and adjacent side QR=4.
Now, using pythagoras theorem in △PQR, we have
PR
2
=PQ
2
+QR
2
=3
2
+4
2
=9+16=25
⇒PR=
25
=5
Therefore, the hypotenuse PR=5.
We know that, in a right angled triangle,
sinθ is equal to opposite side over hypotenuse that is sinθ=
Hypotenuse
Oppositeside
and
cosθ is equal to adjacent side over hypotenuse that is cosθ=
Hypotenuse
Adjacentside
Here, we have opposite side PQ=3, adjacent side QR=4 and the hypotenuse PR=5, therefore, the trignometric ratios of angle A can be determined as follows:
sinA=
Hypotenuse
Oppositeside
=
PR
PQ
=
5
3
cosA=
Hypotenuse
Adjacentside
=
PR
QR
=
5
4
cosec A=
sinA
1
=
5
3
1
=1×
3
5
=
3
5
secA=
cosA
1
=
5
4
1
=1×
4
5
=
4
5
Hence, sinA=
5
3
, cosA=
5
4
, cscA=
3
5
and secA=
4
5
.
Step-by-step explanation:
hope it's helpful to you