in ∆ABC, right angled at C, if tan A= 1/√3 find the value of sin A. cosB+cosA.sinB
Answers
Answer:
Given: parllelogram ABCD circumscribe a circle
To prove: we know that pain syndrome from the extent part are equal in length.
AP=AS-------1
BP=BQ-------2
CR=CQ-------3
DR=DS-------4
Adding 1,2,3 and 4
AP+BP+CR+DR=AS+BQ+CQ+DS
AB+CD=AD+BC-------5
ABCD is a parllelogram
AB=CD
----------6
AD=BC
From 3 and 6 we get
AB+AB=AD+AD
2AB=2AD
AB=AD
If adjacent sides of parllelogram are equal then it becomes rhombus.
Hence ABCD is a Rhombus.
tanA=
3
1
AC
BC
=
3
1
AC=
3
BC
Also, (AC)
2
+(BC)
2
=(AB)
2
3(BC)
2
+(BC)
2
=(AB)
2
AB=2BC
AB:BC:AC=2:1:
3
sinA=
AB
BC
=
2
1
cosA=
AB
AC
=
2
3
sinB=
AB
AC
=
2
3
cosB=
AB
BC
=
2
1
sinAcosB+cosAsinB=
2
1
×
2
1
+
2
3
×
2
3
=1
Also, sinAcosB+cosAsinB=sin(A+B)
In right angled ΔABC.
A+B+C=180
∘
(C=90
∘
)
A+B=90
∘
sin(A+B)=sin90 =1