If the angles A and B of AABC satisfy
the relations 3 sin A +4 cos B =6
and 4 sin B+3 cos A = 1, then the
third angleC is
Answers
Answer:
Third angle ( C ) = 30°
Step-by-step explanation:
Given---> In ΔABC
3SinA + 4CosB = 6 and 4SinB + 3CosA = 1
To find---> Value of third angle C
Solution ---> ATQ,
3SinA + 4 cosA = 6
Squaring both sides , we get
=> ( 3SinA + 4CosB )² = ( 6 )²
We have an identity
( a + b )² = a² + b² + 2ab , applying it here , we get
=> (3SinA )²+(4CosB )² + 2(3SinA )(4CosB ) = 36
=> 9 Sin²A + 16 Cos²B + 24 SinA CosB = 36
............................( 1 )
Now , ATQ,
4SinB + 3CosA = 1
Squaring both sides , we get
( 4SinB + 3 CosA )² = ( 1 )²
Applying formula of ( a + b )² , we get
=> (4SinB)² + (3CosA)² + 24 (SinB) (CosA) = 1
=> 16 Sin²B + 9 Cos²A + 24 CosA SinB = 1 ......... (2)
Adding equation (1) and equation (2)
=> 9Sin²A + 16Cos²B + 24SinA CosB + 16Sin²B
+ 9Cos²A + 24CosASinB = 36 + 1
=> 9 (Sin²A + Cos²A ) + 16 (Sin²B + Cos²B )
+ 24SinACosB + 24 CosASinB= 37
We have an identity , Sin²θ + Cos²θ = 1 , applying it here we get
=> 9 ( 1 ) + 16 ( 1 ) + 24 (SinACosB+CosASinB) = 37
We have a formula
Sin(x + y ) = Sinx Cosy + Cosx Siny
Applying it here we get
=> 9 + 16 + 24 Sin( A + B ) = 37
In triangle ABC, by angle sum property
A + B + C = 180°
A + B = 180° - C , applying this
=> 25 + 24 Sin (180° - C ) =37
We know that , Sin (180° - θ ) = Sinθ , applying it
=> 24 SinC = 37 - 25
=> 24 SinC = 12
=> SinC = 12 / 24
=> SinC = 1 / 2
=> SinC = Sin30°
=> C = 30°
we have to Find value of Angle C ?
- 3sinA+4cosB = 6
- 4sinB + 3cosA = 1
Formula to be used :----
- (a+b)² = a²+b²+2ab
- sin²A+cos²A = 1
- Sin(A+B) = SinAcosB + cosAsinB
- sin30° = 1/2
- sin(180-@) = sin@
Let,
3sinA+4cosB = 6 ------------ Equation (1)
And,
4sinB + 3cosA = 1 ------------ Equation (2)
Squaring both sides of both equation and adding them we get,
(3sinA+4cosB)² + (4sinB + 3cosA)² = 6² + 1²
→ 9sin²A + 16cos²B + 24SinAcosB + 16sin²B + 9cos²A + 24sinBcosA = 36 + 1
Re-arranging them now ,
→ 9(sin²A + cos²A) + 16(cos²B + sin²B) + 24(SinAcosB + sinBcosA) = 37
→ 9 + 16 + 24(sin(A+B)) = 37
→ 24(sin(A+B)) = 37 - 25
→ sin(A+B) = 12/24
→ sin(A+B) = 1/2
Now since sum of all angles of a ∆ = 180°
A + B + C = 180°
A + B = (180-C)
Multiply by sin both sides we get,
Sin(A+B) = sin(180-C) = SinC
so,
Sin(A+B) = 1/2
→ Sin C = 1/2
→ Sin C = sin 30°