Question

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


​

Answer 1

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°

Answer 2

$\underline{{\colorbox{yellow}{Question}}}$ we have to Find value of Angle C ?

$\LARGE\underline{\underline{\sf \red{G}\blue{i}\green{v}\orange{e}\red{n}:}}$

• 3sinA+4cosB = 6
• 4sinB + 3cosA = 1

$\LARGE\underline{\underline{\sf \red{S}\blue{o}\green{l}\orange{u}\pink{t}\purple{i}\orange{o}\red{n}:}}$

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°

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\:</strong><strong>C</strong><strong>=</strong><strong>3</strong><strong>0</strong><strong>°</strong><strong>}}}}}}}}}}$

$\boxed{\begin{minipage}{7 cm}Fundamental Trignometric Identities\\ \\ \sin^2\theta+\cos^2\theta=1\\ \\ 1+\tan^2\theta=\sec^2\theta \\ \\ 1+\cot^2\theta=\text{cosec}^2 \, \theta \end{minipage}}$