Question

when given that $a \: sinB + b \: sinB = c \:$. prove:
$= > ({a \: cosB - b \: sinB}) = \sqrt{ {a}^{2} + {b}^{2} - {c}^{2}.}$​

Answer 1

Solution:

It is given that $a \: sinB + b \: sinB = c \:$

$a \: sinB + b \: sinB = c \: (given)$

Squaring both sides,

$So, \: {(a \: sinB + b \: sinB) }^{2} = {c}^{2} .$

$= > {a}^{2} {sin}^{2} B + {b}^{2} {sin}^{2} B + 2ab \: sinB \: cosB = {c}^{2}$

$= > {a}^{2} (1 - {cos}^{2} B) + {b}^{2} (1 - {sin}^{2} B) + 2ab \: sinB \: cosB = {c}^{2}$

$= > {a}^{2} - {a}^{2} {cos}^{2} + {b}^{2}\: - {b}^{2} {sin}^{2} B + 2ab \: sinB \: cosB = {c}^{2}$

$= > {a}^{2} {cos}^{2} B - 2ab \: sinB \: cosB \: + {b}^{2} {sin}^{2} B= {a}^{2} + {b}^{2} - {c}^{2}. \:$

$= > ({a \: cosB - b \: sinB})^{2} = {a}^{2} + {b}^{2} - {c}^{2}$

$= > ({a \: cosB - b \: sinB}) = \sqrt{ {a}^{2} + {b}^{2} - {c}^{2}.}$

Hence, it is proved.

Answer 2

It is given that a \: sinB + b \: sinB = c \:asinB+bsinB=c

a \: sinB + b \: sinB = c \: (given)asinB+bsinB=c(given)

Squaring both sides,

So, \: {(a \: sinB + b \: sinB) }^{2} = {c}^{2} .So,(asinB+bsinB)

2

=c

2

.

= > {a}^{2} {sin}^{2} B + {b}^{2} {sin}^{2} B + 2ab \: sinB \: cosB = {c}^{2}=>a

2

sin

2

B+b

2

sin

2

B+2absinBcosB=c

2

= > {a}^{2} (1 - {cos}^{2} B) + {b}^{2} (1 - {sin}^{2} B) + 2ab \: sinB \: cosB = {c}^{2}=>a

2

(1−cos

2

B)+b

2

(1−sin

2

B)+2absinBcosB=c

2

= > {a}^{2} - {a}^{2} {cos}^{2} + {b}^{2}\: - {b}^{2} {sin}^{2} B + 2ab \: sinB \: cosB = {c}^{2}=>a

2

−a

2

cos

2

+b

2

−b

2

sin

2

B+2absinBcosB=c

2

= > {a}^{2} {cos}^{2} B - 2ab \: sinB \: cosB \: + {b}^{2} {sin}^{2} B= {a}^{2} + {b}^{2} - {c}^{2}. \:=>a

2

cos

2

B−2absinBcosB+b

2

sin

2

B=a

2

+b

2

−c

2

.

= > ({a \: cosB - b \: sinB})^{2} = {a}^{2} + {b}^{2} - {c}^{2}=>(acosB−bsinB)

2

=a

2

+b

2

−c

2

= > ({a \: cosB - b \: sinB}) = \sqrt{ {a}^{2} + {b}^{2} - {c}^{2}.}=>(acosB−bsinB)=

a

2

+b

2

−c

2

.