Question

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

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

Answer:

Step-by-step explanation:

......................................................................................................................

.Given  asinB +bcosB = c

take    m = acosB -bsinB

         m^2 + c^2 =  [ asinB +bcosB ]^2 +[ acosB -bsinB ]^2

                     = a^2sin^2B +b^2cos^2B +2sinBcosB

                       + a^2cos^2B +b^2sin^2B - 2sinBcosB

                    =  a^2sin^2B +a^2cos^2B + b^2cos^2B +b^2sin^2B

                    = a^2 [sin^2B +cos^2B ]+ b^2 [ cos^2B +sin^2B  ]

                    m^2 + c^2   =  a^2 + cb^2

                    m^2  =  a^2 + b^2  - c^2

                  m  =  √[ a^2 + b^2  - c^2]