Question

$Solution:$

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

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

Answer 1

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ANSWER:------

asinB+BsinB=ca \: sinB + B \: sinB = c \:asinB+BsinB=c

asinB+BsinB=c(given)a \: sinB + B \: sinB = c \: (given)asinB+BsinB=c(given)

we should than Square both sides,

So,(asinB+bsinB)2=c2.So, \: {(a \: sinB + b \: sinB) }^{2} = {c}^{2} .So,(asinB+bsinB)2=c2.

=>a2sin2B+b2sin2B+2absinBcosB=c2= > {a}^{2} {sin}^{2} B + {b}^{2} {sin}^{2} B + 2ab \: sinB \: cosB = {c}^{2}=>a2sin2B+b2sin2B+2absinBcosB=c2

=>a2(1−cos2B)+b2(1−sin2B)+2absinBcosB=c2= > {a}^{2} (1 - {cos}^{2} B) + {b}^{2} (1 - {sin}^{2} B) + 2ab \: sinB \: cosB = {c}^{2}=>a2(1−cos2B)+b2(1−sin2B)+2absinBcosB=c2

=>a2−a2cos2+b2−b2sin2B+2absinBcosB=c2= > {a}^{2} - {a}^{2} {cos}^{2} + {b}^{2}\: - {b}^{2} {sin}^{2} B + 2ab \: sinB \: cosB = {c}^{2}=>a2−a2cos2+b2−b2sin2B+2absinBcosB=c2

=>a2cos2B−2absinBcosB+b2sin2B=a2+b2−c2.= > {a}^{2} {cos}^{2} B - 2ab \: sinB \: cosB \: + {b}^{2} {sin}^{2} B= {a}^{2} + {b}^{2} - {c}^{2}. \:=>a2cos2B−2absinBcosB+b2sin2B=a2+b2−c2.

=>(acosB−bsinB)2=a2+b2−c2= > ({a \: cosB - b \: sinB})^{2} = {a}^{2} + {b}^{2} - {c}^{2}=>(acosB−bsinB)2=a2+b2−c2

=>(acosB−bsinB)=a2+b2−c2.= > ({a \: cosB - b \: sinB}) = \sqrt{ {a}^{2} + {b}^{2} - {c}^{2}.}=>(acosB−bsinB)=a2+b2−c2.

hope it helps:--

T!—!ANKS!!!!!

Answer 2

hey mate!!!

here is your answer.....

hope it helps..

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