Question

If 4 tan A=3 then sinA +CosA is equal to​

Answer 1

Step-by-step explanation:

tan A=3/4

tanA=p/b=3/4

so h²=p²+b²

h²=9+16

h=5

so sinA=p/h=3/5

cos A=4/5

so sinA+cosA=3/5+4/5=7/5

Answer 2

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Q1) If 4tanA=3 then SinA+CosA is equals to

$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

$\implies\boxed{\bf\red{To \:find \to \: the \: value \: of \: SinA+CosA=??}}$

$\ \: \: \boxed{ \sf{draw \: a \: right \: angle \: triangle \: }}$

$\: \: \: \: \mathfrak{ {put \: the \: given \: values \: in \: triangle \triangle}}$

$\: \: \boxed{ \sf{put \: 3 \: in \: opposite \: sides \: and\: 4 \: in \: adjacent \: sides}}$

✪ 3rd sides is not given so use Pythagoras theorem to find hypotenuse

$\implies\boxed{\bf\pink{H^2=B^2+P^2}}$

$\implies\sf{H^2=(4)^2+(3)^2}$

$\implies\sf{H^2=16+9}$

$\implies\sf{H=\sqrt25}$

$\implies\sf{H=5}$

As we know that,

$\bullet\rm{SinA=\dfrac{opposite \: side}{Hypothenus \: side}=\dfrac{3}{5}}$

$\bullet\rm{CosA=\dfrac{Adjacent \: side}{Hypothenus \: side}=\dfrac{4}{5}}$

$\bullet\rm{tanA=\dfrac{Opposite\:side}{Adjacent \: side}=\dfrac{3}{4}}$

$\: \clubsuit\underline{\bf \pink {according \: to \: this \: question}}$

$\mapsto\rm{SinA+CosA............to find }$

$\mapsto\bf{Put \: sides\: of \: right \: angle \: triangle \triangle}$

$\mapsto\rm{SinA+CosA}$

$\mapsto\rm{\dfrac{3}{5}+\dfrac{4}{5}}$

$\mapsto\rm{\dfrac{3+4}{5}}$

$\mapsto\rm{\dfrac{7}{5}}$

$\: \: \: \: \star\boxed { \bf \green{answer \: will \: be \: \frac{7}{5} }}$