Question

15)
If tan A = 3/4 then sinA + COSA=
3\5
7\5
4\5​

Answer 1

Answer:

Option b is correct .

Step-by-step explanation:

tan A = 3/4= P/H

H

${h}^{2} = {p}^{2} + {b}^{2}$

H^2= 3^2+4^2

H^2= 9+16

H^2=25

$h = \sqrt{25}$

H=5

= sin A + cos A

= 3/5+4/5

=7/5

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Answer 2

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Q1) If tanA=3/4 then sinA+cosA=?

$\huge\underline\mathcal\red{Answer}$

$\: \: \heartsuit \boxed{ \bf{in \: right \: angled \: triangle \triangle}}$

$\bullet\bf{SinA=\dfrac{opposite \: side}{Hypotenuse }=\dfrac{3}{5}}$

$\bullet\bf{CosA=\dfrac{Adjacent\: side}{Hypotenuse }=\dfrac{4}{5}}$

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

$\underline{\boxed{\bf{To \: Find \to \: SinA+CosA}}}$

$\implies\displaystyle\sf{SinA+CosA}$

$\implies\displaystyle\sf{Put \: the\: calculated \: values }$

$\implies\displaystyle\sf{\dfrac{3}{5}+\dfrac{4}{5}}$

$\implies\displaystyle\sf{\dfrac{3+4}{5}}$

$\implies\displaystyle\sf{\dfrac{7}{5}}$

$\clubsuit\underline{\boxed{\sf{the \: value\: of\: SinA+CosA=\dfrac{7}{5}}}}$