Question

cosA=3/5 then find the value of sin square A+cos square A​

Answer 1

Step-by-step explanation:

sorry I am in class 9 so I can't understand or able to give answer of this question

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

cos A = 3/5

$\underline{\bf{Explanation\::}}$

Firstly, attachment a figure of right angled Δ A/c to the given question.

As we know that,

$\boxed{\bf{cos\:\theta = \frac{Base}{Hypotenuse} }}$

A/q

$\mapsto\tt{cos\:A = \dfrac{3}{5} = \dfrac{BC}{AC} }$

$\underline{\underline{\tt{By\:\:using\:\:Pythagoras\:\:theorem\::}}}$

⇒ (Hypotenuse)² = (Base)² + (Perpendicular)²

⇒ (AC)² = (BC)² + (AB)²

⇒ (5)² = (3)² + (AB)²

⇒ 25 = 9 + (AB)²

⇒ (AB)² = 25-9

⇒ (AB)² = 16

⇒ AB = √16

⇒ AB = 4 unit

Now,

$\longrightarrow\tt{sin^{2} A + cos^{2} A}$

$\longrightarrow\tt{\bigg(\dfrac{Perpendicular}{Hypotenuse} \bigg) ^{2} + \bigg(\dfrac{Base}{Hypotenuse} \bigg)^{2}}$

$\longrightarrow\tt{\bigg(\dfrac{AB}{AC} \bigg) ^{2} + \bigg(\dfrac{BC}{AC} \bigg)^{2}}$

$\longrightarrow\tt{\bigg(\dfrac{4}{5} \bigg) ^{2} + \bigg(\dfrac{3}{5} \bigg)^{2}}$

$\longrightarrow\tt{\dfrac{16}{25} + \dfrac{9}{25} }$

$\longrightarrow\tt{\dfrac{16+9}{25}}$

$\longrightarrow\tt{\cancel{\dfrac{25}{25}}}$

$\longrightarrow\bf{1}$

Thus,

The sin²A + cos²A will be 1 .