Question

given 15cot A = 8 , find sinA and cosA.............. (show ur intelligency) its so easy btw....​

Answer 1

Given :-

15cotA=8

To Find :-

The value of sinA and cosA

Then ,

$\longmapsto\sf 15cotA=8\\ \\ \longmapsto\sf cotA=\dfrac{8}{15}$

• Now By Pythagoras !

$\boxed{\sf{ Hypotenuse^2= Base^2+Perpendicular^2}}$

We know that

$\sf CotA=\dfrac{Base}{perpendicular}\\ \\ \sf then \ \star base(B) = 8 \ \ \star perpendicular (P)=15$

Now find the Hypotenuse (H)

$\longmapsto\sf H^2= (8)^2+(15)^2\\ \\ \longmapsto\sf H^2= 64+225\\ \\ \longmapsto\sf H=\sqrt{289}\\ \\ \longmapsto\sf H= 17$

Now the value of SinA and CosA

• We know that

$\bigstar{\boxed{\sf {Sin\theta= \dfrac{Perpendicular}{Hypotenuse}}}}$

$\bigstar{\boxed{\sf{Cos\theta =\dfrac{Base}{Hypotenuse}}}}$

We have.

• Base = 8

• Hypotenuse=17

• Perpendicular= 15

$\longmapsto\sf SinA=\dfrac{Perpendicular}{Hypotenuse}\\ \\ \longmapsto\sf SinA=\dfrac{15}{17}$

$\longmapsto\sf CosA=\dfrac{Base}{Hypotenuse}\\ \\ \longmapsto\sf CosA=\dfrac{8}{17}$

$\bigstar{\boxed{\sf{\purple{SinA=\dfrac{15}{17}}}}}$

$\bigstar{\boxed{\sf{\purple{CosA=\dfrac{8}{17}}}}}$

Answer 2

Trigonometric Ratios

Formulation of ratios.

Draw a right-angled triangle ABC (refer to the attachment), whose sides are AB = a, BC = b and CA = h.

AB is the perpendicular, BC is the base and CA is the hypotenuse.

So, h = √(a² + b²)

Let, ∠AOB = θ

Then, AB / CA = a / h = sinθ

BC / CA = b / h = cosθ

AB / BC = a / b = tanθ

CA / AB = h / a = cosecθ

CA / BC = h / b = secθ

BC / AB = b / a = cotθ

Solution.

Given, 15 cotA = 8

or, cotA = 8 / 15 = base / vertical

From the above formulae, we can say

base (b) = 8 and vertical (a) = 15

Then hypotenuse (h) = √(a² + b²)

= √(15² + 8²)

= √(225 + 64)

= √289

= 17

Now, sinA = vertical (a) / hypotenuse (h)

= 15 / 17

and cosA = base (b) / hypotenuse (h)

= 8 / 17

Answer: sinA = 15/17 & cosA = 8/17.