Question

find the distnace between the points (sin20,0) and (0,sin70)​

Answer 1

hope it is correct and help you....

Answer 2

Answer:

1 units.

step-by-step explanation:

We know that if two pts. (x,y) and (a,b) are given

By distance formula,

distance between these pts is given by

√[ ${(x-a)}^{2}$+${(y-b)}^{2}$ ]

here,

x = sin 20

y = 0

a = 0

b = sin 70

so , the distance between these pts.

= √ [ ${(sin20-0)}^{2}$+${(0-sin70)}^{2}$ ]

= √ [ ${sin}^{2}$20 + ${sin}^{2}$70 ] .................(1)

but we know that,

sin (90-@) = cos @

=> sin 70 = sin (90-20) = cos 20

=> ${sin}^{2}$70 = ${cos}^{2}$20

Substituting this value in equation 1,

we get,

= √[ ${sin}^{2}$20 + ${cos}^{2}$20 ]

But we know that,

${sin}^{2}$@ + ${cos}^{2}$@ = 1

so,

√[ ${sin}^{2}$20 + ${cos}^{2}$20 ]

= √1 units

= 1 units.