It is known that α is an acute angle and sinα=k-2. Try to find the range of k.

0 & ltα& lt； 90

0 & ltsinα& lt； 1

0 & ltk-2 & lt； 1

So 2 < k & lt2

It is known that α is an acute angle, and tan α = 3: 4 (that is, three quarters). Try to find the value of cosα.

Let diagonal =3a, and adjacent right angles =4a.

Hypotenuse of solution =5a

So cosα= adjacent right angles/hypotenuse =4/5.

Given that α is an acute angle and cotα=m, try to find the value of sinα.

tanα= 1/cotα= 1/m

The process is the same as above, and the hypotenuse = √ (1+m 2).

Sinα= diagonal/hypotenuse =1√ (1+m 2)

The triangle ABC, ∠ A = 90, AD⊥BC, the vertical foot is D, ∠B is αAC=b to find BD.

BD=b/sinα-b*sinα