given that sin x < cos x and 0° < x < 90°, state a possible value of x. explain your answer clearly.

See below.

Step-by-step explanation:

When x has a value between 0 and 90 the values of sin x ranges from 0, if x = 0 and 1 , if x = 90 degrees.

The cosine of x  has the reverse  value: cos x is 1 for x = 0 and 0 for x = 90 degrees.

At x = 45 the sine and cosine are equal in value.

So for sin x < cos x the value of x must be between 0 and 45  degrees.

So a possible value of x  is 30 degrees.

When x = 30 sin x = 0.5 and cos x = 0.866.

0.84 dang lotta people are looking for this lol or if you round up its 1 but the overall answer to that is 0.84

step-by-step explanation:

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

25.45584