How would I solve this problem? ?

diagonal length= √(60² + 20²)= 20√10

sinΘ= 10/(10√10); Θ≈ 18.4°

angle x= 36.8°

Draw a rectangle from point of intersection to top right corner with length 30cm and width 10cm

this cuts the acute angle in half. label a the other angle (of rectangle) is 90-a

this gives you:

sin(a) = 10/h

sin(90-a) = 30/h

rearrange

h = 10/sin(a)

h = 30/sin(90-a)

10/sin(a) = 30/sin(90-a)

10sin(90-a) = 30sin(a)

Identity: sin(α – β) = sin(α) cos(β) – cos(α) sin(β)

sin(90 – a) = sin(90) cos(a) – cos(90) sin(a) = cos(a)

10cos(a) = 30sin(a)

1/3 = tan(a)

a = arctan(1/3)

a = 18.43

remember this is only half, so double it.

2a = 36.9

36.9