A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 3 2 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall

dα/dt  =  - 0,25  rad/sec

Step-by-step explanation:

The vertical wall, the ground from the vertical wall up to the ladder, and the ladder shape a right triangle. angle α  will be the angle between the ladder and the ground, then:

cos α  = distance between the wall and the down end of the ladder (x)  /  length of the ladder

cos α   =  x/ L

- sin α * dα/dt   =  dx/dt * L /L²

- sin α * dα/dt   =  dx/dt /L

When   the down end of the ladder is 8 ft apart from the wall

cos  α   =  8 / 10   =  0,8  then    cos² α  +  sin²α  = 1

sin² α  =  1 - cos²α        sin²α =  1 - (0.8)²    sin²α  = 0.36

sin α  = 0.6

dα/dt  =  ?

dx/ dt  =  3/2

L  = 10 ft

x  =  8 ft

By substitution

- 0.6*dα / dt   = (3/2) / 10

dα/dt  =  -  1.5 / 6

dα/dt  =  - 0,25  rad/sec

(-) means the angle is decreasing

b.an increase in a causes b to decrease

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.