A climber is ascending the face of a 500-foot high vertical cliff at 5 feet per minute. Her climbing partner is observing from 150 feet away. Find the function that describes the amount of time climbing as a function of the angle of elevation of the observer’s line-of-sight to the climber. Does the viewing angle ever reach 75 degrees?

a) t (α)  =  30 * tan α

b) α =  75 °  is not possible because

Step-by-step explanation:

The vertical cliff  (y),  the observing partner ( 150 ft ) from the point o ( the common point for the cliff and the ground ) between the cliff, and the line of sight from the observer and the climber, shape a right triangle. Hypothenuse is the distance between the partners, the cliff, and distance of the observer to the cliff are the legs.

Then

α is the elevation angle

tan α  =  y / 150           (1)

y(t)  = elevation of the climber

y(t) = speed * time

y(t)  =  5 * t

tan α  =  5 *t / 150

tan α  =  t / 30

t (α)  =  30 * tan α

b) α =  75 °  is not possible because:

tan α  =  y / 150              tan 75°  =  3.73

Then     y  =  3.73 * 150  =  559 ft and the cliff is only 500 ft

351.23

step-by-step explanation:

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116