From tony's seat in the classroom, his eyes are 1.0 m above ground. on the wall 4.2 m away, he can see the top of a blackboard that is 2.1 m above ground. what is the angle of elevation, to the nearest degree, to the top of the blackboard from tony's eyes?
the answer is 27 degrees but i dont know how to get that. can someone show me the steps . will give brainliest.

15 degrees

Step-by-step explanation:

Draw a horizontal segment approximately 4 inches long. Label the right endpoint A and the left endpoint C. Label the length of AC 4.2 meters. That is the horizontal distance between the eye and the blackboard.

At the right endpoint, A, draw a vertical segment going up, approximately 1 inch tall. Label the upper point E, for eye. Label segment EA 1 meter since the eye is 1 meter above ground.

At the left endpoint of the horizontal segment, point C, draw a vertical segment going up approximately 2 inches. Label the upper point B for blackboard. Connect points E and B. Draw one more segment. From point E, draw a horizontal segment to the left until it intersects the vertical segment BC. Label the point of intersection D.

The angle of elevation you want is angle BED.

The length of segment BC is 2.1 meters. The length of segment CD is 1 meter. That means that the length of segment BD is 1.1 meters.

To find the measure of angle BED, we can use the opposite leg and the adjacent leg and the inverse tangent function.

BD = 1.1 m

DE = 4.2 m

tan <BED = opp/adj

tan <BED = 1.1/4.2

m<BED = tan^-1 (1.1/4.2)

m<BED = 15

15 degrees

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x