Jim is designing a seesaw for a children's park. the seesaw should make an angle of 30° with the ground, and the maximum height to which it should rise is 2 meters, as shown below: a see-saw is shown with one end on the ground and the other in the air. the see-saw makes an angle of 30 degrees with the ground. the height of the see-saw from the ground, at the other end, is labeled 2 meters. what is the maximum length of the seesaw? 3.00 meters 3.5 meter 4.00 meters 4.5 meters

given, a seesaw makes an angle with the ground.

the maximum height of the seesaw is 2m.

so to find the length of the seesaw, we will draw a right angled triangle first by using the length, height and angle of the seesaw given.

let, ab = x be the length of the seesaw which is the hypotenuse in the right angled triangle abc.

and ac = 2m be the height of it which is opposite in the right angled triangle.

so here opposite given and we have to find the hypotenuse. we know that the ratio of opposite and hypotenuse is for the trigonometric function sine.

so we can write,

we know the value of is . we will plug it in here to get x.

now to find x, we will cross-multiply. we will get,

so we have got the length of the seesaw.

the required length of the seesaw = 4m.

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)