* L
Fig. la
A projectile is launched with velocity u at an angle a to the horizontal. After time t the particle
reaches point P(x, y) and has velocity v. Let ß be the inclination of the line segment OP to the
horizontal and O be the direction of motion of the particle at P (Fig. 1a).
Write down the expressions for the acceleration, velocity and position of the particle at
any instant of time.
i. Show that tan pheta = tan a -
gt/u cos a
(ii)
Prove that tan beta =0.5 (tan a + tan pheta).

< 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)

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)

b

step-by-step explanation:

$205,000 here you go i had to guess it but i got it correct so yeah