Question

Find the angle of projection at which horizontal range and maximum height are equal

Answer 1

If you remember the following formula :  H / R = tan Ф / 4 ,  then its quick.
  H = R  => Ф = tan⁻¹ 4.
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Derivation:

In two dimensional projection of a particle in Earth's gravitational field, we have the following equations :

  x = u  Cos Ф * t    --- (3)
  y = u Sin Ф * t - 1/2 g t²    --- (1)
  v_x = velocity in horizontal direction :  u * cosФ
  v_y = vertical velocity upwards = u SinФ - g t
  v_y = 0  at  t = u SinФ/ g

  Then by (1) above,  y at t = u SinФ/ g  is equal to  = H
    H = u² Sin²Ф /2g    ---(2)

  In the time duration t = u SinФ/g, the projectile travels, a distance of half the Range.  Using (3) we get :
   R/2 = u² CosФ SinФ / g
   R = u² Sin2Ф / g    -- (4)

From (2) and (4) we get :
           H / R = Tan Ф / 4    ---- (5)

If  H = R are equal,  then the angle of projection =  tan⁻¹ 4 =  75.96 deg.

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sin A = 4 cos A
Sin² A = 16 Cos² A
cos² A = 1/17
Cos A = 1/√17
Sin A = 4/√17