From the top of hill which is 200m high the angle of

Solution

Let CE is a pole of height h meter. AD is a hill of height 200m. Let distance between bottom of hill to pole = x m. According to question ∠XAC=300=∠CB and ∠XAE=600=∠AED From right angled ΔADE tan 600=AD/DE √81=(200/x) X = (200/√3) m From right angled ΔABC Tan300 = (AB/BC) = (AB−BD)/x (1/√3)= (200−h)/x X = √3(200−h) (200/√3)= √3(200−h) (200/3) = (200−h) h = 200−(200/3) = (600−200)/3 = 400/3 = 133.33m Hence, height of pole =133.33m