Urgent:數學大難題(Trigo.)---計數叻,Maths.老師.../自稱數皇的人請進!!!

Let h be the height of the vase
d be required deep
r be the water base radius when water reach required deep

tan 30∘ = 4/h
h = 4√3

volume of vase = (1/3)π(4)^2 h
= (16/3)π(4√3)
= (1/3)(64√3)π
half of volume of vase = (1/3)(32√3)π

4/h = r/d
1/√3 = r/d
r = d/√3

volume of water = (1/3)π(r)^2 d = (1/3)(32√3)π
(1/3)π(d/√3)^2 d = (1/3)(32√3)π
1/9πd^3 = (1/3)(32√3)π
d^3 = 96√3
d = 5.498918548
= 5.50 (3 S.f.)

So, the required deep = 5.50 cm (3 s.f.)

let h be the height of the cone, h/2is the height when the volume of cone s exactly half of the total volume

tan30=4/h
4/tan30=h

therefore, h/2=(4/tan30)/2
h/2=2/tan30
答案你記啦..我相信你識得按計數機的..

點解咁多人唔自己做功課, 要係度問 ?!

問題關鍵1: 體積比例 = (邊長比例) ^ 3
問題關鍵2: height, base radius, 同30度 既關係.

學生應自己計下.... 次次要問答案, 考試點呀 ? 做野點呀 ?