问题1.马戏团的一位艺术家正在攀爬一条20 m长的绳索,该绳索被紧紧地拉紧并从垂直杆的顶部绑到地面。如果绳索与地面的夹角为30°,则求出杆的高度(见图)。
解决方案:
In rt ∆ABC,
AB = pole = ?
AC = rope = 20m
sinθ =
sin30° =
AB = 1/2 * 20
AB = 10m
Height of pole = 10m
问题2。由于暴风雨而折断的树,折断的部分弯曲,使树的顶部与地面接触,与地面成30度角。树木的脚到顶部接触地面的距离为8 m。找到树的高度。
解决方案:
In rt ∆ABC,
BC = 8m
= tan30°
= 1/√3
AB = 8/√3 -(1)
Now,
= cos30°
8/AC = √3/2
√3AC = 16
AC = 16/√3 -(2)
From (1) and (2)
Height of tree = AB + AC
= 8/√3 * 16√3
= 8√3 m
8 * 1.73 = 13.84m
The height of the tree is 13.84
问题3.承包商计划安装两张幻灯片,供孩子们在公园玩耍。对于5岁以下的孩子,她更喜欢滑梯,其顶部的高度为1.5 m,并且与地面倾斜30°,而对于较大的孩子,她希望陡峭的滑梯高度为3 m,并与地面倾斜60°。每种情况下幻灯片的长度应该是多少?
解决方案:
In rt ∆ABC,
AB = 1.5m
AC = side = ?
= sin30°
1.5/AC = 1/2
AC = 1/5 * 2
AC = 3m
In rt ∆PQR,
PQ = 3m
PR = side = ?
= sin60°
3/PR = √3/2
√3 PR = 6
PR = 6/√3
6/√3 * √3/√3
= 2√3
= 2 * 1.73
= 3.46m
问题4.塔的顶部与地面上某个点的仰角为30°,该点与地面的点相距30 m。找到塔的高度。
解决方案:
In rt ∆ABC,
AB = tower = ?
BC = 30m
= tan30°
AB/30 = 1/√3
AB = 30/√3
AB = 30/√3 * √3/√3
= (30√3)/3 = 10√3
= 10 * 1.73
= 17.3m
The height of tower 17.3m
问题5.风筝在离地面60 m的高度飞行。连接到风筝的字符串被暂时绑在地面上的一个点。字符串与地面的倾斜度为60°。查找字符串的长度,假设有字符串中没有松弛。
解决方案:
In rt ∆ABC,
AB = 6Om
AC = string = ?
= sin60°
60/AC = √3/3
√3 AC = 60 * 2
AC = 120/120/(√3) * √3/√3
120/√3 * √3/√3
40 = √3
40 * 1.73 = 69.20m
Length of the string is 69.20m
问题6.一个1.5 m高的男孩正站在30 m高的建筑物的某个距离处。当他走向建筑物时,从他的眼睛到建筑物顶部的仰角从30°增加到60°。找到他走向建筑物的距离。
解决方案:
In fig AB = AE – 1.5
= 30 – 1.5
= 28.5
In rt ∆ABD,
= tan30°
= 28.5/BD = 1/√3
BD = 28.5√3 -(1)
In rt ∆ABC,
= tan60°
28.5/BC*√3
√3 BC = 28.5
BC = 28.5/√3 -(2)
CD = BD − BC
= 28.5√3 – 28.5/√3
= 28.5(2/√3)
57/√3 * √3/√3 = (57√3)/3 = 19√3
19 * 1.73 = 32.87m
The boy walked 32.87m towards the building.
问题7.从地面上的某个点来看,固定在20 m高建筑物顶部的输电塔的底部和顶部的仰角分别为45°和60°。找到塔的高度。
解决方案:
In fig:
AB = tower = ?
BC = building = 20m
In rt ∆BCD
= tan45°
20/CD = 1/1
CD = 20
In rt. ∆ACD,
= tan60°
AC/20 = √3/1
AC = 20√3 -(1)
AB = AC-BC
20√3 – 20
20(√3 – 1)
20(1.732 – 1)
20(0.732)
14.64m
The height of the tower is 14.6m
问题8.一个高1.6 m的雕像站在基座的顶部。从地面上的某个点开始,雕像顶部的仰角为60°,从同一点开始,底座的顶部仰角为45°。找到基座的高度。
解决方案:
In fig: AB = statue = 1.6m
BC = pedestal = ?
In rt ∆ACD
= tan60°
1.6 + BC/CD = √3
√3 CD = 1.6 + BC
CD = 1.6+BC/√3 -(1)
In rt ∆BCD,
= tan45°
= 1/1
CD = BC
From (1)
1.6 + BC/√3 = BC/1
√3 BC = 1.6 + BC
1.732 BC – 1 BC = 1.6
0.732 * BC = 1.6
BC = 1.6/0.732
BC = 16/10 * 100/732 = 1600/732
BC = 2.18m
Height of pedestal is 2.18m
问题9.建筑物顶部相对于塔底高度的仰角为30°,而建筑物顶部相对于塔底高度的仰角为60°。如果塔高为50 m,请找到建筑物的高度。
解决方案:
In fig:
AB = tower = 50m
DC = building = ?
In rt.∆ABC,
= tan60°
√3 BC = 50
BC = 50/√3
In rt. ∆DCB
= tan30°
= 1/√3
DC = 50/√3
DC = 50/√3 * 1/√3
DC = 50/3
DC =
The height of the building is m
问题10.等高的两根电线杆在道路的两侧彼此相对,高度为80 m。从道路上的两点之间,两极顶部的仰角分别为60°和30°。找到极点的高度以及点到极点的距离。
解决方案:
AB and CD on equal poles.
Let their height = h
Let DP = x
Then PB = BD – x
In rt. ∆CDP,
= tan60°
h/x = √3/1
h = √3 x -(1)
In rt. ∆ABP
= tan30°
h/(80 – x) = 1/√3
h = (80 – x)/√3 -(2)
From (1) and (2)
(√3 x)/1 = 80 – x/√3
3x = 80 – x
3x + x = 80
4x = 80
X = 80/4
X = 20
Putting values of X in equation 1
h = √3 x
h = √3(20)
h = 1.732(20)
h = 34.640
Height of each pole = 34.64m
The point is 20m away from first pole and 60m away from second pole.