第 10 类 RD Sharma 解决方案 – 第 12 章三角函数的一些应用 – 练习 12.1 |设置 3

问题 53. 从 60 m 高的建筑物 AB 的顶部观察，垂直灯柱 CD 的顶部和底部的俯角分别为 30° 和 60°。找

(i) AB 和 CD 之间的水平距离。

(ii) 灯柱的高度

(iii) 建筑物与灯柱的高度差。

解决方案：

Let us considered AB is the building and CD is the vertical lamp

From point A, the top of the building angles of depression of C and D are 30° and 60° 

Height of building AB = 60 m

Let us assume that the height of CD = h

Draw CE || DB || AX

∴ ∠ACE = ∠XAC = 30° and ∠ADB = ∠XAD = 60°

EB = CD = h and AE = 60 – h

Let DB = CE = x

Now in ΔACE,

tan 30° = (60 – h)/x

1/√3 = (60 – h) / x

x = √3 (60 – h)        …….(i)

Similarly, in △ADB,

tan 60° = AB/DB = √3 = 60/x

x = 60/√3 = 34.64 m

Hence, 20√3 = √3(60 – h)

h = 40 m

Hence, the height of vertical lamp is 40 m.

(i) Horizontal distance b/w them = DB = x

20√3 m = 3464 m

(ii) Height of vertical lamp = 40 m.

(iii) Difference b/w the heights = 60-40 = 20 m.

问题 54. 两艘船从相反的方向接近中海的一座灯塔。两艘船的灯塔顶部仰角分别为30°和45°。如果两艘船之间的距离是 100 m，求灯塔的高度。

解决方案：

Let us considered LH is the lighthouse and 

A and B are two boats on the opposite directions of the lighthouse 

which are making the angle of elevation of the top L of the lighthouse as 30° and 45°

Given that, AB = 100m

Let us considered LH is h and BH = x,

HA = 100 – x

Now in △LAH,

tan 30° = LH/ HA = h/(100 – x) = 1/√3 = h/(100 – x)

x = 100 – √3h           …….(i)

Similarly, in △LHB,

tan 45° = LH/BH = 1 = h/x

h = x  ……(ii)

From eq(i) and (ii), we get

h = 100 – √3h 

h + √3h = 100

h(√3+1) = 100

h = 100/(√3+1)

On rationalize above term and we get,

h = 50(√3 – 1)

Hence, the height of the lighthouse is 50(√3 – 1) m.

问题55，塔脚的山顶仰角为60°，塔顶与山脚的仰角为30°。如果塔高 50 m，山的高度是多少？

解决方案：

Let us considered TR be the tower, HL is the hill and 

angles of elevation of the top of the hill are 60° and top of the tower is 30

Given, TR = 50 m

Let us considered the height of hill HL = h

and RL = x

Now in △TRL,

tan 30° = TR/RL = 50/x = 1/√3

x = 50√3         …….(i)

Similarly, in △HRL,

tan 60° = HL/RL = h/x

√3 = h/50√3

h = 50√3 x √3 = 150 m

Hence, the height of the hill is 150 m.

问题 56. 从一个 150 m 高的悬崖顶部观察一艘移动的船正在远离悬崖移动。船的俯角在 2 分钟内从 60° 变为 45°。以 m/h 为单位求船的速度。

解决方案：

Let us considered the distance BC be x m and CD be y m.

In △ABC,

tan 60° = AB/BC = 150/x

√3 = 150/x

x = 150/√3      …….(i)

In △ABD,

tan 45° = AB/BD = 150 / (x + y)

x + y = 150

y = 150 – x

Now, using eq(i) we get,

y = 150 – 150/√3 = 150(√3 – 1) / √3

Time is taken to move from point C to point D in 2 min = 2/60 = 1/30 h

Now, Speed = Distance / Time = y / (1/30)

(150(√3-1) / √3) / (1/√3) = 1902 m/h

问题 57. 从一个 120 m 高的塔的前端，一名男子观察到两辆汽车在塔的相对两侧并与塔的底部成一条直线，俯角分别为 60° 和 45°。找出汽车之间的距离。 （取√3 = 1.732）

解决方案：

Let us considered BD be the tower and A and C be the two points on the ground.

Given that, BD, the height of the tower = 120m

∠BAD = 45°, ∠BCD = 60°

tan 45° = BD/BA 

1 = 120 / BA

BA = 120 m              ————–(i)

tan 60° = BD / BC

√3 = 120/BC

BC = 120/√3 = 120√3 / 3 = 40√3 = 69.28 m        —————-(ii)

Distance b/w the two points A and C

AC = BA + BC

120 + 69.28 = 189.28 m

问题 58. A 和 B 两点在塔的同一侧，与塔底在同一条直线上。这些点从塔顶的俯角分别为60°和45°。如果塔的高度为 15 m，则求这些点之间的距离。

解决方案：

Let us considered TR be the tower and 

A, B are two objects which make angles of elevation with the top of the tower as 45° and 60°.

Given that, the height of Tower TR = 15 m

Let us considered, the distance b/w B and A = x m

Now in △TAR,

tan 45° = 15/AR

1 = 15/AR

AR = 15

Similarly, in △TBR,

tan 60° = TR/BR = 15/BR

√3 = 15/BR = BR = 15/√3

Now AB = x = AR – BR,

15 – 15/√3 = 15 – 15 x 1.732 = 6.34 m

So, the distance b/w A and B = 6.34 m.

√3/2 = 12.7/BP = BP = 12.7 x 2/√3

BP = 14.68 km

Hence, P fire station is nearer and its team will reach the building after coming 14.6 km.

问题 59. B 楼发生火灾，通过电话向两个消防站 P 和 Q 报告，在笔直的道路上相距 20 公里。 P 观察到火与道路成 60° 角，Q 观察到它与道路成 45° 角。哪个站应该派出它的团队，这个团队需要旅行多少？

解决方案：

Let us considered B is the building one fire and P and Q the fire stations which are 20 km apart i.e., PQ = 20 km.

Given that, P and Q observes the angles with B, as 60° and 45°

Now, draw BA ⊥ PQ

Let AB = h Now in right ΔBAQ,

tan 45° = h/AQ

1 = h/AQ = AQ = h

Hence, PA = 20 – h

Similarly, in △BAP,

tan 60° = BA/Ap

√3 = h / (20 – h)

h(√3 + 1) = 20√3

On rationalize above term and we get,

h = 20√3(√3 – 1) / 2 = 12.7 km

Hence, AQ = 12.7 km and AP = 20 – 12.7 = 7.3 km

Now sin 45° = P/H

1/√2 = h / BQ

BQ = 12.7 x √2 = 12.7 x 1041

BQ = 17.91 km

Similarly, sin 60° = BA/BP = 12.7 / BP

BP = 12.7 x 2 / √3

BP = 14.68 km.

Hence, P fire station is nearer and its team will reach the building after coming 14.6 km.

问题 60. 船甲板上的人在水面以上 10 m。他观察到悬崖顶部的仰角是 45°，而底部的俯角是 30°。计算悬崖到船的距离和悬崖的高度。

解决方案：

Let us assume M is a man on the deck MN such that MN = 10 m, AB is the cliff

So, from M the angle of elevation of A is 45°

and angle of depression of B is 30°

Now, draw MC || NB

Hence ∠MBN = ∠CMB

Let us considered AB = h, NB = MC = x and AC = h – 10

Now in △AMC,

tan 45° = (h – 10) / x

1 = (h – 10) / x

x = h – 10        …….(i)

Similarly, in △MBN,

tan 30° = MN/BN 

1/√3 = 10/x

x = 10√3   ……(ii)

From eq(i) and (ii), we get

h – 10 = 10√3

h = 10 + 10√3 = 27.32 

and x = h – 10 = 27.32 – 10 = 17.32

Hence, the height of the cliff is 27.32 m and the distance of the ship from the cliff is 17.32 m.

问题 61. 一名男子站在高于水面 8 m 的船甲板上。他观察到山顶的仰角为 60°，山脚的俯角为 30°。计算小山到船的距离和小山的高度。

解决方案：

Let us assume M is the man on the deck MN such that MN = 8 m and AB is the hill 

Now, from M, the angle of elevation of A is 60° and angle of depression of B is 30°.

Draw MC || NB

Hence, ∠MBN = ∠CMB

Let us assume AB = h, NB = MC = x

AC = h – 8

Now in △AMC,

tan 60° = AC/MC = h – 8 / x = √3

x = h – 8 / √3 …….(i)

Similarly, in △MBN,

tan 30° = MN/NB = 8/x

1/√3 = 8/x 

x = 8√3    …….(ii)

From eq(i) and (ii), we get

8√3 = (h – 8)/√3

h = 32

and x = 8√3 or 13.858 m

Hence, the height of the hill = 32 m and the distance between ships = 13.858 m.

问题 62. 有两座寺庙，在河的两岸各一座，正对面。一座寺庙高50 m。从这个太阳穴的顶部看，另一个太阳穴的顶部和脚部的下陷角分别为30°和60°。求河流的宽度和另一座寺庙的高度。

解决方案：

Let us assume AB and CD are two temples on the banks of the river.

Given that, AB = 50 m

Now, from A, the angles of depression of the top and bottom of the other temple are 30° and 60°.

Let us assume CD = h 

So that AE = 50 – h, let BD = EC = x

Now in △ABD,

tan 60° = 50/x = √3 = 50/x

x = 50√3/3     …….(i)

Similarly, in △AEC,

tan 30° = AE/EC = (50 – h)/x

1/√3 = (50 – h)/x

x = √3(50 – h)            …….(ii)

From eq(i) and (ii), we get

50√3/3 = √3(50 – h)

h = 33.33 m

Hence, the height of the second temple is 33.33 m and the width of the river = 50√3/3 = 28.23 m.

问题 63. 飞机从地面一点的仰角是 45°。飞行 15 秒后，仰角变为 30°。如果飞机在 3000 米的高度飞行，求飞机的速度。

解决方案：

 Let us assume A is the plane flying in the sky at its height of 3000 m i.e., AB = 3000 m

Now, P is a point on the ground which from an angle of elevation of 45° at A and 

then after a flight of 15 seconds at A’ the angle of elevation because 30°.

Let us assume PB = y and BB’ = x m

Now in △APB,

tan 45° = AB/PB = 3000/y

y = 3000 m.           …….(i)

Similarly, in △A’PB’

tan 30° = A’B’/PB’ = 1/√3 = 3000/(x + y)

x + y = 3000√3

x = 3000√3 – y

x = 3000(√3 – 1)

= 2196 m.

Hence the distance of 2196 m is covered in 15 seconds.

Speed of the plane = (2196x60x60)/15 = 527040 m/hr = 527.04 km/hr

问题 64。在 60°仰角观察到一架水平飞行在地面以上 1 公里处的飞机。 10 秒后，观察到其仰角为 30°。以公里/小时为单位求飞机的速度。

解决方案：

Let us considered A be the aeroplane and AB is the height which 1 km and 

makes an angle of elevations of 60° from a point P on the ground. 

Now, after moving 10 second’s flight, the angle of elevation becomes 30° from P

Giventhat, A’B’ = AB = 1 km = 1000 m

Let us assume PB = y and BB’ = x

Now in ΔAPB,

tan 60° = AB/PB = 1000/y = √3

y = 1000/√3            …….(i)

Similarly, in △A’PB’,

sin 30° = A’B’/PB’ = 1/√3 = 1000/(x + y)

x + y = 1000√3

x = 1000(√3 – 1)

= 1154.73 m

Now 1154.73 m is covered in 10 seconds.

Hence, speed per hour = (1154.73 x 60 x 60)/10 x 1000 = 415.7 km/hr.

问题 65. 一棵站在水平面上的树向东倾斜。在位于其正西正西的距离为 a 和 b 的两点上，顶部的仰角分别为 a 和 p。证明顶部离地面的高度为((b – a) tanα tanβ) / (tanα – tanβ)。

解决方案：

Let us considered CD is the tree which is leaning towards the east and 

A and B are two points on the West making angles of elevation with top C of the tree as α and β

A and B are at the distance of a and b from the foot of the tree CD, then AD = a, BD = b

Now, draw CL ⊥ BD produce and let DL = x and CL = h

So, in △ CAL,

tan α = h/(a + x)

a + x = h/tanα       …..(i)

Similarly, in △CBL,

tan β = CL/BL = h/(b + x) = h/tanβ      …..(ii)

From eq(i) and (ii), we get

x = h/tanα – α

x = h/tanβ – β

h/tanβ – β = h/tanα – α

h(1/tabβ – 1/tanα) = b – a

h = (β – α) tanα tanβ / (tanα – tanβ)

Hence proved!

问题 66. 文具云从湖上 2500 米处的仰角为 15°，其在湖中反射的俯角为 45°。湖平面以上的云层高度是多少？ （使用棕褐色 15° = 0.268）

解决方案：

Let us considered C is the cloud over a lake LK

Now, from a point, M which, is 2500 m above the lake level, 

the angle of elevation of C is 15° and 

the angle of depression of the reflection of C in the lake which is R is 45°.

Let us assume CK = h then the reflection KR = h

BK = ML = 2500 m then CB = h – 2500

Let MB = LK = x

Now in △CMB,

tan 15° = (h – 2500)/x = x = (h – 2500)/tan 15         …..(i)

Similarly, in △BMR,

tan 45° = BR/MB

1 = (h + 2500) / x

x = h + 2500   ……(ii)

From eq(i) and (ii), we get

h + 2500 = (h – 2500)/tan15

h + 2500 = (h – 2500)/0.2679

h = 4329.67 m

Hence, the height of the cloud is 4329.67 m.

问题 67. 如果云从湖上 h 米点的仰角为 a，在湖中反射的俯角为 p，证明云到点的距离为 2hsecα/tanβ - tanα。

解决方案：

Let us considered C be the cloud and from a point, M which h m is above the lake level 

angle of elevation is α and angle of reflection of the cloud C, is β

Let LK = MN = x

MC = y, height of cloud CK = p

TK = h, CN = p – h and KR = p

Now in △CMN,

cosα = MN/CM = x/y

x = y cosα   ……(i)

and tanα = (p – h)/x = p – h = x tanα

p = h + x tanα           ……(ii)

Similarly, in △MNR,

tanβ = NP/MN = (p + h)/x

p + h = x tanβ     ……(iii) 

From eq(ii) and (iii), we get

h + x tanα + h = x tanβ

x = 2h/(tanβ – tanα)

y cosα = 2h/(tanβ – tanα)

y = 2h secα / (tanβ – tanα)

Hence, distance of cloud from point = 2h secα / (tanβ – tanα)

问题 68。从垂直于水平直线道路上方的飞机，观察到飞机相对两侧的两个连续里程碑的俯角为 a 和 p。证明飞机在道路上方的高度（以英里为单位）由 tanα tanβ / tanα + tanβ 给出。

解决方案：

Let us considered A is aeroplane and C and D are two such points that the 

angles of depression from A are α and β respectively and CD = 1 km

Let us assume the height of the plane be h

XY || CD

Hence ∠C = α, ∠D = β

Let BC = x km then BD = (1 – x) km

Now in △ACB,

tanα = AB/BC = h/x

x = h/tanα

Similarly, in △ABD,

tanβ = AB/BD = h/(1 – x)

h = tanβ – x tanβ

= tanβ – h/tanα x tanβ 

h = (tanβ tanα) / (tanα + tanβ) 

Hence proved

问题 69. PQ 是给定高度 a 的柱子，AB 是一定距离处的塔。如果 a 和 p 是塔顶 B 的仰角，分别在 P 和 Q 处。找出塔的高度及其与柱子的距离。

解决方案：

Let us assume PQ is posted and AB is the tower 

Now, the angles of elevation of B, from P and Q are α and β

PQ = α

Let us assume AB = h and distance between the tower and the post = x

PR || QA

Hence, RA = α and BR = h – α

Now in △ABQ,

tanα = BA/QA = h/x

h = a tanα    ……(i)

Similarly, in △BPR,

tanβ = BR/PR = (h – α) / x

x tanβ = h – α          ……(ii)

From eq(i) and (ii), we get

x(tanα – tanβ) = α

x = α / (tanα – tanβ)

and h = x tanα = α / (tanα – tanβ) x tanα 

= α tanα / (tanα – tanβ)

Hence height of tower is α tanα / (tanα – tanβ) and distance b/w them is α / (tanα – tanβ)

问题 70. 梯子靠在壁上，与水平面成一定角度。它的脚被拉离墙壁一段距离，使其沿着墙壁向下滑动距离 b，与水平面形成角度 p。证明 a / b = cosα - cosβ / sinβ - sinα

解决方案：

 From the figure we conclude that

AC and ED is the same stair, so AC = ED

Now cosα = CB/AC

Similarly, cosβ = DB/ED = (α + CB)/AC

sinα = AB/AC = b + EB / AC

sinβ = EB/ED = EB/AC

Now we solve RHS = cosα − cosβ / sinβ − sinα

= (CB/AC – ((α + CB)/AC)) / (EB/AC – (b + EB / AC))

= (CB – a – CB / EB – b – EB) 

= a/b

LHS = RHS

Hence proved

问题 71. 一座塔在其底部平面中的点 A 处对角，塔脚在 A 上方 b 米处的俯角为 p。证明塔的高度为 b tan α cot β。

解决方案：

Let us considered TR is the tower that subtends angle α at a point A on the same plane

Given that, AB = b and angle of depression of R from B is β

BX || AR

Hence, ∠ARB = ∠XBR = β

Let us assume the height of tower TR = h

and AR = x

Now in △ATR,

tanα = h/x 

x = h/tanα      …….(i)

Similarly, in △ABR,

tanβ = AB/AR = b/x 

x = b/tanβ  …….(ii)

From eq(i) and (ii), we get

h/tanα = b/tanβ

h = b tanα cotβ

Hence, height of the tower is b tanα cotβ.

Hence proved

问题 72. 一个 1.5 m 高的观察者距离一个 30 m 高的塔 28.5 m。从他的眼睛确定塔顶的仰角。

解决方案：

Let us assume TR is the tower and CD is the observer 

Given that CD is 28.5 m away from the tower TR

Height of the tower TR = 30 m

and height of observer CD = 1.5 m

Let us assume θ be the angle of elevation of the top of the tower

from the eye of the observer CD,

Now, draw CE || DR

Hence, CE = DR = 28.5 m and ER = CD = 1.5 m

Hence, TE = 30 – 1.5 = 28.5 m

Now in △TCE,

tanθ – TE/CE = 28.5/28.5 = 1 = tan 45°

Hence, the angle of elevation of the top of the tower from his eye is 45°.

问题 73. 木匠为电工制作凳子，边长为 0.5 m，离地高度为 1.5 m。此外，每条腿与地面成 60° 角倾斜。找出每条腿的长度以及要放在相等距离处的两个台阶的长度。

解决方案：

Let us assume AC be the leg of stool whose top is a square-shaped of side AB 

Given that, AB = 0.5 m

Also, the height of stool AL = 1.5 m, and angle of inclination by the leg of the stool = 60°

Let us assume AC = x m

Now in △ACL,

sin 60° = 1.5/x

√3/2 = 3/2x

x = 1.732 m

Hence, the length of leg = 1.732 m

There are two steps on equal distance,

Distance between two steps = 1.5/3 = 0.5 m

From the given figure 

RS || PQ || CD

Hence ∠R = ∠P = ∠C = 60

In △APM,

tan 60° = AN/RN = √3 = 0.5/RN

RN = 0.288 m 

Hence, RS = 0.5 + 0.2886 + 0.2886 = 1.077 m.

问题 74. 一个男孩站在地上，用 100 m 的字符串在 30° 的高度放风筝。另一个男孩站在一栋 10 m 高的建筑物的屋顶上，在 45° 的高度放风筝。两个男孩都在两个风筝的对面。找出第二个男孩必须有的字符串的长度，以便两只风筝相遇。

解决方案：

Let us considered K be the kite, A and B are two boys flying kites. 

Given that Boy B is standing on a building 10 m high and the string AK of the kite of boy A is 100 m 

Now let us assume h be the height of the kite and x is the length of the string of kite of second boy B

Then, KD = (h – 10) m

Now in △AKT,

sin 30° = h/100 = 1/2 = h/100

h = 50 m

Similarly, in △KDB,

sin 45° = KD/KB = 1/√2 = 40/x

x = 45.6560

Hence, the length of a string of the second kite is 40√2 or 45.656 m.

问题 75. 从灯塔的顶部观察，在灯塔对面的两艘船的俯角分别为 α 和 β。如果灯塔的高度是 h 米，并且连接船只的线穿过灯塔的脚下，则表明船只之间的距离是 h(tanα + tanβ) / tanαtanβ 米。

解决方案：

Let us assume LH be the lighthouse and A and B are two ships that make 

angles of elevation with L are α and β

Height of lighthouse = h m

Let us assume AH = x and BH = y

Now in △LAH,

tanα = LH/AH = h/x

x = h/tanα   …..(i)

Similarly, in △LBH,

tanβ = LH/HB = h/y

y = h/tanβ    ……(ii)

AB = x + y = h/tanα + h/tanβ

h(1/tanα + 1/tanβ) m

AB = h(tanα + tanβ) / tanα tanβ m.

Hence proved

问题 76. 从 h 米高的塔顶，与塔脚在一条直线上的两个物体的俯角分别为 a 和 β（β > α）。求两个物体之间的距离。

解决方案：

Let us considered the distance between two objects is x m and CD = y m

Given that, ∠BAX = α = ∠ABD                              [alternate angle]

∠CAY = β = ∠ACD                                                   [alternate angle]

and the height of tower, AD = h m 

Now, in ΔACD,

tanβ = AD/CD = h/y

y = h/tanβ       …..(i)

In △ABD,

tanα = AD/BD = AD/BC+CD

x + y = h/tanα      

y = h/tanα – x        …..(ii)

From eq(i) and (ii), we get

h/tanβ = h/tanα – x

x = h/tanα – h/tanβ

= h(1/tanα – 1/tanβ) = h(cotα – cotβ)

问题 77. 房子的窗户离地 h 米。从窗户看，位于车道对面的另一栋房屋的顶部和底部的仰角和俯角分别为α和β。证明房子的高度是 h(1 + tan α tan β) 米。

解决方案：

Let us considered the height of the other house = OQ = H and OB = MW = x m

Given that, height of the first house = WB = h = MO

and ∠QWM = α, ∠OWM = β = ∠WOB                                 [alternate angle]

Now, in ΔWOB,

tanβ = WB/OB = h/x

x = h/tanβ    ……..(i)

In △QWM,

tanα = QM/WM = OQ – MO / WM

tanα = (H – h)/x

x = (H – h)/tanα           ……..(ii)

From eq(i) and (ii), we get

h/tanβ  = (H – h)/tanα

h tanα = (H – h) tanβ 

H = h(tanα + tanβ / tanβ)

= h(1 + tanα.cotβ)

Hence, the required height of the other house is h(1 + tanα.cotβ). 

Hence proved

问题 78. 房屋的下窗距地面 2 m 的高度，其上窗垂直于下窗上方 4 m。在某一时刻，从这些窗口观察到气球的仰角分别为 60° 和 30°。求气球离地高度。

解决方案：

Let us considered that the height of the balloon from above the ground is H.

A and OP = W2R = W1Q = x

Now, given that, the height of lower window from above the ground = w2P = 2 m = OR

Height of upper window from above the lower window = W1W2 = 4 m = QR

∴ BQ = OB – (QR + RO)

BQ = H – (4 + 2)

BQ = H – 6

and ∠BW1Q = 30°

∠BW2R = 60°

Now in △BW2R,

tan 60° = BR/W2R = (BQ + QR)/x

√3 = (H – 6) + 4 / x

x = (H – 2)/√3  …….(i)

Now in △BW1Q,

tan 30° = BQ/W1Q = (H – 6)/x = 1/√3

x = √3(H – 6)        …….(ii)

From eq (i) and (ii), we get

√3(H – 6) = (H – 2)/√3

Now solving the above equation, we get

H = 8

Hence, the required height of the balloon from above the ground is 8 m.