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 1

问题 1. 一座塔垂直立于地面。从地面上距塔脚20m处，塔顶仰角为60°。塔的高度是多少？

解决方案：

Given that,

The distance between point of observation and foot of tower(BC) = 20 m 

Angle of elevation of top of tower(θ) = 60° 

And the height of tower (H) = AB

From the figure we conclude that ABC is a right angle triangle

So, tanθ = (AB)/(BC)

AB = 20 tan 60^o

H = 20 x √3 = 20√3

Hence, the height of tower is 20√3 m

问题 2. 梯子靠墙的仰角为 60°，梯脚距墙 9.5 m。求梯子的长度。

解决方案 ：

Given that,

The distance between foot of the ladder and wall(BC) = 9.5 m

Angle of elevation (θ)= 60°

Length of the ladder (L) = AC

From the figure we conclude that ABC is a right angle triangle

So, 

cos 60^o = BC/AC 

AC = 2 x 9.5 = 19m

Hence, the length of ladder is 19m.

问题 3. 梯子沿着房屋的墙壁放置，使其上端接触墙壁顶部。梯子脚距墙2m，梯子与地面成60°角。确定墙的高度。

解决方案：

Given that,

The ladder is making an angle of 60°Distance between foot of the ladder and wall(BC) = 2m 

Angle made by ladder with ground (θ) = 60°

Height of the wall (H) = AB

From the figure we conclude that ABC is a right angle triangle

So, tan 60^o = (AB)/(BC)

√3 = AB/2

AB = 2√3

Hence, the height of wall is 2√3 m

问题 4. 一根电线杆高 10 m。电线杆顶部的一根钢丝固定在地面上的一个点上，以保持杆子直立。如果电线通过杆脚与水平线成 45° 角，求电线的长度。

解决方案：

Given that,

The height of the electric pole (H) = AB = 10 m 

Angle made by steel wire with ground (horizontal) (θ) = 45°

Let us assume length of wire (AC) = L  

From the figure we conclude that ABC is a right angle triangle

So, 

sin 45^o = AB/AC

1/√2 = 10/L

L = 10√2 m

Hence, length of wire is 10√2 m.

问题 5. 一只风筝在距地面 75 米的高度飞行，系在一根与水平面成 60° 角的字符串上。找到最接近米的字符串长度。

解决方案：

Given that,

The height of kite from ground(AB) = 75 m  

Inclination of string with ground (θ) = 60°

And the length of string (L) = AC

From the figure we conclude that ABC is a right angle triangle

So, 

sin60^o = AB/AC

√3/2 = 75/L

L = 50√3 m 

Hence, the length of string is 50√3 m.

问题 6. 风筝与地面上一点之间的字符串长度为 90 米。如果字符串与地面成角度 θ，则 tan θ = 15/8。风筝有多高？假设字符串没有松弛

解决方案：

Given that,

The length of the string between point on ground and kite = 90 m

Angle made by string with ground is θ

And tan θ =15/8

θ = tan⁻¹(15/8)

Height of the kite be ‘H’ m

So, ABC is a right angle triangle

tanθ = AB/BC

15/8 = H/BC

BC = 8H/15    ———–(i)

In ΔABC, 

by using Pythagoras theorem, we get

AC² = BC² + AB²

90² = (8H/15)² + H²

H² = (90×15)²/289

H = 79.41m

Height of the kite from ground

H = 79.41m

Hence, Height of the kite from ground is 79.43 m

问题 7. 竖塔立于水平位置，其上方有竖旗杆。在距塔70米的平面上的一点，观察者注意到距旗杆几米的仰角分别为60°和45°。求旗杆的高度和塔的高度。

解决方案：

Given that, the vertical tower is surmounted by flag staff.

So, the distance between tower and observer(DC) = 70 m  

Angle of elevation of top of tower (α) = 45°

Angle of elevation of top of flag staff (β) = 60°

Height of flag staff (AD) = h 

Height of tower(BC) = H 

From the figure we conclude that ΔACD and ΔBCD are right angle triangles 

So, 

In ΔBCD,

tanθ = (BC)/(DC)

tan45^o = H/70

H = 70

Again, in ΔACD, 

tan θ = AC/DC = (x + H)/70

tan 60^o = (x + H)/70

√3 = (x + 70)/70

x = 51.24 m

Hence, the height of tower is 70 m and height oh flag staff is 51.24 m

问题 8. 一棵垂直笔直的树，高 15 m；被风吹断，其顶部刚好接触地面并与地面成 60° 角。这棵树在离地多高的地方折断了？

解决方案：

Given that, 

The initial height of tree(AB) = H = 15 m 

Let us considered that it is broken at point C.

So, the angle made by broken part with the ground (θ) = 60°

And the height from ground to broken points(BC) = h  

AB = AC + BC

H = AC + h

AC = (H – h) m

From the figure we conclude that ABC is a right angle triangle

So, 

sin60^o = BC/CA

√3/2 = h/(H – h)

√3(15 – h) = 2h

h = (15√3 / 2 + √3) x (2 – √3 / 2 – √3) 

h = 15(2√3 – 3)

Hence, the height of broken points from ground is 15(2√3 – 3) m. 

问题 9. 一个垂直的塔架在一个水平面上，上面有一个 5 米高的垂直旗杆。在平面上的一点，旗杆底部和顶部的仰角分别为30°和60°。求塔的高度。

解决方案：

Given that,

The height of the flag staff(AB) = h = 5 m 

Angle of elevation of the top of flag staff(α) = 60°  

Angle of elevation of the bottom of the flagstaff(β) = 60° 

Let us assume that the height of tower is (H) = AB

From the figure we conclude that ΔACD and ΔBCD are right angle triangles 

So, in ΔBCD,

tanθ = (BC)/(DC)

tan60^o = h/x

√3 = h/x    —————-(i)

So, in ΔACD,

tanθ = (AC)/(DC)

tan30^o = 5 + h/x

1/√3 = 5 + h/x      ——————-(ii)

Equating eqn. (i) and (ii),

h = 2.5 m

Hence, the height of the tower = 2.5 m.

问题 10. 有人观察到塔的仰角为 30°。他沿着平地向塔脚走50 m，发现塔顶仰角为60°。求塔的高度。

解决方案：

Given that,

The angle of elevation of top of tour from first point of elevation(α) = ∠A = 30°

Let us considered that the person walked 50 m from first point (A) to (B) then AB = 50 m

Angle of elevation from second point (β) = ∠C = 60°

From the figure we conclude that ΔABC and ΔABD are right angle triangles 

So, ∠B= 90°

Let us considered the height of the tower is (AB) = h

BC = x m.

In ΔABC, 

tan θ = AB/BC

tan 60^o = h/x

√3 = h/x   ————-(i)

In ΔABD, 

tanθ = (AB)/(BD)

tan30^o = h/(DC + BC)

1/√3 = h/(50 + x)     ——————–(ii)

From eqn. (i) and (ii), we get

h = 25√3

Hence, the height of tower is 25√3 m

问题 11. 当太阳仰角为 45° 时，发现塔的阴影比 60° 时长 10 m。求塔的高度。

解决方案：

Let us considered the length of the shadow of the tower when angle of elevation is 60° be(BC) = x m

Now, the length of the shadow with angle of elevation 45° (BD) = (10 + x)m 

Let us assume that the height of the tower(AB) = H  

From the figure we conclude that ΔABC and ΔABD are right angle triangles 

So, ∠B = 90°

Now, In ΔABC, 

tanα = AB/BC

tan60^o = H/x

x = H/√3  ——————–(i)

Again, In ΔABD 

tanβ = AB/BD

tan45^o = H/(x + 20)

x + 10 = H

x = H – 10       ——————–(ii)

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

H – 10 = H/√3

√3 H – 10√3 = H

H = 10√3/(√3 – 1)

H = 5(3 + √3)

Hence, height of tower is 23.66 m.

问题 12. 降落伞垂直下降，在自己左侧相距 100 m 的两个观测点的仰角分别为 45°和 60°。求他跌落的最大高度以及他跌倒在地面上的点到刚刚观察点的距离。

解决方案：

Let us considered that the parachute at highest point A. 

So, C and D be points which are 100 m apart on ground where from then CD = 100 m

Angle of elevation from point ∠C = 45° = α,

Angle of elevation from point ∠D = 60° = β,

Let’s B be the point just vertically down the parachute

Also, the maximum height of the parachute from the ground (AB) = H m

And the distance of point where parachute falls to just nearest observation point = x m

From the figure we conclude that ΔABC and ΔABD are right angle triangles 

Now, in ΔABC 

tanα = AB/BC

tan45^o = H/x + 100

100+x = H        —————–(i)

In ΔABD

tanβ = AB/BD

tan60^o = H/x

H = √3x         ——————(ii)

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

x = 136.6 m 

Put the value of x in (ii), we get

H = √3 × 136.6 = 236.6m

Hence, the distance between the two points where parachute falls on 

the ground and just the observation is 136.6 m.

问题 13. 在塔的同一侧，有两个物体。从塔顶观察时，它们的俯角分别为 45° 和 60°。如果塔的高度是 150 m，求物体之间的距离。

解决方案：

Given that, the height of the tower(AB) = H = 150 m   

And the angles of depression are 45° and 60°     

In ΔABC 

tan60° = 150/BC

BC = 150/√3              ——————(i)

Now, In ΔABD

tanβ = AB/BD

tan45° = 150 / BC + CD

BC + CD = 150            —————-(ii)

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

BC = 63.4 m

Hence, the distance between object is 63.4 m.

问题 14. 与塔脚在同一水平线上的一点，塔的仰角为 30°。在向塔脚前进 150 米时，塔的仰角变为 60°。证明塔的高度是 129.9 m。

解决方案：

Given that, the angle of elevation of top tower from first point D(α) = 30°              

CD = 150 m,

Angle of elevation of top of tower from second point C(β) = 60°,

Let us assume that the height of tower AB = H m,

From the figure we conclude that ΔABC and ΔABD are right angle triangles 

So, ∠B = 90°,

In ΔABD, 

tanα = AB/BD

tan30^o = H/(150+x)

150 + x = H√3                  ———————-(i)

Now, in ΔABC, 

tanβ = AB/BC

tan60 = H/x

H = x√3                   ————–(ii)

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

H = 129.9 m

Hence, the height of the tower is 129.9 m

问题 15. 从通过塔脚的水平面中的一点观察，塔顶的仰角是 32°。当观察者向塔移动 100 m 时，他发现顶部的仰角为 63°。求塔的高度和第一个位置到塔的距离。 [以 tan 32° = 0.6248 和 tan 63° = 1.9626]

解决方案：

Let’ us considered the height of the tower(AB) = h m, and the distance BC = x m

So, in ΔABC

tan 63^o = AB/BC

1.9626 = h/x

x = h/ 1.9626

x = 0.5095 h …. (i)

Now, in ΔABD

tan 32^o = AB/ BD

0.6248 = h/ (100 + x)

h = 0.6248(100 + x)

h = 62.48 + 0.6248x

h = 62.48 + 0.6248(0.5095 h) ….. [Using eq(i)]

h = 62.48 + 0.3183h

0.6817h = 62.48

h = 62.48/0.6817 = 91.65

Now put the value of h in eq(i), we get

x = 0.5095(91.65) = 46.69

So, the height of the tower is 91.65 m,

Hence, the distance if the first position from the tower = 100 + x = 146.69 m

问题 16. 塔顶从地面 A 点的仰角为 30°。向塔脚移动 20 米的距离到 B 点，仰角增加到 60°。求塔的高度和塔到 A 点的距离。

解决方案：

Given that, the angle of elevation of top of the tower from point A(α) = 30°,

Angle of elevation of top of tower from point B(β) = 60°.

Distance between A and B, AB = 20 m

Let us assume that height of tower CD = ‘h’ m

and the distance between second point B from foot of the tower = ‘x’ m,

The given triangles are right angle triangles 

So, ∠D = 90°

tanα = CD/AD

tan30^o = h/(20 + x)

20 + x = h√3            —————–(i)

tanβ = CD/SD

tan60^o = h/x           

x = h√3                 ——————-(ii)

On Substituting eq(i) in (ii), we get

20 + h/√3 = h√3

h = 10√3

h = 17.31m

x = 10√3/√3

x = 10 m

So, the height of the tower = 17.32 m

Hence, the distance of the tower from point A = (20 + 10) = 30 m

问题 17. 从 15 m 高的建筑物顶部，发现塔顶的仰角为 30°。从同一建筑物的底部，发现塔顶的仰角为 60°。找出塔的高度以及建筑物与塔之间的距离。

解决方案：

Given that, the height of the building is AB = 15 m 

Angle of elevation of top of the tower from top of the building(α) = 30°,

Angle of elevation of top of the tower from bottom of the building(β) = 60°,

Distance between the tower and the building BE = x,

Here, BE = AD

Let us assume that the height of the tower CE = h

Here ABED is a rectangle
BE = AD = x cm

AB= DE = 15 cm

Now, In ΔCDA

tanα = CD/AD

tan30^o = h – 15/x

x = h – 15√3          ——————-(i)

In ΔBEC

tanβ = CE/BE

tan60^o = h/ x

x = h/ √3  ————–(ii) 

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

h/ √3 = h – 15√3

h = 22.5

Now put the value of h in eq(ii), we get

x = 22.5/√3 = 12.990

Hence, height of the tower above ground is 22.5 m and 

the distance between the building and the tower is 12.990

问题 18. 在水平面上有一个垂直的塔，塔的顶部有一个旗杆。在距塔脚 9 m 处，旗杆顶部和底部的仰角分别为 60°和 30°。找出塔的高度和安装在上面的旗杆。

解决方案：

Given that, the distance of the point of observation from foot of the tower BC = 9 m,

Angle of elevation of top of the pole(α) = 60°,

Angle of elevation of bottom of the pole(β) = 30°,

Let us assume that the height of the tower (BC) = h cm 

And the height of the pole(AB) = x cm

From the figure we conclude that ΔACD and ΔBCD are right angle triangles 

So, ∠C = 90°

In ΔACD,

tanα = AC/CD

tan60^o = h + x / 9

h + x = 9√3                 ———————(i)

In ΔBCD

tanβ = BC/CD

tan30^o = h / 9

 h = 9/√3

h = 3√3            ——————-(ii)

h = 5.196 m 

Now put the value of h in eq(1), we get

x = 9√3 – 3√3

x = 6√3

x = 10.392 m

Hence, the height of the tower is 5.196 m, and the height of the tower pole is 10.392 m.

问题 19. 一棵树因暴风雨折断，折断部分弯曲，使树顶与地面接触，与地面成 30°角。树脚与树顶接触地面的距离为 8 m。求树的高度。

解决方案：

Let us considered the initial height of the tree be AC.

And, due to storm the tree is broken at point B.

So, let us assume that the bent portion of the tree be AB = x m and the remaining portion BC = h m

So, the height of the tree AC = (x + h) m

And the distance between the foot of the tree to the point where the top touches the ground (DC) = 8m

Now, in ΔBCD

tan 30° = BC/DC

1/√3 = h/8

h = 8/√3

Again, in ΔBCD

cos30^o = DC/BD

√3/2 = 8/x

x = 16/√3 m

So, x + h = 16/√3 + 8/√3

24/√3 = 8√3

Hence, the height of the tree is 8√3 m.

问题 20. 从地面上的点 P 看，一栋 10 m 高的建筑物的仰角是 30°。大楼顶部悬挂一面旗帜，旗杆与P的仰角为45°。求旗杆的长度和建筑物到点 P 的距离。

解决方案：

Given that, the height of the building(BQ) = 10m

Angle of elevation of top of the building from P (α) = 30°,

Angle of elevation of top of flag staff from P (β) = 45°,

Let height of the flagstaff (AB) = h m,

From the figure we conclude that ΔAQP and ΔBQP are right angle triangles 

So, ∠Q = 90°,

In ΔBQP, 

tanα = BQ/QP

tan30^o = 10/x

x = 10√3

x = 17.32 m ……(i)

In ΔAQP

tanβ = AQ/QP

tan45^o = 10 + h /x

10 + h = x

Put the value of x from eq(i), we get

h = 17.32 – 10 m

h = 7.32 m

So, the height of the flag staff is 7.32 m, and 

the distance between P and foot of the tower is 17.32 m

问题 21. 一个身高 1.6 m 的女孩站在距离灯柱 3.2 m 处，在地上投下 4.8 m 的影子。通过使用找到灯柱的高度

(一) 三角比

(ii) 相似三角形的性质

解决方案：

Given that, A 1.6 m tall girl stands at a distance of 3.2 m

Let us considered AC be the lamp post of height = h cm

The height of the girl(DE) = 1.6m

Distance between the girl and lamp(EC) = 3.2m

and BE = 1.6 m

(i) In ΔBDE,

tan θ = 1.6/4.8

tan θ = 1/3

Next, In ΔABC

tan θ = h/ (4.8 + 3.2)

1/3 = h/8

h = 8/3 m

(ii) By using similar triangles,

ΔBDE and ΔABC are similar, we get

AC/BC = ED/BE

h / 4.8 + 3.2 = 1.6/4.8

h = 8/3 m

Hence, the height of the lamp post is is 8/3 m.

问题 22. 一个 1.5 m 高的男孩站在距离 30 m 高的建筑物不远的地方。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°。找出他走向大楼的距离。

解决方案：

Given that,

The height of the tall boy (AS) = 1.5 m,

The length of the building (PQ) = 30 m.

Let us considered the initial position of the boy be S. 

And, then he walks towards the building and reached at the point T.

So, 

AS = BT = RQ = 1.5 m,

PR = PQ – RQ = (30 – 1.5)m = 28.5 m,

In ΔPAR,

tan 30^o = PR/AR

1/ √3 = 28.5/ AR

AR = 28.5√3

In ΔPRB,

tan 60^o = PR/BR

√3 = 28.5/ BR

BR = 28.5/√3 = 9.5√3

So, ST = AB = AR – BR = 28.5√3 – 9.5√3 = 19√3

Hence, the distance which the boy walked towards the building is 19√3 m.

问题 23. 当太阳高度为 30° 时，发现站在水平地面上的塔的阴影比 60° 时长 40 m。求塔的高度。

解决方案：

Let’ us considered AB be h m and BC be x m.

From the question, DC is 40 m longer than BC.

So, BD = (40 + x) m

And two right triangles ABC and ABD are formed.

In ΔABC,

tan 60^o = AB/ BC

√3 = h/x

x = h/√3              —————(i)

In ΔABD,

tan 30^o = AB/ BD

1/ √3 = h/ (x + 40)

x + 40 = √3h

h/√3 + 40 = √3h                  (using eq(i))

h + 40√3 = 3h

2h = 40√3

h = 20√3

Hence, the height of the tower is 20√3 m.

问题 24. 从地面上的一点看，固定在 20 m 高建筑物顶部的输电塔底部和顶部的仰角分别为 45°和 60°。求输电塔的高度。

解决方案：

Given that,

Height of the building(AB) = 20 m,

Let’ us considered height of tower above building(BC) = h,

Height of tower + building = (h + 20) m [from ground] = OA

Angle of elevation of bottom of tower(α) = 45°

Angle of elevation of top of tower(β) = 60°

Let us considered the distance between tower and observation point = x m

From the figure we conclude that ΔCAO and ΔBAO are right angle triangles 

In ΔBAO

tanα = BA/OA

tan45^o = 20/x

x = 20 m

Next, In ΔCAO 

tanβ = CA/OA

tan60^o = (h + 20) / x

h = 20(√3 – 1)

h = 20(1.73 − 1) = 20 x .732 = 14.64

Hence, the height of the tower is 14.64m

问题 25. 8m 高层建筑的顶部和底部相对于多层建筑顶部的倾角分别为 30°和 45°。求多层建筑的高度和两座建筑之间的距离。

解决方案 ：

Let us considered the height of multistoried building ‘h’ m = AD,

Height of the tall building = 8 m = BE,

Angle of depression of top of the tall building from the multistoried building = 30°,

Angle of depression of bottom of the tall building from the multistoried building = 30°,

And, let the distance between the two buildings = ‘x’ m = ED,

Therefore, BC = x and CD = 8 m                

AD = AC + CD

So, AC = (h – 8) m

We have,

In ΔBCA

tan 30^o = AC/ BC

1/√3 = (h – 8)/x

x = √3(h – 8)              ……..(i)

In ΔADE

tan 45^o = AD/ ED

1 = h/ x

h = x

h = √3(h – 8)                 (using eq(i))

h = √3h – 8√3

(√3 – 1)h = 8√3

h = 8√3/ (√3 -1)

On rationalizing the denominator by (√3 + 1), we get

h = 8√3(√3 + 1)/ (3 – 1)

h = 4√3(√3 + 1)

h = 12 + 4√3

= 4(3 + √3)

So, x = h = 4(3 + √3)

Hence, the height of the building is 4(3 + √3)m and 

the distance between the building is 4(3 + √3)m

问题 26. 一个 1.6 m 高的雕像矗立在一个基座的顶部。从地面上的一点看，雕像顶部的仰角是 60°，从同一点看，基座顶部的仰角是 45°。找到底座的高度。

解决方案：

Let us considered AB be the statue and BC be the pedestal and 

D be the point on ground from where elevation angles are measured.

Given that,

Angle of elevation of the top of statue is 60°,

Angle of elevation of the top of the pedestal is 45°,

We have,

In ΔBCD,

tan 45° = BC/ CD

1 = BC/ CD

CD = BC

Next, In ΔADC

tan 60° = (AB + BC)/ CD

√3 = (AB + BC)/ BC  [Since CD = BC]

BC√3 = AB + BC

AB = (√3 – 1)BC

BC = AB/ (√3 – 1)

BC = 1.6/ (√3 – 1)

On rationalizing the denominator, we get

BC = 1.6(√3 + 1)/ 2

BC = 0.8(√3 + 1) m

Hence, the height of pedestal is 0.8(√3 + 1) m