11类RD Sharma解决方案–第四章角度测量–练习4.1 |套装2

问题11.一条铁路曲线应布置在一个圆上。如果轨道是25^度在40米的距离改变方向应该用什么半径。

解决方案：

Let AB denote the given rail-road. 

We are given ∠AOB = 25^o. We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence 25^o = 25 × π/180 = 5π/36 radians

Also since, θ = Arc/Radius ⇒ ∠AOB = AB/OA ⇒ (5π/36)^c = 40/r ⇒ r = 288/π = 91.64 m 

Hence, the radius of the track is 91.64 m.

问题12。找到在5280m的距离处眼睛对角为1’的长度。

解决方案：

Let θ = 1’ and the length of the arc that subtends θ be l.

Radius = OA = OB = 5280m 

We know, 1’ = 60^o ⇒ 1’ = (1/60)^o.  Since 180^o = π radians = π^c or 1^o = (π/180)^c,

⇒ θ = 1’ = (1/60 × π/180)^c 

Also since, θ = Arc/Radius ⇒ (1/60 × π/180)^c = l/5280 ⇒ l = 1.5365 m

Hence, the length of the arc is 1.5365 m.

问题13.车轮每分钟旋转360转。 1秒内可以旋转多少弧度?

解决方案：

Since the wheel makes 360 revolutions in 1 minute, number of revolutions made by it in 1 second = 360/60 = 6.

Angle made by the wheel in 1 revolution = 360^o

Thus, angle made by the wheel in 6 revolutions = Angle made in 1 second = 360 × 6 = 2160^o

We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence, 2160^o = (2160π/180)^c = 12π radians

Thus, the wheel turns 12π radians in 1 second.

问题14：如果摆长为75厘米，并且笔尖描述了一个长度弧，请找出以弧度为单位摆动的角度：

(i)10厘米

解决方案：

Let OA be the length of the pendulum. ⇒ OA = 75 cm = 0.75 m

Let the arc be denoted by AB. ⇒ AB = 10 cm = 0.1 m

Also since, θ = Arc/Radius = 0.1/0.75 = 2/15 radians

Hence, the angle is 2/15 radians.

(ii)15厘米

解决方案：

Let OA be the length of the pendulum. ⇒ OA = 75 cm = 0.75 m

Let the arc be denoted by AB. ⇒ AB = 15 cm = 0.15 m

Also since, θ = Arc/Radius = 0.15/0.75 = 1/5 radians

Hence, the angle is 1/5 radians.

(iii)21厘米

解决方案：

Let OA be the length of the pendulum. ⇒ OA = 75 cm = 0.75 m

Let the arc be denoted by AB. ⇒ AB = 15 cm = 0.21 m

Also since, θ = Arc/Radius = 0.21/0.75 = 7/25 radians

Hence, the angle is 7/25 radians.

问题15.圆的半径为30厘米。如果圆弧的弦长为30厘米，则求出圆弧的长度。

解决方案：

Let OA = OB = Radius of the circle = 30 cm = 0.3 m, and chord AB = 30 cm = 0.3 m. Let l be the length of the arc AB. 

Since, OA = OB = AB = 0.3 m, the triangle AOB is an equilateral triangle.

∠AOB = 60^o. We know 180^o = π radians = π^c or 1^o = (π/180)^c

Hence, 60^o = 60 × π/180 = π/3 radians

Also since, θ = Arc/Radius ⇒ (0.3π/3) = 0.1 0.1π m = 10π cm

Hence, the length of the arc is 10π cm.

问题16：一列火车在半径150米的圆形曲线上以每小时66公里的速度行驶。它在10秒内转过什么角度?

解决方案：

In the given circular track, OA = OB = r = 150 m

Let θ denote the angle the train turns in 10 seconds.

We are given that speed = 66 km/hr = {66 × 1000/60 × 60} m/sec = 110/6 m/sec

Hence, the train will run 1100/6 m/sec in 10 seconds. ⇒ arc AB = 1100/6 m

Also since, θ = Arc/Radius = 1100/6 × 1500 = 11/90 radians

Hence, the angle is 11/90 radians.

问题17.找到距离眼睛应持有2厘米直径硬币的距离，以便可以隐藏^{角直径为31’的满月吗?}

解决方案：

We are given θ = 31’ and arc AB = 2 cm = 0.02 m

Since, 1’ = 60^o ⇒ 1’ = (1/60)^o.  Since 180^o = π radians = π^c or 1^o = (π/180)^c,

⇒ θ = 31’ = (31/60 × π/180)^c

Also since, θ = Arc/Radius ⇒ (31/60 × π/180) = 0.02/r ⇒ r = 2.217 m

Hence, the coin shall be placed at a distance of 2.217 m from the eye.

问题18：假设太阳在观察者眼中与32°角成对角，则以千米为单位求出太阳的直径。假设太阳的距离为91×10 ⁶ km。

解决方案：

We are given θ = 31’ and r = 91 × 10⁶ km

Since, 1’ = 60^o ⇒ 1’ = (1/60)^o.  Since 180^o = π radians = π^c or 1^o = (π/180)^c,

⇒ θ = 32’ = (32/60 × π/180)^c

Also since, θ = Arc/Radius ⇒ (32/60 × π/180) = (AB/91 × 10⁶) km = 847407.4 km

Hence, the distance of sun is 847407.4 km

问题19：如果两个圆周上相同长度的圆弧^{在中心对角分别为65 o}和110 ^o ，则求出它们的半径比。

解决方案：

Let C₁ and C₂ denote the two given circles with the same arc length l.

Hence, AB = CD = l

Let θ₁ and θ₂ be the angles subtended, and OA = OB = r and OC = OD = R

Given, θ₁ = 65^o = (65π/180)^c and θ₂ = 110^o = (110π/180)^c

Also since, θ = Arc/Radius ⇒ θ₁ = AB/r = l/r or r= l/θ₁     …….(1)

and, θ₂ = CD/R = l/R or R = l/θ₂        …..(2)

From equations (1) and (2), we get,

r/R = l/θ₁ / l/θ₂ = 110π/180 / 65π/180 = 22/13

Hence, the ratio of radii of both circles is 22:13.

问题20.使用π = 22/7 ，找到半径为100 cm的圆的中心与长度为22 cm的弧对向的角度的度数。

解决方案：

Let O denote the centre of the circle and AB denote the arc.

Hence, arc AB = 22 cm, and OA = OB = radius = 100cm

Let θ be the angle subtended by the arc at the centre O by arc AB.

We know, θ = Arc/Radius = 22/100 radians

Since π radians = 180^o or 1 radian = 1^c = (180/π)^o

Hence, 22/100 radians = (22/100 × 180/π)^o = 12.6^o = 12^o36′

Thus, the angle subtended by the arc at the centre of the circle is 12^o36′.