11类RD Sharma解决方案–第四章角度测量–练习4.1 |套装1

问题1.使用π = 22/7查找与以下弧度测量值相对应的度数测量值：

(i) 9π / 5

解决方案：

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

Hence, (9π/5)^c = (9π/5 × 180/π)^o = 324^o

Thus, (9π/5)^c = 324^o

(ii) −5π / 6

解决方案：

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

Hence, (−5π/6)^c = (−5π/6 × 180/π)^o = −150^o

Thus, (9π/5)c = −150^o

(iii) 18π / 5

解决方案：

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

Hence, (18π/5)^c = (18π/5 × 180/π)^o = 648^o

Thus, (18π/5)^c = 648^o

(iv)-3

解决方案：

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

Hence, (−3)^c = (−3 × 180/π)^o = (180 × 7 × −3/22)^o = (−171⁹/₁₁) = −171^o(9 × 60/11)’ = −171^o49’5”

Thus, (−3)^c = −171^o49’5”

(v)11

解决方案：

We know that π radians = 180^o or 1 radian = 1^c = (180/π)⁰

Hence, (11)^c = (11 × 180/π)^o = (11 × 180 × 7/22) = 630^o

Thus, (11)^c =630^o

(六)1

解决方案：

We know that π radians = 180^o or 1 radian = 1^c = (180/π)⁰

Hence, (1)^c = (1 × 180/π)^o = (180 × 7/22) = 57^o(3 × 60/11) = 57^o16¹(4 × 60/11)¹¹ = 57^o16’21”

Thus, (1)^c = 57^o16’21”

问题2。使用以下度数找到与以下度数相对应的弧度数：

^(ⅰ)300Ô

解决方案：

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

Hence, 300⁰ = 300 × π/180 = 5π/3

Thus, 300^o = 5π/3 radians

^(ⅱ)35°

解决方案：

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

Hence, 35^o = 35 × π/180 = 7π/36

Thus, 35^o = 7π/36 radians

(iii) −56 ^o

解决方案：

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

Hence, −56^o = −56^o × π/180 = −14π/45

Thus, −56^o = −14π/45 radians

^(ⅳ)135Ô

解决方案：

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

Hence, 135^o = 135 × π/180 = 3π/4

Thus, 135^o = 3π/4 radians

(v)−300 ^o

解决方案：

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

Hence, −300⁰ = −300 × π/180 = −5π/3

Thus, −300^o = −5π/3 radians

(ⅵ)7 ^O 30′

解决方案：

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

Hence, 7^o30′ = (7 × π/180)^C × (30/60)^o = (7’/₂)^o × (π/180)^C = (15π/360)^c = π/24

Thus, 7^o30′ = π/24 radians

(vii)125 ^o 30′

解决方案：

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

Hence, 125^o30′ = 125^o(30/60)^o = (125’/₂)^o = 251π/360 

Thus, 125^o30′ = 251π/360 radians

(viii)−47 ^o 30′

解决方案：

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

Hence, −47^o30′ = −47^o(30/60)^o = (−47’/₂)^o = (−95/2)^o = (−95/2 × π/180)^o = −19π/72

Thus, −47^o30′ = −19π/72 radians

问题3.直角三角形的锐角之间的差为2π弧度。以度表示角度。

解决方案：

 We know that π rad = 180° ⇒ 1 rad = 180°/ π

Hence, 2π/5 radians = (2π/5 × 180/ π)^o. Substituting the value of π = 22/7, we get

2π/5 radians = (2×22/(7 × 5) × 180/22 × 7) = (2/5 × 180)° = 72°

Let one acute angle be x° and the other acute angle be (90° – x°).

Then, x° – (90° – x°) = 72° ⇒ 2x° – 90° = 72° ⇒ 2x° = 162° ⇒ x° = 81° and

Now, 90° – x° = 90° – 81° = 9°

∴ The angles are 81^o and 9^o.

问题4.三角形的一个角度为2 / 3x坡度，另一个角度为3 / 2x度，而第三个角度为πx / 75弧度。以度表示所有角度。

解决方案：

Given:

One angle of a triangle is 2x/3 grades and another is 3x/2 degree while the third is πx/75 radians.

We know that, 1 grade = (9/10)^o ⇒ 2/3x grade = (9/10 × 2/3x)^o = 3/5x^o

Also since, π radians = 180° ⇒ 1 radian = 180°/π ⇒ πx/75 radians= (πx/75 × 180/π)^o = (12/5x)^o

Since, the sum of the angles of a triangle is 180°.

⇒ 3/5x^o + 3/2x^o + 12/5x^o = 180^o ⇒ (6+15+24)/10x^o = 180^o

Upon cross-multiplication we get, 45x^o = 180^o × 10^o = 180^o ⇒ x^o = 180^o/45^o = 40^o

∴ The angles of the triangle are:

3/5x^o = 3/5 × 40^o = 24^o

3/2x^o = 3/2 × 40^o = 60^o

12/5 x^o = 12/5 × 40^o = 96^o

问题5。找到弧度和内角的大小，以弧度和度为单位：

(i)五角大楼

解决方案：

Since the sum of the interior angles of a polygon = (n – 2)π

And each angle of polygon = sum of interior angles of polygon/number of sides 

Using this rationale,

Number of sides in pentagon = 5

Sum of interior angles of pentagon = (5 – 2) π = 3π radians

Since π radians = 180° ⇒ 1 radian = 180°/ π ⇒ 3π radians = 3π × 180^o/π = 540^o

∴ Each angle of pentagon = 3π/5 × 180^o/π = 108^o

(ii)八角形

解决方案：

Since the sum of the interior angles of a polygon = (n – 2)π

And each angle of polygon = sum of interior angles of polygon/number of sides 

Number of sides in octagon = 8

Sum of interior angles of octagon = (8 – 2)π = 6π

Since π radians = 180° ⇒ 1 radian = 180°/ π ⇒ 6π radians = 6π × 180o/π = 1080^o

∴ Each angle of octagon = 6π/8 × 180^o/π = 135^o 

(iii)七边形

解决方案：

Since the sum of the interior angles of a polygon = (n – 2)π

And each angle of polygon = sum of interior angles of polygon/number of sides

Number of sides in heptagon = 7

Sum of interior angles of heptagon = (7 – 2)π = 5π

Since π radians = 180° ⇒ 1 radian = 180°/ π ⇒ 5π radians = 5π × 180^o/π = 900^o

∴ Each angle of heptagon = 5π/7 × 180^o/ π = 900^o/7 = 128^o34′17”

(iv)二重奏十边形

解决方案：

Since the sum of the interior angles of a polygon = (n – 2)π

And each angle of polygon = sum of interior angles of polygon/number of sides

Number of sides in duo decagon = 12

Sum of interior angles of duo decagon = (12 – 2)π = 10π radians

Since π radians = 180° ⇒ 1 radian = 180°/ π ⇒ 5π radians = 10π × 180^o/π = 1800^o

∴ Each angle of duo decagon = 10π/12 × 180^o/ π = 150^o

问题6.四边形的角度在AP中，最大角度为120 ^o 。用弧度表示角度。

解决方案：

Let the angles of quadrilateral be (a – 3d)°, (a – d)°, (a + d)° and (a + 3d)°.

We know that, the sum of angles of a quadrilateral is 360°.

⇒ (a – 3d + a – d + a + d + a + 3d) = 360° ⇒ 4a = 360° ⇒ a= 90°

Given:

The greatest angle = 120° ⇒ a + 3d = 120° ⇒ 90° + 3d = 120° ⇒ d = 30°/3 = 10^o

∴ The angles are:

(a – 3d)° = 90° – 30° = 60°, (a – d)° = 90° – 10° = 80°, (a + d)° = 90° + 10° = 100° and (a + 3d)° = 120°

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

Using the above rationale, angles of quadrilateral in radians are as follows:

(60 × π/180) radians = π/3, (80 × π/180) radians = 4π/9, (100 × π/180) radians= 5π/9 and (120 × π/180) radians = 2π/3.

Thus, the angles of quadrilateral in radians are π/3, 4π/9, 5π/9 and 2π/3.

问题7.三角形的角度在AP中，最小角度的度数等于平均角度的度数为1：120。找出以弧度为单位的角度。

解决方案：

Let the angles of the triangle be (a – d)°, a° and (a + d)°.

We know that, the sum of the angles of a triangle is 180°.

⇒ (a – d + a + a + d) = 180° ⇒ 3a = 180° ⇒ a = 60°

It is give that, number of degrees in the least angle/number of degrees in the mean angle = 1/120  

⇒ (a-d)/a = 1/120 ⇒ (60-d)/60 = 1/120 ⇒ 120-2d = 1⇒ 2d = 119 ⇒ d = 119/2 = 59.5

∴ The angles (in degrees) are:

(a – d)° = 60° – 59.5° = 0.5°, a° = 60° and (a + d)° = 60° + 59.5° = 119.5°

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

Using the above rationale, angles of quadrilateral in radians are as follows:

(0.5 × π/180) radians = π/360, (60 × π/180) radians = π/3and (119.5 × π/180) radians = 239π/360

Thus, the angles of triangle in radians are π/360, π/3 and 239π/360.

问题8：一个规则多边形与另一个规则多边形的角度为3：2，第一个多边形的边数是第二个多边形的边数的两倍。确定两个多边形的边数。

解决方案：

Let the number of sides in the first polygon be 2x and in the second polygon be x.

We know that, angle of an n-sided regular polygon = [(n-2)/n] π radian  

⇒ The angle of the first polygon = [(2x-2)/2x] π = [(x-1)/x] π radian

⇒ The angle of the second polygon = [(x-2)/x] π radian  

Thus, [(x-1)/x] π / [(x-2)/x] π = 3/2 ⇒ (x-1)/(x-2) = 3/2

Cross multiplying the above we get, 2x – 2 = 3x – 6 ⇒ 3x-2x = 6-2 ⇒ x = 4

∴ Number of sides in the first polygon = 2x = 2(4) = 8

Number of sides in the second polygon = x = 4

问题9.三角形的角度在AP中，以至于最大为最小的5倍。找出以弧度为单位的角度。

解决方案：

Let the angles of the triangle be (a – d)^o, a^o and (a + d)^o.

We know that, the sum of angles of triangle is 180°.

⇒ (a – d + a + a + d) = 180° ⇒ 3a = 180° ⇒ a = 180°/3 = 60^o

We are given that the greatest angle = 5 × least angle

Hence, greatest angle/least angle = 5 ⇒ (a+d)/(a-d) = 5 ⇒ (60+d)/(60-d) = 5

By cross-multiplying we get, (60 + d) = (300 – 5d) ⇒ 6d = 240 ⇒ d = 240/6 = 40

Hence, angles are:

(a – d) ° = 60° – 40° = 20°, a° = 60° and (a + d)° = 60° + 40° = 100°

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

Using the above rationale, angles of quadrilateral in radians are as follows:

(20 × π/180) radians = π/9, (60 × π/180) radians = π/3 and (100 × π/180) radians = 5π/9

Hence, the angles of the triangle in radians are π/9, π/3 and 5π/9.

问题10.两个规则多边形的边数为5：4，它们的角度之差为9 ^o 。找到多边形的边数。

解决方案：

Let the number of sides in the first polygon be 5x and in the second polygon be 4x.

We know that, angle of an n-sided regular polygon = [(n-2)/n] π radian

The angle of the first polygon = [(5x-2)/5x] 180^o

The angle of the second polygon = [(4x-1)/4x] 180^o  

Thus, [(5x-2)/5x] 180^o – [(4x-1)/4x] 180^o = 9 ⇒ 180^o [(4(5x-2) – 5(4x-2))/20x] = 9

Upon cross-multiplication we get, (20x – 8 – 20x + 10)/20x = 9/180 ⇒ 2/20x = 1/20 ⇒ 2/x = 1 ⇒ x = 2

∴Number of sides in the first polygon = 5x = 5(2) = 10

Number of sides in the second polygon = 4x = 4(2) = 8