问题1.在下图中，线l 1和l 2在O处相交，形成如图所示的角度。如果x = 45°。查找x，y，z和u的值。
解决方案：
Given that X = 45°
Find: the value of Y, Z, and u
z = x = 45° [Vertically opposite angles are equal]
z + u = 180° [z and u are angles that are a linear pair]
z = 180° – u
u = 180° – x
u = 180° – 45°
u = 135°
x + y = 180° [x and y angles are a linear pair]
y = 180° – x
y =180° – 45°
y = 135°
Hence, x = 45°, y = 135°, z = 135° and u = 45°
问题2。在下面的图。三条共面线在点O相交，形成如图所示的角度。找到x，y，z和u的值。
解决方案：
From the given figure
∠SOD = z = 90° [Vertically opposite angles are equal]
∠DOF = y = 50°
Now, x + y + z = 180° [Linear pair of angles]
Now put the value of z and y
90° + 50° + x = 180°
x = 180° – 140°
x = 40°
So, x = 40°, y = 50°, z = 90°, u = 40°
问题3.在给定的无花果中，找到x，y和z的值。
解决方案：
From the given figure
y = 25° [Vertically opposite angles are equal]
Now ∠x +∠y = 180° [Linear pair of angles]
x = 180° – 25°
x = 155°
Also,
z = x = 155° [Vertically opposite angles]
y = 25°
z = 155°
So, x = 155°, y = 25°, z = 155°
问题4。在下面的图。找到x的值?
解决方案：
From the figure
AOE = BOF = 5x [Vertically opposite angles are equal]
∠COA + ∠AOE + ∠EOD = 180° [Linear pair angles]
3x + 5x + 2x = 180°
10x = 180°
x = 180°/10
x = 18°
So, the value of x = 18°
问题5.证明一对垂直相对的角的等分线在同一条直线上。
解决方案：
From the figure
Lines AB and CD intersect at point O, such that
∠AOC = ∠BOD [vertically opposite angles are equal] …(1)
Also, OP is the bisector of AOC and OQ is the bisector of BOD
To Prove: POQ is a straight line.
∠AOP = ∠COP [OP is the bisector of ∠AOC]…(2)
∠BOQ = ∠QOD [OQ is the bisector of ∠BOD]…(3)
Now,
∠AOC + ∠BOD + ∠AOP + ∠COP + ∠BOQ + ∠QOD = 360° [sum of all angles around a point is 360°]
∠BOQ + ∠QOD + ∠DOA + ∠AOP + ∠POC + ∠COB = 360°
2∠QOD + 2∠DOA + 2∠AOP = 360° (from eq(1), (2) and (3))
∠QOD + ∠DOA + ∠AOP = 180°
POQ = 180°
Hence proved
问题6.如果两条直线相交，请证明与这样形成的一个角度的等分线相反的光线将垂直相反的角度二等分。
解决方案：
Let us considered AB and CD intersect at a point O
Now draw the bisector OP of AOC
OP = POC … (i)
Let extend OP to Q.
Show that, OQ bisects BOD
Let us considered that OQ bisects BOD,
Prove that POQ is a line.
As we know that,
AOC = DOB …. (ii) [vertically opposite angles.]
AOP = BOQ [vertically opposite angles.]
Similarly, POC = DOQ
AOP + AOD + DOQ + POC + BOC + BOQ = 360° [sum of all angles around a point is 360 degrees ]
2AOP + AOD + 2D0Q + BOC = 360°
2AOP + 2AOD + 2DOQ = 360°
2(AOP + AOD + DOQ) = 360°
AOP + AOD +DOQ = 360°/2
AOP + AOD + DOQ = I80°
Thus, POQ is a straight line.
Hence proved
问题7.由两条相交线形成的四个角度之一是否为直角。然后证明四个角度中的每一个都是直角。
解决方案：
According to question
AB and CD intersecting at O, such that ∠BOC = 90°, ∠AOC = 90 °∠AOD = 90° and ∠BOD = 90°
Given:∠BOC = 90°
∠BOC = ∠AOD = 90° [Vertically opposite angles are equal]
∠AOC + ∠BOC = 180° [Angles in linear pair]
∠AOC + 90° = 180° [Angles in linear pair]
∠AOC = 90°
∠AOC = ∠BOD = 90° [Vertically opposite angles]
Hence, ∠AOC = ∠BOC = ∠BOD = ∠AOD = 90°
问题8.在下图中。射线AB和CD在O处相交。
(i)当x = 60 °时确定y
(ii)当y = 40 °时确定x
解决方案：
(i) Given that x = 60°
∠AOC + ∠BOC = 180° [linear pair of angles]
⟹ 2x + y = 180°
⟹ 2(60°) + y = 180° [since x = 60°]
⟹ y = 60°
Hence, the value of y = 60° when x = 60°
(ii) Given y = 40°
∠AOC + ∠BOC = 180° [linear pair of angles]
⟹ 2x + y = 180°
⟹ 2x + 40° = 180° [since x = 40°]
⟹ 2x =180° – 140°
⟹ 2x = 140°
⟹ x = 70°
Hence, the value of x = 70° when x = 40°
问题9.在下图中。行AB。 CD和EF在O相交。找到∠AOC，∠COF，∠DOE和∠BOF的度量。
解决方案：
From the figure
∠AOE + ∠EOB = 180° [linear pair of angles]
∠AOE + ∠DOE + ∠BOD = 180° [linear pair of angles]
⟹ ∠DOE = 180° – 40° – 35° = 105°
∠DOE = ∠COF = 105° [Vertically opposite angles are equal]
Now, ∠AOE + ∠AOF = 180° [Angles in Linear pair]
∠AOE + ∠AOC + ∠COF = 180°
⟹ 40° + ∠AOC +105° = 180°
⟹ ∠AOC = 180° – 145°
⟹ ∠AOC = 35°
Also, ∠BOF = ∠AOE = 40° [Vertically opposite angles are equal]
Hence, the value of ∠AOC = 35°, ∠COF = 105°, ∠DOE = 105°, and ∠BOF = 40°
问题10. AB，CD和EF是通过点O的三条并发线，使得OF将BOD对等。如果∠BOF=35。找到∠BOC和∠AOD。
解决方案：
Given that OF bisects ∠BOD
∠BOF = 35°
We have to find ∠BOC and ∠AOD
∠BOD = 2 ∠BOF = 70° [since OF bisects ∠BOD]
∠BOD = ∠AOC = 70° [ vertically opposite angles]
Now,
∠BOC + ∠AOC = 180°
∠BOC + 70° = 180°
∠BOC = 110°
∠AOD = ∠BOC = 110° [Vertically opposite angles]
Hence, the value of ∠BOC = 110° and ∠AOD = 110°
问题11.在下图中，线AB和CD在O处相交。如果∠AOC+∠BOE= 70°，而∠BOD= 40°，找到∠BOE和反射∠COE?
解决方案：
Given: AOC + BOE = 70° and BOD = 40°
We have to find ∠BOE and reflex ∠COE
BOD = AOC = 40° [vertically opposite angles]
∠AOC + ∠BOE = 70° [given]
⟹ 40° + ∠BOF = 70°
⟹ ∠BOF = 70° – 40°
⟹ ∠BOE = 30°
⟹ AOC + COF + BOE = 180° [Angles in linear pair]
⟹ COE = 180° – 30° – 40°
⟹ COE = 110°
Reflex ∠COE = 360° – 110° = 250°
Hence, the vale of ∠BOE = 30° and ∠COE =250°
问题12.以下哪个陈述是正确的(T)，哪些是错误的(F)?
(i)形成线性对的角度是互补的。
(ii)如果两个相邻的角度相等，则每个角度为90°
(iii)形成线性对的角度都可以是锐角。
(iv)如果形成线性对的角度相等，则每个角度的大小均为90°
解决方案：
(i) True
(ii) False
(iii) False
(iv) true
问题13.填写公司空白，以使以下说法正确：
(i)如果线性对的一个角度为锐角，则另一个角度为______
(ii)射线在一条直线上站立，则如此形成的两个相邻角度之和为______
(iii)如果两个相邻角度之和为180°，则这两个角度的______臂为相反光线。
解决方案：
(i) Obtuse angle
(ii) 180°
(iii) Uncommon