Download Class 7 Lines and Angles

ID : in-7-Lines-and-Angles [1]
Class 7
Lines and Angles
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Answer t he quest ions
(1)
If AD and BD are bisectors of ∠CAB and ∠CBA respectively, f ind sum of angle x and y.
Choose correct answer(s) f rom given choice
(2)
(3)
If angles of a triangle are in ratio 5:3:8, the triangle is
a. an acute angled triangle
b. a right triangle
c. an isosceles triangle
d. an obtuse angled triangle
If AD and BD are bisectors of ∠CAB and ∠CBA respectively, f ind value of ∠ADB.
a. 143°
b. 128°
c. 135°
d. 138°
(4) T he value of the supplement of the complement of 77° is :
a. 13°
b. 86°
c. 167°
d. none of these
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ID : in-7-Lines-and-Angles [2]
Fill in t he blanks
(5)
(6)
If CD is perpendicular to AB, and CE bisect angle ∠ACB, the angle ∠DCE =
°.
If AP and BP are bisectors of angles ∠CAB and ∠CBD respectively, the angle ∠APB.
°.
(7) Find the angle between
A)
C)
(8)
South-West and South :
West and East :
D)
°
If AB and CD are parallel, value of angle x is
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B)
°
South and North :
North and West :
°
°
°.
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ID : in-7-Lines-and-Angles [3]
(9)
(10)
(11)
If lines AB and CD intersects as shown below, the value of angle x =
If OD is perpendicular to AB, and ∠DOC = 50°, (∠BOC - ∠AOC). =
If AB and CD are parallel, X =
°.
°.
°
(12) If AB and DE are parallel, f ind the value of ∠ACB
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ID : in-7-Lines-and-Angles [4]
(13)
(14)
If two horizontal lines are parallel, value of angle x is
Value of angle X is
°.
°
Check True/False
(15) A triangle can have two obtuse angles.
T rue
False
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ID : in-7-Lines-and-Angles [5]
Answers
(1)
35°
Step 1
It is given that AD and BD are bisectors of ∠CAB and ∠CBA respectively.
T heref ore, x = ∠CAB/2 -----(1)
y = ∠CBA/2 -----(2)
Step 2
In triangle ABC, ∠CAB + ∠CBA + ∠ACB = 180° ...[T he sum of all three angles of a
triangle is 180°]
⇒ ∠CAB + ∠CBA + 110° = 180°
⇒ ∠CAB + ∠CBA = 180° - 110°
⇒ ∠CAB + ∠CBA = 70°
⇒ ∠CAB/2 + ∠CBA/2 = 70/2
⇒ x + y = 35° ...[From equation (1) and (2)]
Step 3
Hence, the sum of the angles x and y is 35°.
(2)
b. a right triangle
Step 1
According to the question, all angles of the triangle are in ratio 5:3:8. We can assume three
angles of the triangle to be 5x, 3x and 8x where x is common f actor.
Step 2
We know that the sum of the three angles of a triangle is 180°.
T heref ore, 5x + 3x + 8x = 180°
⇒ 16x = 180°
⇒x=
180
16
Now, 5x = 5 ×
180
= 56.25,
16
3x = 3 ×
180
= 33.75 and
16
8x = 8 ×
180
= 90.
16
T heref ore, the three angles of the triangle are 56.25°, 33.75° and 90°.
Step 3
Since, one of the angle of the triangle is 90°, the triangle is a right triangle.
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ID : in-7-Lines-and-Angles [6]
(3)
c. 135°
Step 1
It is given that AD and BD are bisectors of ∠CAB and ∠CBA respectively.
T heref ore,
∠BAD = ∠CAB/2------(1)
∠ABD = ∠CBA/2-------(2)
Step 2
In triangle ABC,
∠CAB + ∠CBA + ∠ACB = 180° ...[T he sum of all three angles of a triangle is 180°]
⇒ ∠CAB + ∠CBA + 90° = 180°
⇒ ∠CAB + ∠CBA = 180° - 90°
⇒ ∠CAB + ∠CBA = 90°
⇒ ∠CAB/2 + ∠CBA/2 = 90/2 = 45°-------(3)
Step 3
Now, In triangle ABD,
∠BAD + ∠ABD + ∠ADB = 180°
⇒ ∠CAB/2 + ∠CBA/2 + ∠ADB = 180° ...Using (1) &(2)
⇒ 45° + ∠ADB = 180° ...Using (3)
⇒ ∠ADB = 180° - 45° = 135°
Step 4
Hence, ∠ADB = 135°
(4) c. 167°
Step 1
If you look at the question caref ully, you will notice that f irst of all we have to f ind the
complement of 77° , then f ind the supplement of the complement of 77°.
Step 2
T he sum of the complementary angles is 90°.
T heref ore the complement of 77°
= 90° - 77°
= 13°
Step 3
T he sum of supplementary angles is 180°.
T heref ore, the supplement of 13°
= 180° - 13°
= 167°
Step 4
T heref ore the value of the supplement of the complement of 77° is 167° .
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ID : in-7-Lines-and-Angles [7]
(5)
11
Step 1
It is given that, CE bisect angle ∠ACB.
T heref ore, ∠ACE = ∠ACB/2 -----(1)
Step 2
In triangle ABC, ∠CAB + ∠ABC + ∠ACB = 180° ...[Since the sum of all three angles of a
triangle is 180°]
⇒ 50° + 28° + ∠ACB = 180°
⇒ 78° + ∠ACB = 180°
⇒ ∠ACB = 180° - 78°
⇒ ∠ACB = 102°
⇒ ∠ACB/2 = 102/2
⇒ ∠ACE = 51° ...[From equation (1)]
Step 3
Now in triangle ADC, ∠CAD + ∠ADC + ∠DCA = 180°
⇒ 50° + 90° + ∠DCA = 180°
⇒ ∠DCA = 180° - 90° - 50°
⇒ ∠DCA = 40°
Step 4
Now,
∠DCE = ∠ACE - ∠DCA
⇒ ∠DCE = 51° - 40°
⇒ ∠DCE = 11°
Step 5
Hence, the value of angle ∠DCE is 11°.
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ID : in-7-Lines-and-Angles [8]
(6)
55
Step 1
As per the question ∠CBD is exterior angle of the triangle ABC and we know that an
exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent
interior angles.
T heref ore, ∠CBD = ∠CAB + 110° ----(1)
Step 2
In triangle ABC, ∠CAB + ∠ABC + ∠ACB = 180°
⇒ ∠CAB + ∠ABC = 180° - 110°
⇒ ∠CAB + ∠ABC = 70° -----(2)
Step 3
It is given that AP and BP are bisectors of angles ∠CAB and ∠CBD respectively.
T heref ore, ∠PAB = ∠CAB/2 -----(3)
∠CBP = ∠CBD/2 -----(4)
Step 4
Now, in triangle ABP, ∠PAB + ∠ABP + ∠APB = 180° ...[Since the sum of all three angles
of a triangle is 180°]
⇒ ∠APB = 180° - ∠PAB - ∠ABP
⇒ ∠APB = 180° - ∠CAB/2 - ∠ABP ..[From equation (3), ∠PAB = ∠CAB/2]
⇒ ∠APB = 180° - ∠CAB/2 - (∠ABC + ∠CBP)
⇒ ∠APB = 180° - ∠CAB/2 - (∠ABC + ∠CBD/2) ...[From equation (4), ∠CBP = ∠CBD/2]
⇒ ∠APB = 180° - ∠CAB/2 - {∠ABC + (∠CAB + 110°)/2} ...[From equation (1)]
⇒ ∠APB = 180° - ∠CAB/2 - (∠ABC + ∠CAB/2 + 55°)
⇒ ∠APB = 180° - ∠CAB/2 - ∠ABC - ∠CAB/2 - 55°
⇒ ∠APB = 125° - (∠CAB + ∠ABC)
⇒ ∠APB = 125° - 70° ...[From equation (2)]
⇒ ∠APB = 55°
Step 5
Hence, the value of angle ∠APB is 55°.
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ID : in-7-Lines-and-Angles [9]
(7)
A)
45
Step 1
Let us look at the directions as shown below:
Step 2
We can see that the angle between the South-West and South directions is 45°.
B)
180
Step 1
Let us look at the directions as shown below:
Step 2
We can see that the angle between the South and East directions is 90°.
Also, the angle between the East and North directions is 90°.
Step 3
T heref ore, the angle between the South and North directions is = 90° + 90° =
180°
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ID : in-7-Lines-and-Angles [10]
C)
180
Step 1
Let us look at the directions as shown below:
Step 2
We can see that the angle between the West and South directions is 90°.
Also, the angle between the South and East directions is 90°.
Step 3
T heref ore, the angle between the West and East directions is = 90° + 90° = 180°
D)
90
Step 1
Let us look at the directions as shown below:
Step 2
We can see that the angle between the North and West directions is 90°.
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ID : in-7-Lines-and-Angles [11]
(8)
50
Step 1
It is given that line AB and CD and parallel lines and the third line (say EF) cuts the lines AB
and CD at certain angle as shown in the f igure above.
Let us redraw the f igure as below:
∠a = ∠c (vertically opposite angles)
∠c = ∠e (alternate interior angles)
T heref ore we can write,
∠a = ∠c = ∠e = ∠g
Again,
∠b = ∠d (vertically opposite angles)
∠d = ∠f (alternate interior angles)
T heref ore we can write,
∠b = ∠d = ∠f = ∠h
We know that sum of two adjacent angle is equal to 180°.
T heref ore, f rom the diagram, you can write,
∠a + ∠b = 180°,
∠b + ∠c = 180°,
∠c + ∠d = 180°,
∠d + ∠a = 180°
Step 2
Here, ∠e = 130° and ∠d = x
∠e + ∠h = 180°
⇒ 130° + ∠h = 180°
⇒ ∠h = 180° - 130°
⇒ ∠h = 50°
As ∠h is equal to ∠d, x is 50°.
Step 3
T heref ore, the value of x is 50°.
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ID : in-7-Lines-and-Angles [12]
(9)
120
Step 1
In triangle BCE, ∠CBE + ∠BCE + ∠BEC = 180° ...[Since the sum of all three angles of a
triangle is 180°]
⇒ 35° + 115° + ∠BEC = 180°
⇒ 150° + ∠BEC = 180°
⇒ ∠BEC = 180° - 150°
⇒ ∠BEC = 30°
Step 2
Since ∠AED and ∠BEC are the opposite angles of intersecting lines AB and CD and we
know that the opposite angles are congruent
T heref ore, ∠AED = ∠BEC = 30° ------(1)
Step 3
Now, in triangle ADE, ∠DAE + ∠ADE + ∠AED = 180° ...[Since the sum of all three
angles of a triangle is 180°]
⇒ 30° + x + 30° = 180° ...Using (1)
⇒ 60° + x = 180°
⇒ x = 180° - 60°
⇒ x = 120°
Step 4
Hence, the value of x is 120°.
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ID : in-7-Lines-and-Angles [13]
(11)
21
Step 1
Parallel line AB and CD are intersected by a transversal as shown below,
Here angle ∠P and ∠Q are corresponding angles. i.e.
∠P = ∠Q
Step 2
On comparing given angles with ∠P and ∠Q,
3x + 2x = 105°
⇒ 5x = 105°
⇒ x = 21°
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ID : in-7-Lines-and-Angles [14]
(12)
81
Step 1
If you look at the f igure caref ully, you will notice that the angle ∠BAC = 49°, ∠BDE = 50°.
Step 2
According to question AB and DE are parallel.
T heref ore the angles ∠ABC and ∠BDE are alternate interior angles.
∠ABC = ∠BDE [Alternate interior angles]
⇒ ∠ABC = 50°
Step 3
T he sum of all three angles of a triangle is 180°.
Now in triangle ABC,
∠ABC + ∠BAC + ∠ACB = 180°
⇒ 50° + 49° + ∠ACB = 180° [Since ∠ABC = 50° and ∠BAC = 49°]
⇒ 99° + ∠ACB = 180°
⇒ ∠ACB = 180° - 99°
⇒ ∠ACB = 81°
Step 4
T heref ore the value of ∠ACB is 81°.
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ID : in-7-Lines-and-Angles [15]
(13)
55
Step 1
We know that the angle made by a straight line is 180°.
T heref ore, we can write, 125° + y = 180°
or y = 180° - 125° = 55°
Step 2
When a straight line cuts any two parallel lines, its Corresponding Angles are equal.
Since angles x and y are Corresponding angles.
T heref ore, ∠x = ∠y = 55° ...[Corresponding angles of two parallel lines are equal]
(14)
54
Step 1
If you look at the angles 99° and X + 45°, these are opposite angles
Step 2
We know that opposite angles are equal.
T heref ore X + 45° = 99°
⇒ X = 99° - 45°
⇒ X = 54°
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ID : in-7-Lines-and-Angles [16]
(15) False
Step 1
Let's consider the triangle ABC in the f igure above. Since we know that the sum of all three
angles of a triangle is 180°, in ΔABC:
∠A + ∠B + ∠C = 180°.
Step 2
Let's assume that ∠A of the ΔABC is an obtuse angle.
T hat is, ∠A > 90°.
Now, ∠A + ∠B + ∠C = 180°
⇒ ∠B + ∠C = 180° - ∠A
⇒ ∠B + ∠C < 90° (Since ∠A > 90°)
Step 3
We just saw that the sum of ∠B and ∠C of the ΔABC is less than 90°. T heref ore, we can
say that the ∠B and the ∠C must be acute angles and the statement "A triangle can have
two obtuse angles" is False.
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