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On 4th Sept, 1882, for the first time, New York shone in the brightness of electric light: True.

(d) gramophone.

Edison died on 4th September, 1882: False.

We know that the sum of all the angles of a triangle is equal to 180°.

Therefore, for △ABD:

∠ABD +∠ADB + ∠BAD = 180° (Sum of the angles of △ABD)

50° + x + 50° = 180°

100° + x = 180°

x = 180° – 100°

x = 80°

For △ ABC:

∠ABC + ∠ACB + ∠BAC = 180° (Sum of the angles of △ABC)

50° + z + (50° + 30°) = 180°

50° + z + 50° + 30° = 180°

z = 180° – 130°

z = 50°

Using the same argument for △ADC:

∠ADC + ∠ACD + ∠DAC = 180° (Sum of the angles of △ADC)

y + z + 30° = 180°

y + 50° + 30° = 180° (z = 50°)

y = 180° – 80°

y = 100°

Therefore, we can conclude that the required angles are 80°, 50° and 100°.

We can see that in △ADC, ∠ADC is equal to 90°.

(△ADC is a right triangle)

We also know that the sum of all the angles of a triangle is equal to 180°.

Which means: 45° + 90° + y = 180° (Sum of the angles of △ADC)

135° + y = 180°

y = 180° – 135°.

y = 45°.

We can also say that in △ ABC, ∠ABC + ∠ACB + ∠BAC is equal to 180°.

(Sum of the angles of △ABC)

40° + y + (x + 45°) = 180°

40° + 45° + x + 45° = 180° (y = 45°)

x = 180° –130°

x = 50°

Therefore, we can say that the required angles are 45° and 50°.

 In △ABC and △ADE we have,

∠ ADE = ∠ ABC [corresponding angles]

x = 40°

∠AED = ∠ACB (corresponding angles)

y = 30°

We know that the sum of all the three angles of a triangle is equal to 180°

x + y + z = 180° (Angles of △ADE)

Which means: 40° + 30° + z = 180°

z = 180° – 70°

z = 110°

Therefore, we can conclude that the three angles of the given triangle are 40°, 30° and 110°

Given that one angle of a triangle is equal to the sum of the other two

Let the measure of angles be x, y, z

Therefore we can write above statement as x = y + z

x + y + z = 180°

Substituting the above value we get

x + x = 180°

2x = 180°

x = 180/2

x = 90°

If one angle is 90° then the given triangle is a right angled triangle

In the given figure, side BC of ΔABC is produced to D.

Consider the ΔABC,

We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ ∠ABC + ∠BAC = ∠ACD

= x^o + 32^o = 65^o

= x = 65^o – 32^o

= x = 33^o

Also, we know that the sum of all the angles of a triangles is 180^o.

∴ A + B + C = 180^o

= 32^o + x^o + y^o = 180^o

= 32^o + 33^o + y^o = 180^o

= y + 65^o = 180^o

= y = 180^o – 65^o

= y = 115^o

Hence, the value of x is 33^o and value of y is 115^o.

Given that each angle of a triangle is less than the sum of the other two

Let the measure of angles be x, y and z

From the above statement we can write as

x > y + z

y < x + z

z < x + y

Therefore triangle is an acute triangle.

Given that angles of a triangle are in the ratio: 3: 4: 5

Therefore let the measure of the angles be 3x, 4x, 5x

We know that sum of the angles of a triangle =180°

3x + 4x + 5x = 180°

12x = 180°

x = 180/12

x = 15°

Smallest angle = 3x

= 3 × 15°

= 45°

Therefore smallest angle = 45°

We know that the sum of the angles of a triangle is 180^o.

∴ ∠X + ∠Y + ∠Z = 180^o

= 90^o + ∠Y + 48^o = 180^o

= 138^o + ∠Y = 180^o

= ∠Y = 180^o– 138⁰

= ∠Y = 42^o

Hence, the measures of ∠Y is 42^o.

(i) True

(ii) False

Explanation:

Any three non-parallel line segments can make up a triangle.

(iii) False.

Explanation:

Any three non-parallel line segments can make up a triangle.

(iv) False.

Explanation:

The interior of a triangle is the region enclosed by the triangle and the vertices are not enclosed by the triangle.

(v) True.

Explanation:

The triangular region includes the interior region and the triangle itself.

(vi) False.

Explanation:

The vertices of a triangle are three non-collinear points.

(vii) True.

Explanation:

In an equilateral triangle, any two sides are equal.

(viii) False.

Explanation:

A right triangle can also be an isosceles triangle.

(ix) False.

Explanation:

Each acute triangle is not an equilateral triangle, but each equilateral triangle is an acute triangle.

(x) False.

Explanation:

An isosceles triangle can be an obtuse triangle, a right triangle or an acute triangle

(i) 63° + 37° + 80° = 180°

Angles form a triangle

(ii) 45°, 61°, 73° is not equal to 180°

Therefore not a triangle

(iii) 59°, 72°, 61° is not equal to 180°

Therefore not a triangle

(iv) 45° + 45°+ 90° = 180°

Angles form a triangle

(v) 30°, 20°, 125° is not equal to 180°

Therefore not a triangle

In the given figure, side BC of ΔABC is produced to D.

Consider the ΔABC,

We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ ∠ABC + ∠BAC = ∠ACD

= 68^o + x = 130^o

= x = 130^o – 68^o

= x = 62^o

Also, we know that the sum of all the angles of a triangles is 180^o.

∴ x + y + 68^o = 180^o

= 62^o + y + 68^o = 180^o

= y + 130^o = 180^o

= y = 180^o – 130^o

= y = 50^o

Hence, the value of x is 62^o and value of y is 50^o.

Let the measures of the given angles of the triangles be (4x)^o, (3x)^o and (2x)^o respectively.

Then,

= 4x + 3x + 2x = 180^o … [∵sum of the angles of a triangle is 180^o]

= 9x = 180^o

= x = 180/9

= x = 20

So, the angle measures (4 × 20)^o, (3 × 20)^o, (2 × 20)^o,

i.e., 80^o, 60^o, 40^o.

Hence, the angles of the triangles are 80^o, 60^o, 40^o.

In △ABC and △ADE we have:

∠ADE = ∠ABC (Corresponding angles)

y = 50°

Also, ∠AED = ∠ACB (Corresponding angles)

z = 40°

We know that the sum of all the three angles of a triangle is equal to 180°.

We can write as x + 50° + 40° = 180° (Angles of △ADE)

x = 180° – 90°

x = 90°

Therefore, we can conclude that the required angles are 50°, 40° and 90°.

Given one of the angles of a triangle is 130°

Also given that remaining two angles are equal

So let the second and third angle be x

We know that sum of all the angles of a triangle = 180°

130° + x + x = 180°

130° + 2x = 180°

2x = 180° – 130°

2x = 50°

x = 50/2

x = 25°

Therefore the two other angles are 25° each

Given two angles of a triangle are of measures 150° and 30°

Let the required third angle be x

We know that sum of all the angles of a triangle = 180°

105° + 30° + x = 180°

135° + x = 180°

x = 180° – 135°

x = 45°

Therefore the third angle is 45°

In the given figure, side BC of ΔABC is produced to D.

Consider the ΔABC,

We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ ∠ACD = ∠ABC + ∠BAC

∠ACD = 45^o + 75^o

∠ACD = 120^o

Hence, the measures of ∠ACD is 120^o.

Given that one of the angles of the given triangle is 60°.

Also given that the other two angles of the triangle are in the ratio 1: 2.

Let one of the other two angles be x.

Therefore, the second one will be 2x.

We know that the sum of all the three angles of a triangle is equal to 180°.

60° + x + 2x = 180°

3x = 180° – 60°

3x = 120°

x = 120°/3

x = 40°

2x = 2 × 40°

2x = 80°

Hence, we can conclude that the required angles are 40° and 80°.

Given that one of the angles of the given triangle is 100°.

Also given that the other two angles are in the ratio 2: 3.

Let one of the other two angles be 2x.

Therefore, the second angle will be 3x.

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

100° + 2x + 3x = 180°

5x = 180° – 100°

5x = 80°

x = 80/5

x = 16

2x = 2 ×16

2x = 32°

3x = 3×16

3x = 48°

Thus, the required angles are 32° and 48°.

Let the other two equal angles be x.

Then,

= x + x + 100^o = 180^o … [∵sum of the angles of a triangle is 180^o]

= 2x = 180^o – 100^o

= 2x = 80^o

= x = 80^o/2

= x = 40^o

Hence, the other two equal angles are 40^o and 40^o.

Given that three angles of a triangle are equal to one another

So let the each angle be x

We know that sum of all the angles of a triangle = 180°

x + x + x = 180°

3x = 180°

x = 180/3

x = 60°

Therefore angle is 60° each

Given angles of the triangle are in the ratio 1: 2: 3

So take first angle as x, second angle as 2x and third angle as 3x

We know that sum of all the angles of a triangle = 180°

x + 2x + 3x = 180°

6x = 180°

x = 180/6

x = 30°

2x = 30° × 2 = 60°

3x = 30° × 3 = 90°

Therefore the first angle is 30°, second angle is 60° and third angle is 90°.

Given the angles of a triangle are (x − 40)°, (x − 20)° and (1/2 − 10)°.

We know that sum of all the angles of a triangle = 180°

(x − 40)° + (x − 20)° + (1/2 − 10)° = 180°

x + x + (x/2) – 40° – 20° – 10° = 180°

x + x + (x/2) – 70° = 180°

(5x/2) = 180° + 70°

(5x/2) = 250°

x = (2/5) × 250°

x = 100°

Hence the value of x is 100°

∠BAE = ∠EDC = 52° [Alternate angles]

Sum of all angles of a triangle = 180° 

x = 180° – 40° – 52° 

= 180° − 92° 

= 88°

The angles of a triangle are (x − 40)°, (x − 20)° and (1/2 x − 10)° 

Sum of all angles of triangle = 180° 

(x − 40)° + (x − 20)° + (1/2 x − 10)° = 180° 

5/2 x – 70° = 180°

5/2 x = 180° + 70° 

5x = 2(250)° 

x = 500/5 

x = 100°

∠ABC = 180° – 120° = 60° [Linear pair] 

∠ACB = 180° – 110° = 70° [Linear pair] 

Sum of all angles of a triangle = 180° 

x = ∠BAC = 180° − ∠ABC − ∠ACB 

= 180° – 60° – 70° 

= 50°

Angles of a triangle are in the ratio 1:2:3

Let the angles be x, 2x, 3x

Sum of all angles of triangles = 180°

x + 2x + 3x = 180° 

6x = 180° 

x = 180/6 

x = 30°

Answer: x = 30° 

2x = 2(30)° = 60° 

3x = 3(30)° = 90°

Triangle is a three-sided polygon that consists of three edges and three vertices. The most important property of a triangle is that the sum of the internal angles of a triangle is equal to 180 degrees.

Let ∠BAD = y, ∠BAC = 3y 

∠BDA = ∠BAD = y (As AB = DB)

Now, ∠BAD + ∠BAC + 108° = 180° [Linear Pair] 

y + 3y + 108° = 180° 

4y = 72° 

or y = 18°

Now, In ΔADC 

∠ADC + ∠ACD = 108° [Exterior Angle Property] 

x + 18° = 180° 

x = 90°

C) (45°, 45°)

Let angles are x° and x°.

(\(\because\) Given that both angles are equal)

\(\therefore\) x° + x° = 90° (\(\because\) Given angles are complementary to each other)

⇒ 2x° = 90°

⇒ x° = 90°/2 = 45°

Hence, required angles are 45° and 45°.

Correct option is D) 45°

Length of Hamida’s rectangular field = 20 cm
and breadth = 15 cm

Base of Kali’s triangular field = 25 cm

Area of triangluar = Area of Hamida’s rectangular shape field

⇒ 1/2 x base x height = length x breadth

⇒ 1/2 x 25 x height = 20 x 15

⇒ height = (20 x 15 x 2)/25  = 24 cm

Correct option is (B) 91°

Supplementary angle to \(89^\circ\) \(=180^\circ-89^\circ=91^\circ.\)

\((\because\) Sum of supplementary angles is \(180^\circ)\)

Correct option is  B) 91°

If the sum of any two angles is 90°, then the angles are called complementary angles.

Complementary angle of 2° is (90 – 2) = 88°

In ∆ABC, AB = AC = 7.6 cm

Base BC = 9.5 cm

Height o f triangle AD = 4 cm

Area of ∆ABC = 1/2 x Base x Height

= 1/2 x BC x AD = 1/2 x 9.5 x 4

= 9.5 x 2 = 19 sq. cm

In isosceles triangle ABC

AC = 7.6 cm and BE = ?

Area of ∆ABC = 19 sq. cm

⇒ 1/2 x AC x BE = 19

⇒ 1/2 x 7.6 x BE = 19

7.6 x BE = 19 x 2

⇒ BE = 38/7.6 = 5 cm

If the sum of any two angles is 90°, then the angles are called complementary angles.

Complementary angle of k° is (90 – k)°

The base of a triangle is 12 cm and height is 5 cm then area of triangle will be 30 sq. cm.

Number of boxes covered by the triangles are as follows:

Number of boxes covered by ∆A’BC = 155 sq. unit

Number of boxes covered by ∆A’BC = 155 sq. unit

Number of boxes covered by ∆A”BC = 155 sq. unit

So, if is clear that the area of all three triangles are equal.

Let base of parallelogram = x cm

Height of parallelogram = x/4

Area of parallelogram = 144 sq. cm

⇒ Base x Height = 144 x 4

x × x/4 = 144

x² = 144 × 4

x = \(\sqrt{144 \times 4}\)

x = 12 x 2 = 24 cm

Base of parallelogram = 24 cm

Height of parallelogram = x/4 = 24/4 = 6 cm.

If the sum of any two angles is 90°, then the angles are called complementary angles.

Complementary angle of 43 is (90 – 43) = 47°

Yes. 

Each angle in any pair of complementary angles is always acute.

As sum of two angles is 90°, each of the angle must be less than 90°.

C) 54°, 126°

Correct option is (C) 18°

\(\because\) \(\frac x4+90^\circ+x=180^\circ\)

\(\Rightarrow\) \(x+\frac x4=180^\circ-90^\circ=90^\circ\)

\(\Rightarrow\) \(\frac{5x}4=90^\circ\)

\(\Rightarrow\) \(x=90^\circ\times\frac45=72^\circ\)

\(\therefore\) \(\angle DOC\) \(=\frac x4=\frac{72^\circ}4=18^\circ\)

Correct option is  C) 18°

Let the angle be x° 

Then Its complement = (900- x°) 

By problem, x° = 90° – x° 

x° + x°= 90° 

2x = 90°

x = 90°/2

x = 45°

Let one of supplement be x°

Another supplement be = 180° – x°

According to the question = x° = 180° – x°

x° + x° = 180°

2x° = 180°

x° = 180/2 = 90°

(d) 150^o

Sum of all angle about a point given in the figure are equal to 360^o.

Then, 100^o + 46^o + 64^o + x = 360^o

210^o + x = 360^o

x = 360^o – 210^o

x = 150^o

(d) ∠a = ∠d

∠a ≠ ∠d

∠a = ∠b [because vertically opposite angles]

∠d = ∠c [because vertically opposite angles]

∠a + ∠d = 180^o

(c) 80^o

We know that, ∠APR = ∠PRD … [because interior alternate angles]

∠APQ + ∠QPR = 130^o

50^o + ∠QPR = 130^o

∠QPR = 130^o – 50^o

∠QPR = 80^o