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Measure of each exterior angle = \(\frac { { 360 }^{ 0 } }{ n }\)

n = \(\frac { 360 }{ 45 }\) = 8

Hence number of sides in the polygon = 8.

Answer is (B) (2n - 4)/n right angle

Here, it is given that the measure of each interior angle of a regular polygon =110° (if possible)

Hence, measure of each exterior angle 

= 180° – 110° = 70°.

Suppose number of sides of the polygon = n
But sum of all the n exterior angles 

= 360°

⇒ n x 70° = 360°

⇒ n = \(\frac { 360^o }{ 70^o }\)

= 5\(\frac { 1 }{ 7 }\) ≠ a whole number.

Hence, a regular polygon having measure of each interior angle 110° can not exist.

Answer is (B) 108°

Answer is (C) 8 right angles

Answer is (C) an octagon

Answer is (B) 52(1/2)°

It is given that the measure of each interior angle of a regular polygon = 137° (if possible)

Suppose number of sides of the polygon = n

Sum of all then exterior angles = 360°

⇒ n x 43° = 360°

⇒ n = 360°/43°

= 8(16/43) ≠ a whole number

Hence, a regular polygon having a measure of each interior angle 137° cannot exist.

∠A + ∠B + ∠C = 180° …(i) (by angle sum property of a triangle)

But ∠A + ∠B = 2∠C (given)

2∠C + ∠C = 180°

⇒ 3∠C = 180°

⇒ ∠C = 60°

From (i), ∠A + ∠B + 60° = 180°

⇒ ∠A + ∠B = 120°

But ∠A = 2 ∠B (given)

2∠B + ∠B = 120°

⇒ 3∠B = 120°

⇒ ∠B = 40°

∠A = 2 x 40 = 80°

Hence angles of a ΔABC are 80°, 40°, 60°.

Let two angles are x and y, then according to question

∠x + ∠y = 123° …(i)

and ∠x – ∠y = 13° …(ii)

Adding (i) and (ii), we get

2∠x = 136° ⇒ ∠x = 68°

Now substituting ∠x = 68° in equation (i), we get

68° + ∠y = 123°

⇒ ∠y = 123° – 68° = 55°

Third angle = 180° – 123° = 57°

Hence angles of a triangle are 68°, 55°, 57°.

Answer is (C) 115°

At point B,

100° + ∠ABC = 180° (linear pair angles)

∠ABC = 80°

At point C,

∠DCB + 95° = 180°

⇒ ∠DCB = 85°

Similarly at D,

82° + ∠CDA = 180° (linear pair axiom)

∠CDA = 180° – 82° = 98°

Now in quadrilateral ABCD

∠A + 80° + 85° + 98° = 360° (by angle sum property of a quadrilateral)

⇒ ∠A = 360° – 263°

⇒ ∠A = 97°.

∠ACB = 180° – 120° = 60° (due to linear pair)

Exterior ∠B = ∠A + ∠ACB

112° = ∠A + 60°

⇒ ∠A = 52°

∠QPR = 180° – (50° + 45°) 

= 180° – 95° = 85°

∠A + ∠B + ∠C = 180°

⇒ ∠A + 60° + 40° = 180°

⇒ ∠A = 80°

Given: The side BC of ΔABC is produced to D.

The bisector of ∠A meets BC at P.

To prove: ∠ABC + ∠ACD = 2∠APC

Proof: ∠APC = ∠ABP + ∠BAP (the exterior angle is  equal to sum of opposite interior angles)
∠APC = ∠B + \(\frac { 1 }{ 2 }\) ∠A …(i)

Now in ΔABC

∠ABC + ∠BAC = ∠ACD …(ii) (reason as above)

Adding ∠ABC to both side of equation (ii), we get

∠ABC + ∠ACD = ∠ABC + ∠ABC + ∠BAC

= 2 ∠ABC + ∠BAC

= 2 ∠B + ∠A

= 2(∠B + \(\frac { 1 }{ 2 }\) ∠A)

= 2 ∠APC [from equation (i)]

∠ABC + ∠ACD = 2 ∠APC.

Given: In quad. ABCD the bisectors of ∠A and ∠B intersect at O.

To prove: 2∠AOB = ∠C + ∠D

Proof: In quad. ABCD

∠A + ∠B + ∠C + ∠D = 360° …(i) (angle sum property of a quad.)

Also in ΔAOB

∠OAB + ∠OBA + ∠AOB = 180° (by angle sum property of a Δ)

⇒ \(\frac { 1 }{ 2 }\) ∠A + \(\frac { 1 }{ 2 }\) ∠B + ∠AOB = 180°

⇒ ∠A + ∠B + 2∠AOB = 360°…(ii)

From (i) and (ii), we get

∠A + ∠B + ∠C + ∠D = ∠A + ∠B + 2∠AOB

⇒ 2∠AOB = ∠C + ∠D

Answer is (D) 155°

Here ∠DCE = 53°

∴∠ACB = 53°

(Vertically opposite angle)

In ∆ABC

∠ABC + ∠ACB = ∠CAF

(By exterior angle property)

⇒ ∠ABC + 53°= 112°

⇒ ∠ABC = 112° – 53°

⇒ ∠ABC = 59°

Now in ∆ABC

∠BAC + ∠ABC + ∠BCA = 180°

(by angle sum property)

⇒ ∠BAC + 59° + 53° = 180°

⇒ ∠BAC = 68°

Hence, ∠BAC = 68°, ∠ABC = 59° and ∠ACB = 53°.

\(\frac { 3x }{ 2 }\) + x = 180° (Linear pair axiom)

⇒ 5x = 360°

⇒ x = 72°

Sum of the four angles of a quadrilateral is 50° + 60° + 112° + 130° = 352°

But sum of all the four angles of a quadrilateral is always 360°.

Hence given statement is false.

No. of diagonals = \(\frac { 10(10-1) }{ 2 }\) – 10 

= 45 – 10 = 35

According to question

x + x + 96° = 180° (by angle sum property of a triangle)

⇒ 2x + 96° = 180°

⇒ 2x = 84°

⇒ x = 42°

Measures of each equal angles of an isosceles triangle is 42°.

BA || FE

∠x = 42° (alternate angle)

∠ACD = ∠x + 66° (ext. angle is equal to sum of opposite interior angles)

∠ACD = 42° + 66° = 108°

But ∠y + ∠ACD = 180°

∠y = 180° – 108° (linear pair axiom)

∠y = 72°

∠x = 42° and ∠y = 72°

∵ ∆ABC is an equilateral triangle

∴AB = BC = CA

⇒ ∠ABC = ∠ACB = ∠BAC = 60°

60° = x + 22°

(exterior angle property)

x = 38°

Also

38° + 22° + ∠z = 180°

(by angle sum property of a ∆)

⇒ ∠z = 180° – 60° = 120°

Again by exterior angle property

⇒ ∠ACB = ∠y + 38°

⇒ 60° = ∠y + 38°

⇒ ∠y = 22°

Hence, ∠x = 38°, ∠y = 22° and ∠z = 120°

In ∆BCE

88° = 22° + ∠z (by ext. angle property)

∠z = 88° – 22°

⇒ ∠z = 66°

Also AB || DC

88° + ∠y = 180° (interior angle property)

∠y = 180° – 88°

⇒ ∠y = 92°

Again, 102° + ∠DAB = 180° (linear pair axiom)

∠DAB = 180° – 102° = 78°

AB || DC, then x + 78° = 180° (The sum of the interior angles on the same of a transversal is 180°)

⇒ x = 102°

Hence ∠x = 102°, ∠y = 92°, ∠z = 66°

According to figure

∠x + ∠y = 120° …(i) (by exterior angle property)

and ∠x – ∠y = 10° (given) …(ii)

2∠x = 130°

⇒ ∠x = 65°

Now, 65° + ∠y = 120°

⇒ ∠y = 120° – 65°

⇒ ∠y = 55°

Hence, ∠x = 65° and ∠y = 55°.

In ∆ACE

∠A + ∠C + ∠E = 180°

(by angle sum property of a triangle)

⇒ 31° + 75° + ∠E = 180°

⇒ ∠E = 180° – 106°

⇒ ∠E = 74°

Answer is (C) 60°

Let equal angles be x each and third angle 

= x + 21°

Then according to question, we have

x + x + (x + 21°) = 180° (angle sum property)

⇒ 3x + 21° = 180°

⇒ 3x = 159°

⇒ x = 53°

Measures of each angles are 53°, 53° and 74°.

Answer is (D) 60°

Suppose number of sides of the polygon be n.

According to problem

2 x 90° + (n – 2) x 150°=(2n – 4) x 90°

⇒ 180 + 150n – 300 = 180n – 360°

⇒ 360 – 300 + 180 = 180n – 150n

⇒ 240 = 30n

⇒ n = 8

Answer is (A) 66°

We are given that AB || DE

and ∠BAC = 35° and ∠CDE = 53°

AB || DE (given)

⇒ ∠BAE = ∠AED = 35°

(alternate angles)

Now in ∆CDE,

∠CDE + ∠E + ∠DCE = 180°

(angle sum property of a triangle)

⇒ 53° + 35° + ∠DCE = 180°

⇒ ∠DCE = 180° – 88° = 92°

⇒ ∠DCE = 92°

Hence, ∠DCE = 92°.

Answer is (D) right-angled triangle

Answer is (A) acute angles

No, if a triangle have two right angles, then sum of two angles will be 180° which is not possible.

Let angles of a triangle are 2x, 3x, 4x
2x + 3x + 4x = 180°

⇒ 9x = 180°

⇒ x = 20°

Angles are

2x = 2 x 20° = 40°,

3x = 3 x 20° = 60°,

4x = 4 x 20° = 80°

The sum of the interior angles of a polygon = (2n – 4) x 90°

⇒ 2160° = 180°n – 360°

⇒ 2160° + 360° = 180°n

∴ n = 2520°/180° = 14

Hence, number of sides of required polygon is 14.

Answer is (B) 100°

Answer is (C) 18°

Answer is (A) 80°

Number of diagonals of a polygon of n sides 

= \(\frac { n(n-1) }{ 2 }\) – n

⇒ 14 = \(\frac { n(n-1) }{ 2 }\) – n

⇒ 28 = n² – n – 2n

⇒ n² – 3n – 28 = 0

⇒ n² – 7n + 4n – 28 = 0

⇒ n(n – 7) + 4(n – 7) = 0

⇒ (n – 7) (n + 4) = 0

⇒ n = 7

Suppose base angles are B and C

According to problem

∠B + ∠C = 130° (given)

∠B – ∠C = 40° (given)

⇒ 2 ∠B = 170°

⇒ ∠B = 85°

∠C = 130° – 85° = 45°

Hence, ∠B = 85°, ∠C = 45° and ∠A = 50°.

We know that

∠BOC = 90° + \(\frac { 1 }{ 2 }\) ∠A

∵ ∠A = 80°

⇒ ∠BOC = 90° + \(\frac { 1 }{ 2 }\) x 80° 

= 90° + 40° = 130°

∠C = 180° – (58° + 67°) 

= 180° – 125° = 55°