Triangles

A triangle is possible only if the sum of its three angles is 180°.

Solving,

⇒ 20° + 70° + 90° = 180°.

Since the sum is 180°, such a triangle is possible.

Hence, a triangle with angles 20°, 70° and 90° can be formed.

A triangle is possible only if the sum of its three angles is 180°.

Solving,

⇒ 40° + 130° + 20° = 190° ≠ 180°.

Since the sum is not 180°, such a triangle is not possible.

Hence, a triangle with angles 40°, 130° and 20° cannot be formed.

A triangle is possible only if the sum of its three angles is 180°.

Solving,

⇒ 60° + 60° + 50° = 170° ≠ 180°.

Since the sum is not 180°, such a triangle is not possible.

Hence, a triangle with angles 60°, 60° and 50° cannot be formed.

A triangle is possible only if the sum of its three angles is 180°.

Solving,

⇒ 125° + 40° + 15° = 180°.

Since the sum is 180°, such a triangle is possible.

Hence, a triangle with angles 125°, 40° and 15° can be formed.

Let each angle of the triangle be x.

By the angle sum property of a triangle,

⇒ x + x + x = 180°

⇒ 3x = 180°

⇒ x = $\frac{180^{\circ}}{3}$

⇒ x = 60°

Hence, each angle of the triangle is 60°.

By the angle sum property of a triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 45° + 75° + ∠C = 180°

⇒ ∠C = 180° − 120° = 60°

Hence, ∠C = 60°.

Given ∠P = 60° and ∠Q = ∠R. Let ∠Q = ∠R = x.

By the angle sum property of a triangle,

⇒ ∠P + ∠Q + ∠R = 180°

⇒ 60° + x + x = 180°

⇒ 60° + 2x = 180°

⇒ 2x = 180° − 60°

⇒ 2x = 120°

⇒ x = $\frac{120^{\circ}}{2}$

⇒ x = 60°

Hence, ∠R = 60°.

By the angle sum property of a triangle,

⇒ x° + 90° + 30° = 180°

⇒ x° = 180° − 120° = 60°

Hence, x° = 60°.

By the angle sum property of a triangle,

⇒ y° + 80° + 20° = 180°

⇒ y° = 180° − 100° = 80°

Hence, y° = 80°.

By the angle sum property of a triangle,

⇒ a° + 90° + 40° = 180°

⇒ a° = 180° − 130° = 50°

Hence, a° = 50°.

By the angle sum property of a triangle,

⇒ 5x° + 4x° + x° = 180°

⇒ 10x° = 180°

⇒ x° = $\frac{180^{\circ}}{10}$

⇒ x° = 18°

∴ ∠A = 5x° = 5 × 18° = 90°, ∠B = 4x° = 4 × 18° = 72° and ∠C = x° = 18°.

Hence, ∠A = 90°, ∠B = 72° and ∠C = 18°.

By the angle sum property of a triangle,

⇒ x° + 2x° + 2x° = 180°

⇒ 5x° = 180°

⇒ x° = $\frac{180^{\circ}}{5}$

⇒ x° = 36°

∴ ∠A = x° = 36°, ∠B = ∠C = 2x° = 2 × 36° = 72°.

Hence, ∠A = 36° and ∠B = 72° = ∠C.

From the figure, two angles of the triangle are each equal to b° and the third angle is 50°. By the angle sum property,

⇒ b° + 50° + b° = 180°

⇒ 2b° + 50° = 180°

⇒ 2b° = 180° − 50°

⇒ 2b° = 130°

⇒ b° = $\frac{130^{\circ}}{2}$

⇒ b° = 65°

Hence, b° = 65°.

From the figure, two angles of the triangle are each equal to x° and the third angle is 90°. By the angle sum property,

⇒ x° + 90° + x° = 180°

⇒ 2x° + 90° = 180°

⇒ 2x° = 180° − 90°

⇒ 2x° = 90°

⇒ x° = $\frac{90^{\circ}}{2}$

⇒ x° = 45°

Hence, x° = 45°.

Here all three angles of the triangle are equal to k°. By the angle sum property,

⇒ k° + k° + k° = 180°

⇒ 3k° = 180°

⇒ k° = $\frac{180^{\circ}}{3}$

⇒ k° = 60°

Hence, k° = 60°.

The three angles of the triangle are (m° − 5°), 60° and (m° + 5°). By the angle sum property,

⇒ (m° − 5°) + 60° + (m° + 5°) = 180°

⇒ 2m° + 60° = 180°

⇒ 2m° = 180° − 60°

⇒ 2m° = 120°

⇒ m° = $\frac{120^{\circ}}{2}$

⇒ m° = 60°

Hence, m° = 60°.

In the given figure, AB is parallel to CD. Let AD and BC intersect each other at the point O.

Since AB ∥ CD and BC is the transversal,

⇒ ∠BCD = ∠ABC = ∠b (alternate angles are equal)

In △OCD, the side DO is produced to A, so ∠AOC is the exterior angle at O.

⇒ ∠a = ∠AOC

By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles. Here the two interior opposite angles are ∠OCD (= ∠BCD) and ∠ODC (= ∠ADC).

∴ ∠a = ∠BCD + ∠ADC

⇒ ∠a = ∠b + ∠c

Hence, proved that ∠a = ∠b + ∠c.

(i) Solving,

⇒ ∠a = ∠b + ∠c = 60° + 50° = 110°.

Hence, ∠a = 110°.

(ii) Solving,

⇒ ∠a = ∠b + ∠c

⇒ 100° = 55° + ∠c

⇒ ∠c = 100° − 55° = 45°

Hence, ∠c = 45°.

(iii) Solving,

⇒ ∠a = ∠b + ∠c

⇒ 108° = ∠b + 48°

⇒ ∠b = 108° − 48° = 60°

Hence, ∠b = 60°.

Let the three angles be 4x, 5x and 6x.

By the angle sum property of a triangle,

⇒ 4x + 5x + 6x = 180°

⇒ 15x = 180°

⇒ x = $\frac{180^{\circ}}{15}$

⇒ x = 12°

∴ The angles are 4x = 4 × 12° = 48°, 5x = 5 × 12° = 60° and 6x = 6 × 12° = 72°.

Hence, the angles of the triangle are 48°, 60° and 72°.

Let the other two angles be 5x and 7x.

By the angle sum property of a triangle,

⇒ 60° + 5x + 7x = 180°

⇒ 60° + 12x = 180°

⇒ 12x = 180° − 60°

⇒ 12x = 120°

⇒ x = $\frac{120^{\circ}}{12}$

⇒ x = 10°

∴ The two angles are 5x = 5 × 10° = 50° and 7x = 7 × 10° = 70°.

Hence, the two angles are 50° and 70°.

The ratio of the other two angles is

$1\frac{1}{2} : 1\frac{1}{3} = \frac{3}{2} : \frac{4}{3}$

L.C.M. of 2 and 3 = 6

Multiplying both terms by 6,

$\frac{3}{2} × 6 : \frac{4}{3} × 6 = 9 : 8$

Let the two angles be 9x and 8x.

By the angle sum property of a triangle,

⇒ 61° + 9x + 8x = 180°

⇒ 61° + 17x = 180°

⇒ 17x = 180° − 61°

⇒ 17x = 119°

⇒ x = $\frac{119^{\circ}}{17}$

⇒ x = 7°

∴ The two angles are 9x = 9 × 7° = 63° and 8x = 8 × 7° = 56°.

Hence, these angles are 63° and 56°.

By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Here, 110° is the exterior angle, and x° and 30° are its two interior opposite angles.

⇒ x° + 30° = 110°

⇒ x° = 110° − 30°

⇒ x° = 80°

Hence, x° = 80°.

By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Here, 120° is the exterior angle, and y° and 60° are its two interior opposite angles.

⇒ y° + 60° = 120°

⇒ y° = 120° − 60°

⇒ y° = 60°

Hence, y° = 60°.

By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Here, 122° is the exterior angle, and k° and 35° are its two interior opposite angles.

⇒ k° + 35° = 122°

⇒ k° = 122° − 35°

⇒ k° = 87°

Hence, k° = 87°.

By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Here, 135° is the exterior angle, and a° and 73° are its two interior opposite angles.

⇒ a° + 73° = 135°

⇒ a° = 135° − 73°

⇒ a° = 62°

Hence, a° = 62°.

By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Here, 125° is the exterior angle at the vertex of b°, and a° and c° are its two interior opposite angles.

⇒ a° + c° = 125° ....(1)

Also, 140° is the exterior angle at the vertex of c°, and a° and b° are its two interior opposite angles.

⇒ a° + b° = 140° ....(2)

Adding (1) and (2),

⇒ 2a° + b° + c° = 125° + 140°

⇒ 2a° + b° + c° = 265° ....(3)

By the angle sum property,

⇒ a° + b° + c° = 180° ....(4)

Subtracting (4) from (3),

⇒ a° = 265° − 180°

⇒ a° = 85°

Substituting a° = 85° in (1),

⇒ 85° + c° = 125°

⇒ c° = 125° − 85°

⇒ c° = 40°

Substituting a° = 85° in (2),

⇒ 85° + b° = 140°

⇒ b° = 140° − 85°

⇒ b° = 55°

Hence, a° = 85°, b° = 55° and c° = 40°.

By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Here, 112° is the exterior angle, and y° and 63° are its two interior opposite angles.

⇒ y° + 63° = 112°

⇒ y° = 112° − 63°

⇒ y° = 49°

Now x° and 112° form a linear pair.

⇒ x° + 112° = 180°

⇒ x° = 180° − 112°

⇒ x° = 68°

Hence, x° = 68° and y° = 49°.

By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Here, 120° is the exterior angle, and a° and a° are its two interior opposite angles.

⇒ a° + a° = 120°

⇒ 2a° = 120°

⇒ a° = $\frac{120^{\circ}}{2}$

⇒ a° = 60°

Hence, a° = 60°.

By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

The interior angle at the right vertex and 140° form a linear pair.

⇒ Interior angle at the right vertex = 180° − 140° = 40°

Here, 4m is the exterior angle at the top-left vertex, and 2m and 40° are its two interior opposite angles.

⇒ 4m = 2m + 40°

⇒ 4m − 2m = 40°

⇒ 2m = 40°

⇒ m = $\frac{40^{\circ}}{2}$

⇒ m = 20°

Hence, m = 20°.

By the exterior angle property, an exterior angle of a triangle is equal to the sum of its two interior opposite angles.

The triangle is isosceles, so its two base angles are each equal to b°.

Here, 105° is the exterior angle, and b° and b° are its two interior opposite angles.

⇒ b° + b° = 105°

⇒ 2b° = 105°

⇒ b° = $\frac{105^{\circ}}{2}$

⇒ b° = 52·5° = 52°30′

Now a° and 105° form a linear pair.

⇒ a° + 105° = 180°

⇒ a° = 180° − 105°

⇒ a° = 75°

Hence, a° = 75° and b° = 52·5° = 52°30′.

The triangle is isosceles with the two equal sides, so the base angles x and y are equal. The vertical angle is 80°. By the angle sum property,

⇒ x + y + 80° = 180°

⇒ x + x + 80° = 180°

⇒ 2x + 80° = 180°

⇒ 2x = 180° − 80°

⇒ 2x = 100°

⇒ x = $\frac{100^{\circ}}{2}$

⇒ x = 50°

∴ x = y = 50°.

Hence, x = 50° = y.

The triangle is isosceles with the two equal sides, so the base angles 40° and b are equal, giving b = 40°. By the angle sum property,

⇒ a + 40° + b = 180°

⇒ a + 40° + 40° = 180°

⇒ a + 80° = 180°

⇒ a = 180° − 80°

⇒ a = 100°

Hence, a = 100° and b = 40°.

The two equal sides meet at the right angle (90°), so the base angles x and y are equal. By the angle sum property,

⇒ 90° + x + y = 180°

⇒ 90° + 2x = 180°

⇒ 2x = 180° − 90°

⇒ 2x = 90°

⇒ x = $\frac{90^{\circ}}{2}$

⇒ x = 45°

∴ x = y = 45°.

Hence, x = y = 45°.

The two equal sides meet at the 80° vertex, so the base angles a and b are equal. By the angle sum property,

⇒ 80° + a + b = 180°

⇒ 80° + 2a = 180°

⇒ 2a = 180° − 80°

⇒ 2a = 100°

⇒ a = $\frac{100^{\circ}}{2}$

⇒ a = 50°

∴ a = b = 50°.

Now x is the exterior angle at the top vertex, so it forms a linear pair with the base angle b.

⇒ x = 180° − b = 180° − 50° = 130°

Hence, x = 130° and a = 50° = b.

In the given isosceles triangle,

∠BDC and 86° form a linear pair.

⇒ ∠BDC + 86° = 180°

⇒ ∠BDC = 180° − 86° = 94°

Since, CD = BD

⇒ ∠BCD = ∠CBD [Angles opposite to equal sides in a triangle are equal]

By angle sum property,

⇒ ∠BDC + ∠BCD + ∠CBD = 180°

⇒ 94° + ∠BCD + ∠BCD = 180°

⇒ 2∠BCD = 180° − 94°

⇒ 2∠BCD = 86°

⇒ ∠BCD = $\frac{86^{\circ}}{2}$ = 43°

⇒ ∠BCD = ∠CBD = 43°

Now p and ∠CBD form a linear pair, so

⇒ p + ∠CBD = 180°

⇒ p = 180° − ∠CBD

⇒ p = 180° − 43° = 137°

Hence, p = 137°.

In isosceles △ACD,

AC = CD

⇒ m = 35° [Angles opposite to equal sides in a triangle are equal]

Now, in △ABD,

∠A = 60° + 35° = 95°

By angle sum property,

⇒ ∠A + ∠B + ∠D = 180°

⇒ 95° + n + m = 180°

⇒ 95° + n + 35° = 180°

⇒ n + 130° = 180°

⇒ n = 180° − 130° = 50°

Hence, n = 50° and m = 35°.

In the given figure,

Since AB is parallel to DC,

⇒ x = 60° (Alternate angles are equal)

As, AB = AC

⇒ ∠ABC = ∠ACB [Angles opposite to equal sides in a triangle are equal]

By angle sum property,

⇒ ∠ABC + ∠A + ∠ACB = 180°

⇒ ∠ABC + x + ∠ABC = 180°

⇒ 2∠ABC + 60° = 180°

⇒ 2∠ABC = 180° − 60°

⇒ 2∠ABC = 120°

⇒ ∠ABC = $\frac{120^{\circ}}{2}$ = 60°

As y and ∠ABC form a linear pair,

⇒ y + ∠ABC = 180°

⇒ y + 60° = 180°

⇒ y = 120°

Hence, x = 60° and y = 120°.

In isosceles △ADC, AC = DC

⇒ ∠CAD = ∠CDA = 70° (base angles of an isosceles triangle are equal)

By the angle sum property,

⇒ ∠C + ∠CAD + ∠CDA = 180°

⇒ x + 70° + 70° = 180°

⇒ x = 180° − 70° − 70° = 40°.

Now, ∠BDA + ∠CDA = 180° (linear pair)

⇒ ∠BDA = 180° − 70° = 110°

In isosceles △ADB, BD = AD

⇒ ∠DAB = ∠DBA = y (base angles of an isosceles triangle are equal)

By the angle sum property,

⇒ ∠DBA + ∠BAD + ∠BDA = 180°

⇒ y + y + 110° = 180°

⇒ 2y + 110° = 180°

⇒ 2y = 180° − 110°

⇒ 2y = 70°

⇒ y = $\frac{70^{\circ}}{2}$

⇒ y = 35°

Hence, x = 40° and y = 35°.

In isosceles △ABD, AB = AD

⇒ ∠ABD = ∠ADB (base angles of an isosceles triangle are equal)

By angle sum property,

⇒ ∠A + ∠ABD + ∠ADB = 180°

⇒ 100° + 2∠ABD = 180°

⇒ ∠ABD = $\frac{180^{\circ} − 100^{\circ}}{2} = 40^{\circ}$

In equilateral △BCD, BC = CD = BD

⇒ ∠DCB = ∠CBD = ∠BDC

By angle sum property,

⇒ ∠DCB + ∠CBD + ∠BDC = 180°

⇒ 3∠DCB = 180°

⇒ ∠DCB = $\frac{180^{\circ}}{3}$

⇒ ∠DCB = 60°

∴ x = ∠ABD + ∠CBD = 40° + 60° = 100°

and

y = ∠ADB + ∠BDC = 40° + 60° = 100°.

Hence, x = 100° = y.

From the figure,

Since the horizontal lines are parallel. p and 50° are interior alternate angles.

⇒ p = 50° [Alternate interior angles are equal]

From figure,

⇒ ∠ACB + 130° = 180° (linear pair)

⇒ ∠ACB = 50°

In the △ABC, by the angle sum property,

⇒ ∠BAC + ∠ACB + ∠ABC = 180°

⇒ x + 50° + p = 180°

⇒ x + 50° + 50° = 180°

⇒ x + 100° = 180°

⇒ x = 80°

50°, x and y lie on a straight line.

⇒ 50° + x + y = 180°

⇒ 50° + 80° + y = 180°

⇒ y = 180° − 50° − 80° = 50°

Hence, x = 80°, p = 50° and y = 50°.

In △ABD, AB = AD

⇒ ∠ABD = ∠ADB [Angles opposite to equal sides in a triangle are equal]

By the angle sum property,

⇒ ∠A + ∠ABD + ∠ADB = 180°

⇒ 30° + 2∠ABD = 180°

⇒ ∠ABD = $\frac{180^{\circ} − 30^{\circ}}{2}$ = 75°

In △BDC,

⇒ BD = CD

⇒ ∠DBC = ∠DCB [Angles opposite to equal sides in a triangle are equal]

By the angle sum property,

⇒ ∠DCB + ∠DBC + ∠BDC = 180°

⇒ 2∠DBC + 90° = 180°

⇒ ∠DBC = $\frac{180^{\circ} − 90^{\circ}}{2}$ = 45°

∴ x = ∠ABD + ∠DBC = 75° + 45° = 120° and y = 45°.

Hence, x = 120° and y = 45°.

In △ABC, AB = AC

⇒ ∠ABC = ∠ACB [Angles opposite to equal sides in a triangle are equal]

By the angle sum property,

⇒ ∠ACB + ∠ABC + ∠A = 180°

⇒ 2∠ABC + 40° = 180°

⇒ ∠ABC $= \frac{180^{\circ} − 40^{\circ}}{2} = 70^{\circ}$

x and y are the exterior angles at the other two vertices, each forming a linear pair with a base angle :

⇒ x = 180° − 70° = 110° and y = 180° − 70° = 110°

Hence, x = 110° = y.

∠ABC + 120° = 180° (linear pair)

⇒ ∠ABC = 180° − 120° = 60°

In △ABC, AB = AC

⇒ ∠ABC = ∠ACB = 60° [Angles opposite to equal sides in a triangle are equal]

By the angle sum property,

⇒ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 60° + 60° + x = 180°

⇒ x = 180° − 60° − 60° = 60°

Now y and ∠ACB form a linear pair, so

⇒ y = 180° − 60° = 120°

In △ACD, by the angle sum property,

⇒ z + y + 25° = 180°

⇒ z + 120° + 25° = 180°

⇒ z = 35°

Hence, x = 60°, y = 120° and z = 35°.

In an isosceles triangle the two base angles are equal. Let each base angle be x.

By the angle sum property of a triangle,

⇒ 100° + x + x = 180°

⇒ 100° + 2x = 180°

⇒ 2x = 180° − 100°

⇒ 2x = 80°

⇒ x = $\frac{80^{\circ}}{2}$

⇒ x = 40°

Hence, the base angles are 40° and 40°.

In an isosceles triangle the two base angles are equal, so the other base angle is also 52°. Let the vertex angle be x.

By the angle sum property of a triangle,

⇒ x + 52° + 52° = 180°

⇒ x = 180° − 104° = 76°

Hence, the angle of vertex is 76°.

Let the vertical angle be x. Then each base angle is 4x.

By the angle sum property of a triangle,

⇒ x + 4x + 4x = 180°

⇒ 9x = 180°

⇒ x = $\frac{180^{\circ}}{9}$

⇒ x = 20°

∴ The vertical angle is 20° and each base angle is 4 × 20° = 80°.

Hence, the angles of the triangle are 20°, 80° and 80°.

Let each base angle be b. Then the vertical angle is (b + 15°).

By the angle sum property of a triangle,

⇒ b + b + (b + 15°) = 180°

⇒ 3b + 15° = 180°

⇒ 3b = 180° − 15°

⇒ 3b = 165°

⇒ b = $\frac{165^{\circ}}{3}$

⇒ b = 55°

∴ Each base angle is 55° and the vertical angle is 55° + 15° = 70°.

Hence, the angles of the triangle are 55°, 55° and 70°.

Let the vertical angle be v. Then each base angle is (v + 15°).

By the angle sum property of a triangle,

⇒ (v + 15°) + (v + 15°) + v = 180°

⇒ 3v + 30° = 180°

⇒ 3v = 180° − 30°

⇒ 3v = 150°

⇒ v = $\frac{150^{\circ}}{3}$

⇒ v = 50°

∴ The vertical angle is 50° and each base angle is 50° + 15° = 65°.

Hence, the angles of the triangle are 65°, 65° and 50°.

Let each base angle be b. The sum of the base angles is 2b, so the vertical angle is 3 × 2b = 6b.

By the angle sum property of a triangle,

⇒ b + b + 6b = 180°

⇒ 8b = 180°

⇒ b = $\frac{180^{\circ}}{8}$

⇒ b = 22·5° = 22°30′

∴ Each base angle is 22°30′ and the vertical angle is 6 × 22·5° = 135°.

Hence, the angles of the triangle are 22°30′, 22°30′ and 135°.

Let the base angle be x. Then the vertical angle is 4x. As the triangle is isosceles, the other base angle is also x.

By the angle sum property of a triangle,

⇒ x + x + 4x = 180°

⇒ 6x = 180°

⇒ x = $\frac{180^{\circ}}{6}$

⇒ x = 30°

∴ Each base angle is 30° and the vertical angle is 4 × 30° = 120°.

Hence, the angles of the triangle are 30°, 30° and 120°.

From the figure,

∠A = 40°

In △ABC, by the angle sum property,

⇒ ∠A + ∠ABC + ∠ACB = 180°

⇒ 40° + ∠ABC + ∠ACB = 180°

⇒ ∠ABC + ∠ACB = 180° − 40°

⇒ ∠ABC + ∠ACB = 140° ....(1)

Since BI is the bisector of ∠ABC,

⇒ ∠IBC = $\frac{1}{2}$ ∠ABC ....(2)

Since CI is the bisector of ∠ACB,

⇒ ∠ICB = $\frac{1}{2}$ ∠ACB ....(3)

Adding (2) and (3),

⇒ ∠IBC + ∠ICB = $\frac{1}{2}$ ∠ABC + $\frac{1}{2}$ ∠ACB

⇒ ∠IBC + ∠ICB = $\frac{1}{2}$ (∠ABC + ∠ACB)

Substituting the value of (∠ABC + ∠ACB) from (1),

⇒ ∠IBC + ∠ICB = $\frac{1}{2}$ × 140°

⇒ ∠IBC + ∠ICB = 70° ....(4)

In △BIC, by the angle sum property,

⇒ ∠BIC + ∠IBC + ∠ICB = 180°

Substituting the value of (∠IBC + ∠ICB) from (4),

⇒ ∠BIC + 70° = 180°

⇒ ∠BIC = 180° − 70°

⇒ ∠BIC = 110°

Hence, ∠BIC = 110°.

In the triangle ABC, the two marked sides BA and CB are equal, so the base angles ∠BAC and ∠BCA are equal.

Since ∠ACE and ∠BCA are on the same line, they form a linear pair.

⇒ ∠ACE + ∠BCA = 180°

⇒ b + ∠BCA = 180°

⇒ ∠BCA = 180° − b

As both base angles are equal,

⇒ ∠BAC = 180° − b

Since ∠DBC and ∠ABC are on the same line, they form a linear pair.

⇒ ∠DBC + ∠ABC = 180°

⇒ a + ∠ABC = 180°

⇒ ∠ABC = 180° − a

By the angle sum property of a triangle,

⇒ ∠ABC + ∠BCA + ∠BAC = 180°

⇒ (180° − a) + (180° − b) + (180° − b) = 180°

⇒ 540° − a − 2b = 180°

⇒ a = 360° − 2b

Hence, a = 360° − 2b.

From the figure, ∠BAC = 60°.

In △ABC, AB = AC

⇒ ∠ABC = ∠ACB [Angles opposite to equal sides in a triangle are equal]

By the angle sum property,

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

Substituting ∠ACB = ∠ABC,

⇒ 60° + ∠ABC + ∠ABC = 180°

⇒ 2∠ABC = 180° − 60°

⇒ 2∠ABC = 120°

⇒ ∠ABC = $\frac{120^{\circ}}{2}$

⇒ ∠ABC = 60°

Since BP bisects ∠ABC,

⇒ ∠PBC = $\frac{1}{2}$∠ABC

⇒ ∠PBC = $\frac{60^{\circ}}{2}$

⇒ ∠PBC = 30°

Since AP ∥ BC and BP is the transversal, x and ∠PBC are alternate angles.

⇒ x = ∠PBC (alternate angles are equal)

⇒ x = 30°

Hence, x = 30°.

In △ADB, DA = DB

⇒ ∠DAB = ∠DBA [Angles opposite to equal sides in a triangle are equal]

Given, ∠ADB = 70°.

By the angle sum property,

⇒ ∠DAB + ∠DBA + ∠ADB = 180°

Substituting ∠DAB = ∠DBA,

⇒ ∠DBA + ∠DBA + 70° = 180°

⇒ 2∠DBA = 180° − 70°

⇒ 2∠DBA = 110°

⇒ ∠DBA = $\frac{110^{\circ}}{2}$

⇒ ∠DBA = 55°

Since BD bisects ∠ABC,

⇒ ∠DBC = ∠DBA

⇒ ∠DBC = 55°

In △DBC, DB = DC

⇒ ∠DCB = ∠DBC [Angles opposite to equal sides in a triangle are equal]

⇒ ∠DCB = 55°

From the figure, x = ∠DCB.

⇒ x = 55°

Hence, x = 55°.

Since ABCD is a square, each of its angles = 90° and all its sides are equal.

Since △BEC is an equilateral triangle, each of its angles = 60° and BE = BC = EC.

(i) From the figure, E lies outside the square, so ∠ABE = ∠ABC + ∠CBE.

⇒ ∠ABE = 90° + 60°

⇒ ∠ABE = 150°

Hence, ∠ABE = 150°.

(ii) Since AB = BC and BC = BE, we get AB = BE.

⇒ ∠BAE = ∠BEA [Angles opposite to equal sides in a triangle are equal]

In △ABE, by the angle sum property,

⇒ ∠BAE + ∠BEA + ∠ABE = 180°

⇒ 2∠BAE + 150° = 180°

⇒ 2∠BAE = 180° − 150°

⇒ 2∠BAE = 30°

⇒ ∠BAE = $\frac{30^{\circ}}{2}$

⇒ ∠BAE = 15°

Hence, ∠BAE = 15°.

Since ABCD is a square, each of its angles = 90° and all its sides are equal.

Since △BEC is an equilateral triangle, each of its angles = 60° and BE = BC = EC.

(i) From the figure, E lies inside the square, so ∠ABE = ∠ABC − ∠EBC.

⇒ ∠ABE = 90° − 60°

⇒ ∠ABE = 30°

Hence, ∠ABE = 30°.

(ii) Since AB = BC and BC = BE, we get AB = BE.

⇒ ∠BAE = ∠BEA [Angles opposite to equal sides in a triangle are equal]

In △ABE, by the angle sum property,

⇒ ∠BAE + ∠BEA + ∠ABE = 180°

⇒ 2∠BAE + 30° = 180°

⇒ 2∠BAE = 180° − 30°

⇒ 2∠BAE = 150°

⇒ ∠BAE = $\frac{150^{\circ}}{2}$

⇒ ∠BAE = 75°

Hence, ∠BAE = 75°.

Since AB = BC, the base angles are equal :

⇒ ∠BAC = ∠BCA

By the angle sum property of ΔABC,

⇒ ∠BAC + ∠BCA + 54° = 180°

⇒ 2∠BCA + 54° = 180°

⇒ 2∠BCA = 180° − 54°

⇒ 2∠BCA = 126°

⇒ ∠BCA = $\frac{126^{\circ}}{2}$

⇒ ∠BAC = ∠BCA = 63°

Now a is the exterior angle at C and b is the exterior angle at A, each forming a linear pair with a base angle :

⇒ a = 180° − ∠BCA = 180° − 63° = 117°

⇒ b = 180° − ∠BAC = 180° − 63° = 117°

Hence, a = b = 117°.

In the figure,

∠BAC = 80° (Vertically opposite angles are equal)

In triangle ABC, by the angle sum property,

⇒ 80° + 50° + x = 180°

⇒ 130° + x = 180°

⇒ x = 180° − 130°

⇒ x = 50°

Hence, option 2 is the correct option.

From figure,

∠BAC = 84° (Vertically opposite angles are equal)

Since ABC is an isosceles triangle, its base angles are equal.

∴ ∠ABC = ∠ACB

By angle sum property,

⇒ ∠BAC + ∠ABC + ∠ACB = 180°

⇒ ∠BAC + 2∠ACB = 180°

⇒ 84° + 2∠ACB = 180°

⇒ 2∠ACB = 180° − 84°

⇒ 2∠ACB = 96°

⇒ ∠ACB = $\frac{96^{\circ}}{2}$ = 48°

As x and ∠ACB form a linear pair,

⇒ x + ∠ACB = 180°

⇒ x + 48° = 180°

⇒ x = 180° − 48°

⇒ x = 132°

Hence, option 3 is the correct option.

Let the other two angles be x and 2x.

⇒ 60° + x + 2x = 180°

⇒ 60° + 3x = 180°

⇒ 3x = 180° − 60°

⇒ 3x = 120°

⇒ x = $\frac{120^{\circ}}{3}$

⇒ x = 40°

∴ The two angles are 40° and 80°, and the bigger one is 80°.

Hence, option 4 is the correct option.

Let the angles be 2x, 3x and x.

⇒ 2x + 3x + x = 180°

⇒ 6x = 180°

⇒ x = $\frac{180^{\circ}}{6}$

⇒ x = 30°

∴ The smallest angle is x = 30°.

Hence, option 1 is the correct option.

∠AED and ∠CED form a linear pair, so ∠AED = 180° − 95° = 85°.

Since DE ∥ BC, ∠ADE = ∠ABC = 50° (corresponding angles).

In ΔADE, by the angle sum property,

⇒ x + 85° + 50° = 180°

⇒ x + 135° = 180°

⇒ x = 180° − 135°

⇒ x = 45°

Hence, option 2 is the correct option.

In the triangle ABC, the two marked sides CA and CB are equal, so the base angles ∠CAB and ∠CBA are equal.

Since ∠ABD and ∠CBA are on the same line, they form a linear pair.

⇒ ∠ABD + ∠CBA = 180°

⇒ 124° + ∠CBA = 180°

⇒ ∠CBA = 180° − 124°

⇒ ∠CBA = 56°

⇒ ∠CAB = ∠CBA = 56°

By the angle sum property,

⇒ ∠ACB + ∠CBA + ∠CAB = 180°

⇒ x + 56° + 56° = 180°

⇒ x + 112° = 180°

⇒ x = 180° − 112°

⇒ x = 68°

Hence, option 2 is the correct option.

In △ABC, AB = AC

⇒ ∠B = ∠ACB

As ∠BAC = 90°,

By angle sum property,

⇒ ∠B + ∠ACB + ∠BAC = 180°

⇒ 2∠B + 90° = 180°

⇒ ∠B = $\frac{180^{\circ} − 90^{\circ}}{2}$ = 45°

∠ACB + ∠ACD = 180°

45° + ∠ACD = 180°

∠ACD = 135°

In △ACD, AC = CD

⇒ ∠D = ∠CAD = x

By angle sum property,

⇒ ∠D + ∠ACD + ∠CAD = 180°

⇒ x + 135° + x = 180°

⇒ 2x = 180° − 135° = 45°

⇒ x = $\frac{45^{\circ}}{2}$ = 22·5° = 22°30′

Hence, option 4 is the correct option.

The two sides marked equal are AC and BC, so

⇒ x + 38 = 2x + 15

⇒ 38 − 15 = 2x − x

⇒ 23 = x

⇒ x = 23

From the figure, AB = 3x + 4.

Substituting x = 23,

⇒ AB = 3 × 23 + 4

⇒ AB = 69 + 4

⇒ AB = 73 unit

Hence, option 4 is the correct option.

In △ADB, by angle sum property:

⇒ ∠A + ∠B + ∠D = 180°

⇒ 50° + ∠B + 90° = 180°

⇒ ∠B = 180° − 140° = 40°

In △BCE, by angle sum property:

⇒ ∠E + ∠B + ∠C = 180°

⇒ 85° + 40° + x = 180°

⇒ 125° + x = 180°

⇒ x = 180° − 125° = 55°

Hence, option 3 is the correct option.

In isosceles triangle ADE,

DA = DE

⇒ ∠DEA = ∠DAE [Base angles of an isosceles triangle are equal]

By angle sum property,

⇒ ∠DAE + ∠DEA + ∠ADE = 180°

⇒ ∠DAE + ∠DAE + ∠ADE = 180°

⇒ 2∠DAE + 72° = 180°

⇒ ∠DAE = $\frac{180^{\circ} − 72^{\circ}}{2} = 54^{\circ}$

Since DE is parallel to BC,

⇒ ∠B = ∠ADE = 72° (corresponding angles are equal)

Now, in triangle ABC,

By angle sum property,

⇒ ∠A + ∠B + ∠ACB = 180°

⇒ 54° + 72° + ∠ACB = 180°

⇒ ∠ACB = 180° − 126° = 54°

As, x is the reflex angle, so

⇒ x = 360° − ∠ACB = 360° − 54° = 306°

Hence, option 4 is the correct option.

Since a triangle can have only one obtuse angle, the other two angles must be acute.

Thus, statement 1 is true.

A triangle can have only one obtuse angle, not "one or more"; an obtuse-angled triangle is a triangle with exactly one obtuse angle.

Thus, statement 2 is false.

Hence, option 3 is the correct option.

By the exterior angle property, the opposite exterior angle = 60° + 50° = 110°, not 120°. So, the Assertion (A) is false.

The Reason states the exterior angle property correctly, so the Reason (R) is true.

Hence, option 2 is the correct option.

In triangle ABC,

AB = AC

∠ABC = ∠ACB

By angle sum property,

⇒ ∠ABC + ∠ACB + ∠A = 180°

⇒ 2∠ABC + 40° = 180°

⇒ 2∠ABC = 180° − 40° = 140°

⇒ ∠ABC = 70°

As, BI and CI are the bisectors of ∠ABC and ∠ACB.

⇒ ∠IBC + ∠ICB + ∠BIC = 180°

⇒ 35° + 35° + ∠BIC = 180°

⇒ ∠BIC = 180° − 70° = 110°, which is obtuse. So △BIC is an obtuse angled triangle, and the Assertion (A) is true.

The Reason correctly states the isosceles triangle property, so the Reason (R) is also true.

Hence, option 3 is the correct option.

As, the given triangle is equilateral. Each angle is of 60°.

⇒ a = 180° − 60° = 120°

⇒ b = 180° − 60° = 120°

⇒ a = b

So it is possible to express a in terms of b, hence the Assertion (A) is false.

An equilateral triangle is always isosceles, but an isosceles triangle need not be equilateral. So, the reason (R) is true.

Hence, option 2 is the correct option.

∠ABC = 145° is the greatest angle of the triangle, and the side opposite it is AC. Hence AC is the longest side, so the Assertion (A) is true.

The Reason correctly states that the greater angle has a greater side opposite to it, so the Reason (R) is also true.

Hence, option 3 is the correct option.