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Triangles: Angle Properties, Sides and Pythagoras

By ExamAtlas · 10/9/2026

Triangles: Angle Properties, Sides and Pythagoras

Every triangle question rests on four facts: the angles add to 180°, an exterior angle equals the two opposite interior angles, any two sides together exceed the third, and in a right triangle the square of the hypotenuse equals the sum of the squares of the other sides. NTPC asks isosceles angle problems directly; the rest appears inside mensuration and height questions.

1. Properties Box

PropertyStatement
Angle sum∠A + ∠B + ∠C = 180°
Exterior angleExterior angle = sum of the two opposite interior angles
IsoscelesEqual sides face equal angles
Side-angle orderThe largest side faces the largest angle
Triangle inequalityEach side < sum of the other two and > their difference
Pythagoras (right angle at C)c² = a² + b²
Acute or obtuse test (c longest)c² < a² + b² acute; c² > a² + b² obtuse
Bisectors of ∠B and the exterior angle at CMeet at angle ∠A/2
Internal bisectors of ∠B and ∠CMeet at angle 90° + ∠A/2

2. Pythagorean Triplets

TripletMultiples seen in papers
3, 4, 56-8-10, 9-12-15, 15-20-25
5, 12, 1310-24-26
7, 24, 2514-48-50
8, 15, 1716-30-34
9, 40, 4118-80-82
20, 21, 2940-42-58

Isosceles right triangle: sides a, a, a√2 30-60-90 triangle: sides 1 : √3 : 2

✗ Sides 6 and 10, so the third side can be 4 to 16  |  ✓ Strictly between: 4 < x < 16, so 5 to 15 for integers

✗ Exterior angle at C = ∠A + ∠C  |  ✓ It equals the two opposite interior angles: ∠A + ∠B

हिंदी नोट: त्रिभुज के कोणों का योग 180° और बाह्य कोण दोनों अंतः सम्मुख कोणों के योग के बराबर होता है। किन्हीं दो भुजाओं का योग तीसरी से बड़ा होना चाहिए। समकोण त्रिभुज में कर्ण² = आधार² + लंब²।

Exam Pointer: Verified NTPC pattern: base angles of an isosceles triangle from its vertex angle (June 2025 Graduate CBT-1). Exterior angle, triangle inequality and Pythagoras questions had no verified NTPC shift and are tagged syllabus-based.

Pariksha Pattern: Every Way NTPC Asks This Topic

Pattern 1: Angle sum and isosceles triangles

[PYQ: NTPC Graduate CBT-1 21-Jun-2025 Shift-3]

EXAM LEVEL

Q. The vertex angle of an isosceles triangle is 50°. Find each base angle.

The base angles are equal and share 180° − 50° = 130°, so each is 65°.

Answer: 65°

EXAMATLAS LEVEL

Q. In an isosceles triangle each base angle is 15° more than twice the vertex angle. Find all the angles.

Let the vertex angle be v. Then v + 2(2v + 15) = 180 gives 5v = 150, so v = 30° and each base angle is 75°.

Answer: 30°, 75°, 75°

Pattern 2: Exterior angle theorem and bisector angles

[Pattern: syllabus-based, PYQ-style]

EXAM LEVEL

Q. An exterior angle of a triangle is 110° and one of the opposite interior angles is 45°. Find the other two angles of the triangle.

The other opposite angle = 110° − 45° = 65°, and the angle next to the exterior angle = 180° − 110° = 70°.

Answer: 65° and 70°

EXAMATLAS LEVEL

Q. In triangle ABC, BC is extended to D so that ∠ACD = 120°, and ∠A : ∠B = 3 : 1. Find the angles of the triangle, and the angle at which the bisector of ∠B meets the bisector of ∠ACD.

∠A + ∠B = 120° in the ratio 3 : 1, so ∠A = 90°, ∠B = 30° and ∠C = 60°. The two bisectors meet at ∠A/2 = 45°.

Answer: 90°, 30°, 60°; 45°

Pattern 3: Triangle inequality and the acute or obtuse test

[Pattern: syllabus-based, PYQ-style]

EXAM LEVEL

Q. Can sides 4 cm, 7 cm and 12 cm form a triangle?

4 + 7 = 11, which is less than 12, so no.

Answer: No

EXAMATLAS LEVEL

Q. Two sides of a triangle are 6 cm and 10 cm, and the third side is a whole number. How many triangles are possible, and how many of them are obtuse?

4 < x < 16 gives x = 5 to 15, eleven triangles. If x is the longest side, it is obtuse when x² > 36 + 100 = 136: x = 12, 13, 14, 15. If 10 is the longest, it is obtuse when 100 > 36 + x², that is x² < 64: x = 5, 6, 7. Obtuse = 7.

Answer: 11 triangles; 7 obtuse

Pattern 4: Pythagoras: ladders, poles and diagonals

[Pattern: syllabus-based, PYQ-style]

EXAM LEVEL

Q. A 13 m ladder reaches a window 12 m high. How far is its foot from the wall?

√(169 − 144) = 5 m (5-12-13 triplet).

Answer: 5 m

EXAMATLAS LEVEL

Q. A 25 m ladder rests against a wall with its top 24 m up. The foot slips 8 m further from the wall. How far does the top slide down?

The foot is first √(625 − 576) = 7 m away, then 15 m. The new height is √(625 − 225) = 20 m, so the top slides 4 m.

Answer: 4 m

CBT-2 LEVEL

Pattern 5: Angle bisector theorem and the median length

[Pattern: syllabus-based, PYQ-style]

EXAM LEVEL

Q. In triangle ABC, AB = 6 cm, AC = 9 cm and BC = 10 cm. The bisector of ∠A meets BC at D. Find BD and DC.

BD : DC = AB : AC = 2 : 3, so BD = 4 cm and DC = 6 cm.

Answer: 4 cm and 6 cm

EXAMATLAS LEVEL

Q. In triangle ABC, AB = 8 cm, AC = 6 cm and BC = 10 cm, and AD is the median to BC. Find AD by Apollonius' theorem and check it another way.

AB² + AC² = 2(AD² + BD²) gives 100 = 2(AD² + 25), so AD = 5 cm. Check: 6-8-10 is right-angled at A, and the median to the hypotenuse is half of it, 5 cm.

Answer: 5 cm

60-Second Revision

  • Angles add to 180°; the exterior angle is the sum of the two opposite interior angles.
  • The third side lies strictly between the difference and the sum.
  • Compare c² with a² + b² to decide acute, right or obtuse.
  • Learn the triplets 3-4-5, 5-12-13, 7-24-25, 8-15-17.

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