NDA Mathematics · 44 questions · 2017 to 2026
NDA Properties of Triangles previous year questions with solutions
Properties of Triangles gives one to four questions in most Mathematics papers, about two marks in a hundred, and has grown since 2023, when linked pairs on one triangle became the rule. Most of it is the sine rule. Every question UPSC has set on it since 2017, solved. Grouped by the method each needs, newest first; open a question for its answer and worked solution.
15.1 The sine rule: a/sin A = b/sin B = c/sin C = 2R (17)15.2 The cosine rule and the projection rule (8)15.3 Half-angle formulas, area, the incircle and the circumcircle (6)15.4 Identities among the angles of a triangle (A + B + C = π) (5)15.5 The nature of a triangle from a trigonometric relation (8)
15.1 The sine rule: a/sin A = b/sin B = c/sin C = 2R
1. NDA 2026 (II)
For the next two (02) items that follow:
The angles A, B and C of a triangle ABC are in the ratio 1 : 2 : 7.
The angles A, B and C of a triangle ABC are in the ratio 1 : 2 : 7.
What is the ratio of the side of greatest length to the side of least length?
Answer and solution
Answer: (c) (3 + √5): 2
Method. angles from the ratio, then sine rule with sin 18° and cos 36°
1 : 2 : 7 gives A = 18°, B = 36°, C = 126°.
Greatest side is opposite C, least is opposite A: ratio = sin 126°sin 18° = sin 54°sin 18° = cos 36°sin 18°.
sin 18° = √5 − 14, cos 36° = √5 + 14.
Ratio = √5 + 1√5 − 1 = (√5 + 1)²4 = 6 + 2√54 = 3 + √52, i.e. (3 + √5): 2.
2. NDA 2026 (II)
What is sin A · cos B equal to?
Answer and solution
Answer: (a) 1/4
Method. use sin 18° and cos 36° (A = 18°, B = 36°)
From the previous item A = 18°, B = 36°.
sin A cos B = sin 18° cos 36° = √5 − 14·√5 + 14 = 5 − 116 = 14.
Shortcut. sin 18° cos 36° = (sin 18° cos 18°)(cos 36°)cos 18° = sin 36° cos 36°2 cos 18° = sin 72°4 cos 18° = 14.
3. NDA 2026 (II)
For the next two (02) items that follow:
In a triangle ABC, a + b13 = b + c11 = c + a12.
In a triangle ABC, a + b13 = b + c11 = c + a12.
What is sin A: sin B: sin C equal to?
Answer and solution
Answer: (b) 7 : 6 : 5
Method. set the ratios equal to k and solve for a, b, c; sine rule
Let a + b = 13k, b + c = 11k, c + a = 12k. Adding: a + b + c = 18k.
Then c = 18k − 13k = 5k, a = 18k − 11k = 7k, b = 18k − 12k = 6k.
sin A: sin B: sin C = a: b: c = 7 : 6 : 5.
4. NDA 2026 (II)
What is cos A: cos B: cos C equal to?
Answer and solution
Answer: (a) 7 : 19 : 25
Method. cosine rule with a = 7k, b = 6k, c = 5k
Take k = 1: a = 7, b = 6, c = 5.
cos A = 36 + 25 − 4960 = 1260 = 15.
cos B = 49 + 25 − 3670 = 3870 = 1935.
cos C = 49 + 36 − 2584 = 6084 = 57.
Over the common denominator 35: 735: 1935: 2535 = 7 : 19 : 25.
5. NDA 2026 (I)
The angles A, B and C of a triangle are in the ratio 1 : 1 : 4. If the longest side of the triangle is 3 units, then what is the perimeter of the triangle?
Answer and solution
Answer: (b) 3 + 2√3 units
Method. angles from the ratio, then the sine rule
1 : 1 : 4 gives A = B = 30°, C = 120°; the longest side c = 3 is opposite C.
csin 120° = 3√3/2 = 2√3, so a = b = 2√3·sin 30° = √3.
Perimeter = 3 + √3 + √3 = 3 + 2√3 units.
6. NDA 2025 (I)
Consider the following for the two (02) items that follow:
In a triangle ABC, two sides BC and CA are in the ratio 2 : 1 and their opposite corresponding angles are in the ratio 3 : 1.
In a triangle ABC, two sides BC and CA are in the ratio 2 : 1 and their opposite corresponding angles are in the ratio 3 : 1.
One of the angles of the triangle is
Answer and solution
Answer: (b) 30°
Method. sine rule with A = 3B, using sin 3B = 3 sin B − 4 sin³B
BC = a is opposite A and CA = b is opposite B, so ab = sin Asin B = 2 and A = 3B.
sin 3B = 3 sin B − 4 sin³B, so sin 3Bsin B = 3 − 4 sin²B = 2, giving sin²B = 14.
B = 30° (150° would make A = 450°, impossible). Then A = 90° and C = 180° − 120° = 60°.
The angles are 30°, 60°, 90°, so one of them is 30°: option (b).
7. NDA 2025 (I)
Consider the following statements:
I. The triangle is right-angled.
II. One of the sides of the triangle is 3 times the other.
III. The angles A, C and B of the triangle are in AP.
Which of the statements given above is/are correct?
I. The triangle is right-angled.
II. One of the sides of the triangle is 3 times the other.
III. The angles A, C and B of the triangle are in AP.
Which of the statements given above is/are correct?
Answer and solution
Answer: (c) I and III only
Method. sides in ratio sin A: sin B: sin C with A, B, C = 90°, 30°, 60°
From the previous item A = 90°, B = 30°, C = 60°, so a: b: c = sin 90° : sin 30° : sin 60° = 1 : 12: √32 = 2 : 1 : √3.
I: right-angled (A = 90°): correct.
III: the angles in the order A, C, B are 90°, 60°, 30°, a common difference of −30°: in AP, correct.
II: as printed, one side is 3 times another. The sides are in the ratio 1 : √3 : 2 and no pair has ratio 3 (√3, 2 and 2√3 are the only ratios), so II is false as printed.
As printed the answer is I and III only: option (c). If the root sign was dropped (one side is √3 times another, the evident intent), II is also correct and the answer would be option (d).
As printed, statement II says one side is 3 times the other; the data give a: b: c = 2 : 1 : √3, so no side is 3 times another and II is false, while I and III hold: option (c). If UPSC meant '√3 times' (b: c, a dropped root sign), II would be true and the answer (d).
8. NDA 2024 (II)
In a triangle ABC, ∠A = 75° and ∠B = 45°. What is 2a − b equal to?
Answer and solution
Answer: (b) √2c
Method. angle C from the angle sum, then the sine rule
C = 180° − 75° − 45° = 60°.
2a − b = k(2 sin 75° − sin 45°) = k[√6 + √22 − √22] = k·√62.
c = k sin 60° = k·√32.
Ratio: 2a − bc = √6√3 = √2, so 2a − b = √2 c.
9. NDA 2023 (II)
Consider the following for the next two (02) items that follow:
The angles A, B and C of a triangle ABC are in the ratio 3 : 5 : 4.
The angles A, B and C of a triangle ABC are in the ratio 3 : 5 : 4.
What is the value of a + b + √2 c equal to?
Answer and solution
Answer: (c) 3b
Method. sine rule with the angles fixed by the ratio 3 : 5 : 4
A + B + C = 180° and A: B: C = 3 : 5 : 4, so A = 45°, B = 75°, C = 60°.
By the sine rule a = k sin 45°, b = k sin 75°, c = k sin 60°.
sin 45° = √22, sin 75° = √6 + √24, sin 60° = √32.
a + b + √2 c = k[√22 + √6 + √24 + √62] = k·3√2 + 3√64 = 3k·√6 + √24 = 3k sin 75° = 3b.
10. NDA 2023 (II)
What is the ratio of a2 : b2 : c2?
Answer and solution
Answer: (a) 2 : 2 + √3 : 3
Method. sine rule: a² : b² : c² = sin²A: sin²B: sin²C
From the previous item A = 45°, B = 75°, C = 60°.
sin²45° = 12, sin²75° = 2 + √34, sin²60° = 34.
Multiplying all three by 4: a² : b² : c² = 2 : 2 + √3 : 3.
11. NDA 2023 (I)
Consider the following for the next two (02) items that follow:
The perimeter of a triangle ABC is 6 times the AM of sine of angles of the triangle. Further BC = √3 and CA = 1.
The perimeter of a triangle ABC is 6 times the AM of sine of angles of the triangle. Further BC = √3 and CA = 1.
What is the perimeter of the triangle?
Answer and solution
No correct option. This question is defective as printed in UPSC's paper.
Method. law of sines; circumradius from the perimeter condition
Sides: a = BC = √3, b = CA = 1. Perimeter = 6 x AM of sines = 2(sinA + sinB + sinC) = a + b + cR, so R = 1.
sinA = a2R = √32, sinB = b2R = 12. Since b < a, B < A, so B = 30°, and A = 60° or 120°.
A = 60°: C = 90°, c = 2, perimeter = √3 + 3.
A = 120°: C = 30°, c = 1, perimeter = √3 + 2.
Both triangles satisfy every given condition (R = 1 in both), and √3 + 2 and √3 + 3 are both options.
The data (BC = √3, CA = 1, perimeter = 6 x AM of sines, i.e. R = 1) fit two triangles, 30-60-90 (perimeter √3 + 3) and 30-30-120 (perimeter √3 + 2), and both perimeters are options; the evident intended reading is the right-angled triangle, which would yield (c) √3 + 3.
12. NDA 2023 (I)
Consider the following statements:
1. ABC is right angled triangle
2. The angles of the triangle are in AP
Which of the statements given above is/are correct?
1. ABC is right angled triangle
2. The angles of the triangle are in AP
Which of the statements given above is/are correct?
Answer and solution
No correct option. This question is defective as printed in UPSC's paper.
Method. classify both triangles that fit the data of the previous item
The data give two triangles (see the previous item): 30-60-90 and 30-30-120.
30-60-90: right-angled, and 30, 60, 90 are in AP, so both statements are true (option c).
30-30-120: no right angle, and 30, 30, 120 are not in AP, so neither statement is true (option d).
The question's words (BC = √3, CA = 1, perimeter = 6 x AM of sines) do not exclude either triangle, so the answer depends on which case is meant.
Inherits the SSA ambiguity of the previous item: the 30-60-90 triangle gives (c) both statements true, the 30-30-120 triangle gives (d) neither; the evident intended reading is the right-angled triangle, which would yield (c).
13. NDA 2020 (I)
Directions for the following two (02) items:
Read the following information and answer the two items that follow:
ABCD is a trapezium such that AB and CD are parallel and BC is perpendicular to them. Let ∠ADB = θ, ∠ABD = α, BC = p and CD = q.
Read the following information and answer the two items that follow:
ABCD is a trapezium such that AB and CD are parallel and BC is perpendicular to them. Let ∠ADB = θ, ∠ABD = α, BC = p and CD = q.
Consider the following:
1. AD sin θ = AB sin α
2. BD sin θ = AB sin (θ + α)
Which of the above is/are correct?
1. AD sin θ = AB sin α
2. BD sin θ = AB sin (θ + α)
Which of the above is/are correct?
Answer and solution
Answer: (c) Both 1 and 2
Method. law of sines in triangle ABD
In triangle ABD, ∠ADB = θ, ∠ABD = α, so ∠DAB = π − (θ + α) and sin∠DAB = sin(θ + α).
Law of sines: ABsinθ = ADsinα = BDsin(θ + α).
Statement 1: AD sinθ = AB sinα follows from the first equality. Correct.
Statement 2: BD sinθ = AB sin(θ + α) follows from the first and third. Correct.
14. NDA 2020 (I)
What is AB equal to?
Answer and solution
Answer: (a) (p2 + q2) sin θp cos θ + q sin θ
Method. law of sines in triangle ABD, with BD and the angle α read from right triangle BCD
AB and CD are parallel, BC is perpendicular to both, and A, D lie on the same side of BC (convex trapezium ABCD). In right triangle BCD, BD = √(p2 + q2).
∠ABD = α is the angle between BA (parallel to CD) and BD, so it equals ∠BDC: sinα = pBD, cosα = qBD.
From the previous item, AB = BD sinθsin(θ + α) = BD sinθsinθ cosα + cosθ sinα.
Substitute: AB = BD2 sinθq sinθ + p cosθ = (p2 + q2) sinθp cosθ + q sinθ.
15. NDA 2019 (II)
If the angles of a triangle ABC are in AP and b: c = √3 : √2, then what is the measure of angle A?
Answer and solution
Answer: (d) 75°
Method. angles in AP give B = 60°; law of sines gives C
Angles in AP: A + C = 2B, so 3B = 180° and B = 60°.
bc = sinBsinC = √3√2, so sinC = √2√3 x √32 = 1√2.
C = 45° (135° is impossible, since B + C would exceed 180°).
A = 180° - 60° - 45° = 75°.
16. NDA 2019 (I)
If the angles of a triangle ABC are in the ratio 1 : 2 : 3, then the corresponding sides are in the ratio
Answer and solution
Answer: (c) 1 : √3 : 2
Method. angle ratio 1 : 2 : 3 fixes the angles; sides are proportional to sines
Angles are 30°, 60°, 90° (x + 2x + 3x = 180°, x = 30°).
Sides are proportional to sin30° : sin60° : sin90° = 12: √32: 1 = 1 : √3 : 2.
Shortcut. Recognise the 30-60-90 triangle.
17. NDA 2018 (I)
In a triangle ABC if a = 2, b = 3 and sin A = 2/3, then what is angle B equal to?
Answer and solution
Answer: (b) π/2
Method. law of sines
By the law of sines, sinBb = sinAa, so sinB = b sinAa = 3 x (2/3)2 = 1.
Hence B = π2.
Check: cosA = √53, c = b cosA = √5, and a2 + c2 = 4 + 5 = 9 = b2.
15.2 The cosine rule and the projection rule
18. NDA 2026 (I)
In a triangle ABC, if a, b and c are the lengths of the sides opposite to the angles A, B and C respectively, then what is sin(A − B)sin(A + B) equal to?
Answer and solution
Answer: (c) a2 − b2c2
Method. sine rule and sin²A − sin²B = sin(A + B) sin(A − B)
sin²A − sin²B = sin(A + B) sin(A − B).
In a triangle sin(A + B) = sin C, so sin(A − B)sin(A + B) = sin(A − B) sin(A + B)sin²(A + B) = sin²A − sin²Bsin²C.
With a = k sin A, b = k sin B, c = k sin C this is a² − b²c².
19. NDA 2025 (I)
Consider the following for the three (03) items that follow:
The sides of a triangle ABC are AB = 3 cm, BC = 5 cm and CA = 7 cm.
The sides of a triangle ABC are AB = 3 cm, BC = 5 cm and CA = 7 cm.
Consider the following statements:
I. The triangle is obtuse-angled triangle.
II. The sum of acute angles of the triangle is also acute.
Which of the statements given above is/are correct?
I. The triangle is obtuse-angled triangle.
II. The sum of acute angles of the triangle is also acute.
Which of the statements given above is/are correct?
Answer and solution
Answer: (c) Both I and II
Method. cosine rule for the largest angle
The longest side CA = 7 is opposite B.
cos B = AB² + BC² − CA²2·AB·BC = 9 + 25 − 4930 = −12, so B = 120°.
I: B is obtuse, so the triangle is obtuse-angled: correct.
II: the acute angles are A and C, and A + C = 180° − 120° = 60°, which is acute: correct.
Both are correct.
20. NDA 2025 (I)
What is ∠B equal to?
Answer and solution
Answer: (c) 120°
Method. cosine rule
cos B = AB² + BC² − CA²2·AB·BC = 9 + 25 − 492.3.5 = −1530 = −12.
So ∠B = 120°.
21. NDA 2025 (I)
What is the area of the triangle?
Answer and solution
Answer: (a) 15√3 /4 square cm
Method. area = 12 ac sin B with B = 120°
Area = 12·AB·BC·sin B = 12·3.5·sin 120° = 152·√32 = 15√34 square cm.
Check by Heron: s = 7.5, area = √(7.5.4.5.2.5.0.5) = √(42.1875) = 6.495 = 15√34.
22. NDA 2023 (I)
Consider the following for the next two (02) items that follow:
In the triangle ABC,
a2 + b2 + c2 = ac + √3bc
In the triangle ABC,
a2 + b2 + c2 = ac + √3bc
What is the nature of the triangle?
Answer and solution
Answer: (c) Right angled triangle
Method. complete the squares
Rewrite a2 + b2 + c2 - ac - √3 bc = 0 as (a − c2)2 + (b − √3 c2)2 + c2 - c24 − 3c24 = 0.
The last three terms cancel, so (a − c2)2 + (b − √3 c2)2 = 0, giving a = c2 and b = √3 c2.
Sides are in the ratio 1 : √3 : 2, with a2 + b2 = c2: right-angled (and scalene, so (d) is excluded).
23. NDA 2023 (I)
If c = 8, what is the area of the triangle?
Answer and solution
Answer: (c) 8√3
Method. sides from the previous item, then area = 12ab
With c = 8: a = c2 = 4 and b = √3 c2 = 4√3, and the angle between a and b is 90° (c is the hypotenuse).
Area = 12 x 4 x 4√3 = 8√3.
24. NDA 2022 (II)
In a triangle ABC, a = 4, b = 3, c = 2. What is cos3C equal to?
Answer and solution
Answer: (a) 7128
Method. cosine rule for C, then cos3C = 4cos3C − 3cosC
cosC = a2 + b2 - c22ab = 16 + 9 − 424 = 78.
cos3C = 4 x 343512 − 3 x 78 = 343128 − 336128 = 7128.
25. NDA 2021 (I)
The sides of a triangle are m, n and √(m2 + n2 + mn). What is the sum of the acute angles of the triangle?
Answer and solution
Answer: (b) 60°
Method. cosine rule gives the largest angle
Let c = √(m2 + n2 + mn) be the largest side, opposite angle C.
cosC = m2 + n2 - c22mn = -mn2mn = -12, so C = 120°.
The other two angles are acute and sum to 180° - 120° = 60°.
15.3 Half-angle formulas, area, the incircle and the circumcircle
26. NDA 2023 (I)
Consider the following for the next two (02) items that follow:
In a triangle PQR, P is the largest angle and cosP = 13. Further the in-circle of the triangle touches the sides PQ, QR and RP at N, L and M respectively such that the lengths PN, QL and RM are n, n + 2, n + 4 respectively where n is an integer.
In a triangle PQR, P is the largest angle and cosP = 13. Further the in-circle of the triangle touches the sides PQ, QR and RP at N, L and M respectively such that the lengths PN, QL and RM are n, n + 2, n + 4 respectively where n is an integer.
What is the value of n?
Answer and solution
Answer: (c) 8
Method. tangent lengths from a vertex are equal
Tangent lengths: PM = PN = n, QN = QL = n + 2, RL = RM = n + 4.
PQ = 2n + 2, QR = 2n + 6, RP = 2n + 4. QR is the longest side and is opposite P, consistent with P being the largest angle.
cosP = PQ2 + PR2 - QR22 PQ PR = 4n2 - 162(2n + 2)(2n + 4) = (n − 2)(n + 2)2(n + 1)(n + 2) = n − 22(n + 1).
Set equal to 13: 3n − 6 = 2n + 2, so n = 8.
Check: sides 18, 20, 22 give 324 + 400 − 484720 = 13.
27. NDA 2023 (I)
What is the length of the smallest side?
Answer and solution
Answer: (d) 18
Method. sides from the previous item
With n = 8: PQ = 2n + 2 = 18, RP = 2n + 4 = 20, QR = 2n + 6 = 22.
The smallest side is PQ = 18.
28. NDA 2022 (I)
Let a, b, c be the lengths of sides BC, CA, AB respectively of a triangle ABC. If p is the perimeter and q is the area of the triangle, then what is p(p − 2a) tan(A2) equal to?
Answer and solution
Answer: (d) 4q
Method. p = 2s, half-angle formula tan(A/2) = Δs(s − a)
p = a + b + c = 2s, so p − 2a = b + c − a = 2(s − a).
p(p − 2a) = 4s(s − a).
tan(A2) = √((s − b)(s − c)s(s − a)) = Δs(s − a), where Δ = √(s(s − a)(s − b)(s − c)) = q.
Product = 4s(s − a) x qs(s − a) = 4q.
29. NDA 2021 (I)
A chord subtends an angle 120° at the centre of a unit circle. What is the length of the chord?
Answer and solution
Answer: (d) √3 units
Method. isosceles triangle with two radii as equal sides
Chord = 2 r sin(θ2) = 2 x 1 x sin60° = 2 x √32 = √3 units.
(Or by the cosine rule: c2 = 1 + 1 − 2cos120° = 3.)
30. NDA 2021 (I)
What is the area of the triangle ABC with sides a = 10 cm, c = 4 cm and angle B = 30°?
Answer and solution
Answer: (c) 10 cm2
Method. area = 12 a c sinB
Area = 12 a c sinB = 12 x 10 x 4 x sin30° = 12 x 10 x 4 x 12 = 10 cm2.
31. NDA 2017 (II)
In a triangle ABC, a − 2b + c = 0. The value of cot(A2)cot(C2) is
Answer and solution
Answer: (b) 3
Method. half-angle formula tan(A/2)tan(C/2) = (s − b)/s
a − 2b + c = 0 gives a + c = 2b, so s = a + b + c2 = 3b2 and s − b = b2.
tan(A2)tan(C2) = s − bs = b/23b/2 = 13.
Hence cot(A2)cot(C2) = 3.
Shortcut. Test the 3-4-5 triangle: 3 + 5 = 2 x 4, and cot(A/2)cot(C/2) = 3.
15.4 Identities among the angles of a triangle (A + B + C = π)
32. NDA 2024 (II)
In a triangle ABC
tanA + tanB + tanC = k
What is the value of cotA cotB cotC?
tanA + tanB + tanC = k
What is the value of cotA cotB cotC?
Answer and solution
Answer: (b) 1/k
Method. in a triangle tan A + tan B + tan C = tan A tan B tan C
A + B = 180° − C gives tan(A + B) = −tan C, so tan A + tan B = −tan C(1 − tan A tan B), i.e. tan A + tan B + tan C = tan A tan B tan C.
So tan A tan B tan C = k, and cot A cot B cot C = 1tan A tan B tan C = 1k.
33. NDA 2024 (I)
ABC is a triangle such that angle C = 60°, then what is cosA + cosBcos(A − B2) equal to?
Answer and solution
Answer: (c) 1
Method. sum-to-product formula, then A + B = 120°
cos A + cos B = 2 cosA + B2 cosA − B2.
So the expression = 2 cosA + B2.
C = 60° gives A + B = 120°, so A + B2 = 60° and the value is 2 cos 60° = 2·12 = 1.
34. NDA 2024 (I)
In a triangle ABC, AB = 16 cm, BC = 63 cm and AC = 65 cm. What is the value of cos 2A + cos 2B + cos 2C?
Answer and solution
Answer: (a) -1
Method. converse of Pythagoras: right angle at B, then cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C
AB² + BC² = 16² + 63² = 256 + 3969 = 4225 = 65² = AC², so ∠B = 90°.
In any triangle cos 2A + cos 2B + cos 2C = −1 − 4 cos A cos B cos C.
cos B = cos 90° = 0, so the value is −1.
Check directly: cos 2B = −1, and 2A + 2C = 180° so cos 2C = −cos 2A, the two cancel.
Shortcut. Right angle at B gives 2A + 2C = 180°, so cos 2A + cos 2C = 0 and the sum is cos 2B = −1.
35. NDA 2024 (I)
Consider the following statements:
1. In a triangle ABC, if cotA·cotB·cotC > 0, then the triangle is an acute angled triangle.
2. In a triangle ABC, if tanA·tanB·tanC > 0, then the triangle is an obtuse angled triangle.
Which of the statements given above is/are correct?
1. In a triangle ABC, if cotA·cotB·cotC > 0, then the triangle is an acute angled triangle.
2. In a triangle ABC, if tanA·tanB·tanC > 0, then the triangle is an obtuse angled triangle.
Which of the statements given above is/are correct?
Answer and solution
Answer: (a) 1 only
Method. sign of the product of tangents or cotangents in a triangle
At most one angle of a triangle is obtuse or right.
Statement 1: cot A cot B cot C > 0 needs an even number of negative factors (0 or 2). Two obtuse angles are impossible, and a right angle makes the product 0, so no angle is obtuse or right: the triangle is acute. Statement 1 is correct.
Statement 2: in a triangle tan A + tan B + tan C = tan A tan B tan C. If the triangle is obtuse, exactly one tangent is negative, so the product is negative. A positive product therefore means all three tangents are positive, i.e. the triangle is acute, not obtuse. Statement 2 is false.
Example: 40°, 60°, 80° gives a positive product and an acute triangle.
36. NDA 2019 (II)
If angle C of a triangle ABC is a right angle, then what is tan A + tan B equal to?
Answer and solution
Answer: (d) c2ab
Method. definition of tan in a right triangle, then Pythagoras
With C = 90°, tanA = ab and tanB = ba.
tanA + tanB = a2 + b2ab = c2ab, since a2 + b2 = c2.
15.5 The nature of a triangle from a trigonometric relation
37. NDA 2026 (I)
In a triangle ABC, sin A = cos B + cos C then what is tan (B2) + cot (B2) equal to?
Answer and solution
No correct option. This question is defective as printed in UPSC's paper.
Method. sum-to-product on cos B + cos C
cos B + cos C = 2 cosB + C2 cosB − C2 = 2 sinA2 cosB − C2, and sin A = 2 sinA2 cosA2.
The condition becomes cosA2 = cosB − C2, so A = B − C or A = C − B, i.e. B = A + C(B = 90°) or C = A + B (C = 90°).
The expression is tanB2 + cotB2 = 2sin B.
If B = 90° it equals 2: option (d). But if C = 90° the condition also holds (sin A = cos B, cos C = 0) with B free, and 2sin B > 2 takes any value above 2, equal to none of the options.
The intended reading is that the right angle is at B, which gives option (d).
The condition sin A = cos B + cos C is satisfied when B = 90° or when C = 90° (with B free); the asked value 2/sin B is 2 only in the first case and exceeds 2 and is not fixed in the second, so it is not determined by the data; the intended reading (right angle at B, or the question asking for the value when B = 90°) yields option (d). Since tan(B/2) + cot(B/2) = 2/sin B ≥ 2 for every B, options (a)–(c) are impossible in any triangle; (d) is the only attainable option, reached exactly when B = 90°.
38. NDA 2024 (II)
In a triangle ABC
acosA = bcosB = ccosC
What is the area of the triangle if a = 6 cm?
acosA = bcosB = ccosC
What is the area of the triangle if a = 6 cm?
Answer and solution
Answer: (a) 9√3 square cm
Method. sine rule turns the ratios into tan A = tan B = tan C
By the sine rule a = k sin A, b = k sin B, c = k sin C, so acos A = k tan A, and similarly for B and C.
The equal ratios give tan A = tan B = tan C, so A = B = C = 60°: the triangle is equilateral with side 6 cm.
Area = √34·6² = 9√3 square cm.
39. NDA 2024 (II)
The roots of the equation 7x2 - 6x + 1 = 0 are tan α and tan β, where 2α and 2β are the angles of a triangle. Which one of the following is correct?
Answer and solution
Answer: (c) The triangle is right-angled
Method. tan(α + β) from the sum and product of roots
tan α + tan β = 67, tan α tan β = 17.
tan(α + β) = 6/71 − 1/7 = 6/76/7 = 1, and α, β are acute (both roots positive), so α + β = 45°.
Then 2α + 2β = 90°, so the third angle is 180° − 90° = 90°: the triangle is right-angled.
The roots are 3 ± √27, which are unequal, so α ≠ β and 2α ≠ 2β; the triangle is not isosceles. So only option (c) is correct.
40. NDA 2022 (I)
In a triangle ABC,
sinA − cosB − cosC = 0.
What is angle B equal to?
sinA − cosB − cosC = 0.
What is angle B equal to?
Answer and solution
No correct option. This question is defective as printed in UPSC's paper.
Method. factorise sinA = cosB + cosC with sum-to-product
A = π − (B + C), so sinA = 2sin(A2)cos(A2) with cos(A2) = sin(B + C2); and cosB + cosC = 2cos(B + C2)cos(B − C2) = 2sin(A2)cos(B − C2).
The condition becomes cos(A2) = cos(B − C2), so A = B − C or A = C − B, i.e. B = A + C(B = 90°) or C = A + B (C = 90°).
So the triangle is right-angled at B or at C. If C = 90°, B can be any acute angle (e.g. A = 60°, B = 30°, C = 90° satisfies sinA = cosB + cosC), so B is not determined.
The condition only forces a right angle at B or at C, so angle B is not determined and … so angle B is not determined: the data are consistent with every one of the four options (B = 30, 45, 60 with C = 90, or B = 90 itself); the evident intended reading is 'right-angled at B′, which would yield (d) pi/2.; the evident intended reading is 'which angle is the right angle' or 'right-angled at B′, which would yield (d) π/2.
41. NDA 2020 (I)
Consider the following statements:
1. If ABC is a right-angled triangle, right-angled at A and if sin B = 13, then cosec C = 3.
2. If b cos B = c cos C and if the triangle ABC is not right-angled, then ABC must be isosceles.
Which of the above statements is/are correct?
1. If ABC is a right-angled triangle, right-angled at A and if sin B = 13, then cosec C = 3.
2. If b cos B = c cos C and if the triangle ABC is not right-angled, then ABC must be isosceles.
Which of the above statements is/are correct?
Answer and solution
Answer: (b) 2 only
Method. sine/cosine of complementary angles; double-angle equation sin2B = sin2C
Statement 1: A = 90°, so C = 90° - B and cosecC = 1sinC = 1cosB. With sinB = 13, cosB = 2√23, so cosecC = 32√2, which is not 3. Incorrect.
Statement 2: b cosB = c cosC with b = 2R sinB, c = 2R sinC gives sinB cosB = sinC cosC, i.e. sin2B = sin2C.
So 2B = 2C, or 2B = π − 2C.
The second case gives B + C = π2, so A = π2, which is excluded. Hence B = C and the triangle is isosceles. Correct.
42. NDA 2020 (I)
Consider the following statements:
1. If in a triangle ABC, A = 2B and b = c, then it must be an obtuse-angled triangle.
2. There exists no triangle ABC with A = 40°, B = 65° and ac = sin 40° cosec 15°.
Which of the above statements is/are correct?
1. If in a triangle ABC, A = 2B and b = c, then it must be an obtuse-angled triangle.
2. There exists no triangle ABC with A = 40°, B = 65° and ac = sin 40° cosec 15°.
Which of the above statements is/are correct?
Answer and solution
Answer: (b) 2 only
Method. angle sum and law of sines
Statement 1: A = 2B and b = c gives B = C. Then A + B + C = 2B + B + B = 180°, so B = 45° and A = 90°. The triangle is right-angled, not obtuse. Incorrect.
Statement 2: A = 40°, B = 65° give C = 75°, and then ac = sin40°sin75°.
The given ratio is sin40° cosec15° = sin40°sin15°, and sin15° is not sin75°. So no such triangle exists. Correct.
43. NDA 2018 (II)
If A, B and C are the angles of a triangle and 1 1 1 1 + sin A 1 + sin B 1 + sin C sin A + sin2 A sin B + sin2 B sin C + sin2 C = 0, then which one of the following is correct?
Answer and solution
Answer: (a) The triangle ABC is isosceles
Method. row operations on the determinant, then a Vandermonde determinant
Let x = sinA, y = sinB, z = sinC. The rows are (1, 1, 1), (1 + x, 1 + y, 1 + z), (x + x2, y + y2, z + z2).
R2 -> R2 - R1 gives (x, y, z); then R3 -> R3 - R2 gives (x2, y2, z2).
The determinant is the Vandermonde determinant (y − x)(z − x)(z − y) = 0.
So two of sinA, sinB, sinC are equal, say sinA = sinB. A + B = π is impossible in a triangle, so A = B and the triangle is isosceles.
Equilateral is only a special case and is not forced; scalene is excluded.
44. NDA 2017 (II)
In triangle ABC, if sin2 A + sin2 B + sin2 Ccos2 A + cos2 B + cos2 C = 2, then the triangle is
Answer and solution
Answer: (a) right-angled
Method. identities for the sums of sin2 and cos2 of the angles of a triangle
In any triangle: sin2 A + sin2 B + sin2 C = 2 + 2cosA cosB cosC and cos2 A + cos2 B + cos2 C = 1 − 2cosA cosB cosC.
Let P = cosA cosB cosC. The condition is 2 + 2P = 2(1 − 2P), so 6P = 0 and P = 0.
So one of cosA, cosB, cosC is zero: one angle is 90° and the triangle is right-angled.
Shortcut. The 3-4-5 triangle gives ratio 2/1 = 2.