Question 1

In triangle ABC, a = 7, b = 8, c = 9. The value of cos(B/2) is:

Accepted Answer

We use the half-angle formula for cosine in terms of sides of a triangle.

Given: a = 7, b = 8, c = 9

Step 1: Calculate the semi-perimeter s.
s = (a + b + c)/2 = (7 + 8 + 9)/2 = 24/2 = 12

Step 2: Apply the half-angle formula for cosine.
cos(B/2) = √[s(s-b)/(ac)]

where s - b = 12 - 8 = 4

Step 3: Substitute values.
cos(B/2) = √[12 × 4/(7 × 9)]
          = √[48/63]
          = √[16/21]
          = 4/√21
          = 4√21/21

Step 4: Verify the simplification.
48/63 = 16/21 (dividing by 3)
√(16/21) = 4/√21 = 4√21/21 ≈ 0.873

This is consistent with angle B being acute (since b is the middle-length side).

Question 2

In triangle ABC, if a = 4, b = 5, c = 6, then tan(C/2) is equal to:

Accepted Answer

We use the half-angle formula for tangent in terms of sides of a triangle.

Given: a = 4, b = 5, c = 6

Step 1: Calculate the semi-perimeter s.
s = (a + b + c)/2 = (4 + 5 + 6)/2 = 15/2

Step 2: Apply the half-angle formula for tangent.
tan(C/2) = r/(s-c)

where r is the inradius given by r = √[s(s-a)(s-b)(s-c)/s] = √[(s-a)(s-b)(s-c)/s]

Alternatively, use: tan(C/2) = √[(s-a)(s-b)/(s(s-c))]

Step 3: Calculate the required values.
s - a = 15/2 - 4 = 7/2
s - b = 15/2 - 5 = 5/2
s - c = 15/2 - 6 = 3/2

Step 4: Substitute into the formula.
tan(C/2) = √[(7/2 × 5/2)/(15/2 × 3/2)]
          = √[(35/4)/(45/4)]
          = √[35/45]
          = √[7/9]
          = √7/3

Step 5: Verify.
35/45 = 7/9 (dividing by 5)
√(7/9) = √7/3 ≈ 0.882

Question 3

In triangle ABC, if sin A = 3/5 and sin B = 5/13, then sin C is equal to:

Accepted Answer

We use the sine rule and the angle sum property of triangles.

Given: sin A = 3/5, sin B = 5/13

Step 1: Find cos A and cos B using the Pythagorean identity.
sin²A + cos²A = 1
(3/5)² + cos²A = 1
9/25 + cos²A = 1
cos²A = 16/25
cos A = 4/5 (taking positive value since A is an angle in a triangle)

Similarly, sin²B + cos²B = 1
(5/13)² + cos²B = 1
25/169 + cos²B = 1
cos²B = 144/169
cos B = 12/13 (taking positive value)

Step 2: Use the angle sum property C = π - A - B.
sin C = sin(π - A - B) = sin(A + B)

Step 3: Apply the sine addition formula.
sin(A + B) = sin A cos B + cos A sin B
           = (3/5)(12/13) + (4/5)(5/13)
           = 36/65 + 20/65
           = 56/65

Step 4: Verify that this is a valid sine value.
56/65 ≈ 0.862, which is between 0 and 1. ✓

Question 4

In triangle ABC, a = 5, b = 6, c = 7. Using the cosine rule, find cos C and then determine whether angle C is acute or obtuse.

Accepted Answer

Using the cosine rule: c² = a² + b² - 2ab·cos C. Rearranging: cos C = (a² + b² - c²)/(2ab). Substituting: cos C = (25 + 36 - 49)/(2·5·6) = 12/60 = 1/5 = 0.2. Since cos C = 0.2 > 0, angle C is acute.

Question 5

In triangle ABC, if a = 13, b = 14, c = 15, then the value of sin(A/2) is:

Accepted Answer

We use the half-angle formula for sine in terms of sides of a triangle.

Given: a = 13, b = 14, c = 15

Step 1: Calculate the semi-perimeter s.
s = (a + b + c)/2 = (13 + 14 + 15)/2 = 42/2 = 21

Step 2: Apply the half-angle formula:
sin(A/2) = √[s(s-a)/(bc)]

Step 3: Substitute values.
s - a = 21 - 13 = 8
sin(A/2) = √[21 × 8/(14 × 15)]
         = √[168/210]
         = √[4/5]
         = 2/√5
         = 2√5/5

Step 4: Verify by simplifying the fraction under the radical.
168/210 = 4/5 (dividing numerator and denominator by 42)
√(4/5) = 2/√5 = 2√5/5 ✓

Question 6

In triangle ABC, if a/sin A = b/sin B = c/sin C = k, then k is equal to:

Accepted Answer

This question tests the sine rule (law of sines) for triangles.

Step 1: State the sine rule.
The sine rule states: a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius of the triangle.

Step 2: Identify what k represents.
From the given condition, a/sin A = b/sin B = c/sin C = k.

Step 3: Compare with the standard form.
The constant k in the sine rule is the diameter of the circumcircle.
Therefore, k = 2R, where R is the circumradius.

Step 4: Conclusion.
The value of k is 2R (the diameter of the circumscribed circle).

Question 7

In triangle ABC, if cos A + cos B + cos C = 1 + 4sin(A/2)sin(B/2)sin(C/2), then the triangle is:

Accepted Answer

We verify the given trigonometric identity for triangles.

Step 1: Recall the standard identity for sum of cosines in a triangle.
For any triangle ABC: cos A + cos B + cos C = 1 + 4sin(A/2)sin(B/2)sin(C/2)

Step 2: Verify this is a known identity.
This is a standard identity that holds for ALL triangles (not just special ones).

Step 3: Derive to confirm.
Using A + B + C = π, we have C = π - (A + B).
cos A + cos B + cos C = cos A + cos B + cos[π - (A + B)]
                      = cos A + cos B - cos(A + B)
                      = cos A + cos B - (cos A cos B - sin A sin B)
                      = cos A(1 - cos B) + cos B + sin A sin B

After algebraic manipulation using half-angle formulas and the constraint A + B + C = π, this simplifies to:
1 + 4sin(A/2)sin(B/2)sin(C/2)

Step 4: Conclusion.
Since the identity holds for ANY triangle, the triangle is a general triangle (no special property is required).

Question 8

In triangle ABC, a = 4, b = 5, c = 6. The value of cos(B/2) is:

Accepted Answer

We use the half-angle formula for cosine in terms of the sides of a triangle.

Given: a = 4, b = 5, c = 6

Step 1: Calculate the semi-perimeter s.
s = (a + b + c)/2 = (4 + 5 + 6)/2 = 15/2

Step 2: Apply the half-angle formula for cosine.
cos(B/2) = √[s(s-b)/(ac)]

Step 3: Calculate s - b.
s - b = 15/2 - 5 = 15/2 - 10/2 = 5/2

Step 4: Substitute into the formula.
cos(B/2) = √[(15/2)(5/2)/(4 × 6)]
         = √[(75/4)/24]
         = √[75/(4 × 24)]
         = √[75/96]
         = √[25/32]
         = 5/(4√2)
         = 5√2/8

Step 5: Verify the simplification.
75/96: GCD(75, 96) = 3, so 75/96 = 25/32 ✓
√(25/32) = 5/√32 = 5/(4√2) = 5√2/8 ✓

Question 9

In triangle ABC, if a² + b² - c² = ab, then angle C is equal to:

Accepted Answer

We use the law of cosines to find angle C.

Step 1: Recall the law of cosines.
For any triangle: c² = a² + b² - 2ab cos C

Step 2: Rearrange to express cos C.
cos C = (a² + b² - c²)/(2ab)

Step 3: Substitute the given condition.
Given: a² + b² - c² = ab
Therefore: cos C = ab/(2ab) = 1/2

Step 4: Find angle C.
cos C = 1/2
C = 60° or C = π/3 radians

Step 5: Verify.
If C = 60°, then cos 60° = 1/2 ✓
And the law of cosines gives: c² = a² + b² - 2ab(1/2) = a² + b² - ab ✓
Which matches the given condition: a² + b² - c² = ab ✓

Question 10

In triangle ABC, if a/sin A = b/sin B = c/sin C = 2R (where R is the circumradius), and a = 6, b = 8, then the value of sin A : sin B is:

Accepted Answer

We use the sine rule for triangles.

Given: a/sin A = b/sin B = c/sin C = 2R (sine rule)
Also given: a = 6, b = 8

Step 1: Apply the sine rule.
From the sine rule: a/sin A = b/sin B

Step 2: Rearrange to find the ratio of sines.
sin A/sin B = a/b

Step 3: Substitute the given values.
sin A/sin B = 6/8 = 3/4

Step 4: Express as a ratio.
sin A : sin B = 3 : 4

Step 5: Verify.
This is consistent with the sine rule, which states that sides are proportional to the sines of opposite angles.

Question 11

In triangle ABC, if a = 13, b = 14, c = 15, find cos B.

Accepted Answer

We use the cosine rule: b² = a² + c² - 2ac·cos B. Rearranging: cos B = (a² + c² - b²)/(2ac). Substituting values: cos B = (13² + 15² - 14²)/(2·13·15) = (169 + 225 - 196)/(390) = 198/390 = 33/65.

Question 12

In triangle ABC, a = 7, b = 8, c = 9. Find sin A using the sine rule and the area formula.

Accepted Answer

First, find the area using Heron's formula. Semi-perimeter s = (7 + 8 + 9)/2 = 12. Area K = √[s(s-a)(s-b)(s-c)] = √[12·5·4·3] = √720 = 12√5. Using the sine rule: a/sin A = 2K/sin A·sin B·sin C is not direct. Instead, use K = (1/2)bc·sin A. So 12√5 = (1/2)·8·9·sin A = 36·sin A. Therefore sin A = 12√5/36 = √5/3.

Question 13

In triangle ABC, if a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius, and a = 10, A = 60°, find R.

Accepted Answer

Using the sine rule: a/sin A = 2R. Substituting a = 10 and A = 60°: 10/sin 60° = 2R. Since sin 60° = √3/2, we have 10/(√3/2) = 2R, which gives 20/√3 = 2R. Therefore R = 10/√3 = 10√3/3.

Question 14

In triangle ABC, if A = 45°, B = 60°, and a = 2√2, find side b using the sine rule.

Accepted Answer

Using the sine rule: a/sin A = b/sin B. Substituting a = 2√2, A = 45°, B = 60°: (2√2)/sin 45° = b/sin 60°. Since sin 45° = 1/√2 and sin 60° = √3/2, we have (2√2)/(1/√2) = b/(√3/2). Simplifying left side: 2√2 · √2 = 4. So 4 = b/(√3/2), which gives b = 4 · (√3/2) = 2√3.

Question 15

In triangle ABC, if a = 13, b = 14, c = 15, then sin(A/2) is equal to:

Accepted Answer

We use the half-angle formula for sine in terms of sides of a triangle.

Given: a = 13, b = 14, c = 15

Step 1: Calculate the semi-perimeter s.
s = (a + b + c)/2 = (13 + 14 + 15)/2 = 42/2 = 21

Step 2: Apply the half-angle formula.
sin(A/2) = √[s(s-a)/(bc)]

where s - a = 21 - 13 = 8

Step 3: Substitute values.
sin(A/2) = √[21 × 8/(14 × 15)]
         = √[168/210]
         = √[4/5]
         = 2/√5
         = 2√5/5

Step 4: Verify by simplifying the fraction under the radical.
168/210 = 4/5 (dividing numerator and denominator by 42)
√(4/5) = 2/√5 = 2√5/5 ≈ 0.894

This is consistent with angle A being acute (since a is the smallest side).

Question 16

In triangle ABC, if sin A + sin B + sin C = 1 + cos A + cos B + cos C, then one of the angles is:

Accepted Answer

We rearrange and use sum-to-product identities.

Given: sin A + sin B + sin C = 1 + cos A + cos B + cos C

Step 1: Rearrange the equation.
sin A + sin B + sin C - cos A - cos B - cos C = 1
(sin A - cos A) + (sin B - cos B) + (sin C - cos C) = 1

Step 2: Use the identity sin θ - cos θ = √2 sin(θ - π/4).
For each term:
sin A - cos A = √2 sin(A - π/4)
sin B - cos B = √2 sin(B - π/4)
sin C - cos C = √2 sin(C - π/4)

Step 3: Substitute.
√2[sin(A - π/4) + sin(B - π/4) + sin(C - π/4)] = 1
sin(A - π/4) + sin(B - π/4) + sin(C - π/4) = 1/√2 = √2/2

Step 4: Use sum-to-product on the first two terms.
sin(A - π/4) + sin(B - π/4) = 2 sin[(A + B - π/2)/2] cos[(A - B)/2]
                             = 2 sin[(A + B)/2 - π/4] cos[(A - B)/2]

Step 5: Since A + B + C = π, we have A + B = π - C.
sin[(π - C)/2 - π/4] = sin[π/2 - C/2 - π/4] = sin[π/4 - C/2] = cos(C/2 + π/4)

This becomes complex. Let me try a direct approach.

Step 6: Alternative approach — test specific angles.
If C = π/2 (right angle):
sin A + sin B + sin(π/2) = 1 + cos A + cos B + cos(π/2)
sin A + sin B + 1 = 1 + cos A + cos B + 0
sin A + sin B = cos A + cos B
tan A + tan B = 1 + tan A tan B (dividing by cos A cos B)

For a right triangle with C = π/2: A + B = π/2, so B = π/2 - A.
sin A + sin(π/2 - A) = cos A + cos(π/2 - A)
sin A + cos A = cos A + sin A ✓

This is always true! So C = π/2 satisfies the equation.

Step 7: Verify this is the only solution by checking if other angles work.
The equation is satisfied when one angle equals π/2.

Question 17

In triangle ABC, if a² + b² - c² = ab, then angle C is equal to:

Accepted Answer

We use the cosine rule to relate the sides and angles.

Given: a² + b² - c² = ab

Step 1: Recall the cosine rule.
c² = a² + b² - 2ab cos C

Rearranging:
a² + b² - c² = 2ab cos C

Step 2: Substitute the given condition.
ab = 2ab cos C

Step 3: Divide both sides by ab (assuming a, b ≠ 0).
1 = 2 cos C

cos C = 1/2

Step 4: Find C.
C = arccos(1/2) = π/3 (or 60°)

Step 5: Verify.
If C = π/3, then cos C = 1/2.
From the cosine rule: a² + b² - c² = 2ab cos C = 2ab(1/2) = ab ✓

Therefore, C = π/3.

Question 18

In triangle ABC, if a² + b² - c² = ab, then angle C equals:

Accepted Answer

Using the Law of Cosines:
c² = a² + b² - 2ab·cos(C)

Rearranging:
a² + b² - c² = 2ab·cos(C)

Step 1: Substitute the given condition
Given: a² + b² - c² = ab
Therefore: ab = 2ab·cos(C)

Step 2: Solve for cos(C)
Dividing both sides by ab (assuming a, b ≠ 0):
1 = 2·cos(C)
cos(C) = 1/2

Step 3: Find angle C
C = arccos(1/2) = 60° or π/3 radians

Step 4: Verify
If C = 60°, then cos(60°) = 1/2 ✓
And a² + b² - c² = 2ab(1/2) = ab ✓

Question 19

In triangle ABC, if a = 13, b = 14, c = 15, then the value of sin(A/2) is:

Accepted Answer

We use the half-angle formula for sine in terms of sides of a triangle.

Given: a = 13, b = 14, c = 15

First, calculate the semi-perimeter:
s = (a + b + c)/2 = (13 + 14 + 15)/2 = 42/2 = 21

The half-angle formula states:
sin(A/2) = √[((s-b)(s-c))/(bc)]

Calculate the required terms:
s - b = 21 - 14 = 7
s - c = 21 - 15 = 6

Substitute into the formula:
sin(A/2) = √[(7 × 6)/(14 × 15)]
         = √[42/210]
         = √[1/5]
         = 1/√5
         = √5/5

Verification: This is consistent with the standard half-angle formula sin(A/2) = √[((s-b)(s-c))/(bc)] derived from the cosine rule and half-angle identities.

Question 20

In triangle ABC, a = 7, b = 8, c = 9. The value of cos(B/2) is:

Accepted Answer

We use the half-angle formula for cosine in terms of sides of a triangle.

Given: a = 7, b = 8, c = 9

First, calculate the semi-perimeter:
s = (a + b + c)/2 = (7 + 8 + 9)/2 = 24/2 = 12

The half-angle formula for cosine states:
cos(B/2) = √[(s(s-b))/(ac)]

Calculate the required terms:
s = 12
s - b = 12 - 8 = 4

Substitute into the formula:
cos(B/2) = √[(12 × 4)/(7 × 9)]
         = √[48/63]
         = √[16/21]
         = 4/√21
         = 4√21/21

Verification: The formula cos(B/2) = √[(s(s-b))/(ac)] is derived from the half-angle identity cos(B/2) = √[(1+cos B)/2] combined with the cosine rule.

NDA Solution of Triangles

Solution of Triangles— Rules & Concept

Key Points to Remember

Exam-Specific Tips

Solution of Triangles — Practice Questions

60-Second Revision — Solution of Triangles