Questions & explanations
1. In triangle ABC, a=5, b=6, c=7, A≈44.4°, B≈57.1°, C≈78.5°. Which Mollweide formula check fails?
- (a+b)/c = cos((A−B)/2)/sin(C/2)
- (a−b)/c = sin((A−B)/2)/cos(C/2)
- Both hold
- Both fail
Answer: Both hold
Mollweide's formulas: (a+b)/c = cos((A−B)/2)/sin(C/2) and (a−b)/c = sin((A−B)/2)/cos(C/2). Compute LHS: (5+6)/7=11/7≈1.5714; RHS: cos((44.4-57.1)/2)/sin(78.5/2)=cos(-6.35°)/sin(39.25°)≈0.994/0.632≈1.572. LHS: (5-6)/7=-1/7≈-0.1429; RHS: sin(-6.35°)/cos(39.25°)≈-0.1106/0.775≈-0.1427. Both match within rounding, so both hold.
2. In a triangle with sides a, b, c and semi-perimeter s, which of the following is the correct expression for sin(A/2)?
- √[(s−b)(s−c)/(bc)]
- √[s(s−a)/(bc)]
- √[(s−b)(s−c)/(s(s−a))]
- √[(s−a)(s−b)/(ab)]
Answer: √[(s−b)(s−c)/(bc)]
Using the identity cos A = 1 − 2 sin²(A/2), we get sin²(A/2) = (1 − cos A)/2. Substituting cos A from the cosine rule and simplifying using s = (a+b+c)/2 gives sin²(A/2) = (s−b)(s−c)/(bc). Taking positive square root (since A/2 is acute) yields sin(A/2) = √[(s−b)(s−c)/(bc)].
3. In triangle ABC, b=7, c=5, A=60°. Find B−C in degrees.
- 32.2°
- −32.2°
- 16.1°
- −16.1°
Answer: −32.2°
Using Napier's analogy: tan((B−C)/2) = ((b−c)/(b+c)) cot(A/2). Here b−c=2, b+c=12, so (b−c)/(b+c)=1/6. cot(30°)=√3≈1.732. RHS = (1/6)*1.732 ≈ 0.2887. Thus (B−C)/2 = arctan(0.2887) ≈ 16.1°, so B−C ≈ 32.2°. Since b<c, B<C, so B−C is negative: −32.2°.
4. In a triangle, s = 14 and R = 65/8. If the sides are in the ratio 13:14:15, then the sides are
- 26/3, 28/3, 10
- 13, 14, 15
- 26, 28, 30
- 13/2, 7, 15/2
Answer: 26/3, 28/3, 10
Let sides be 13k, 14k, 15k. s = (13k+14k+15k)/2 = 21k = 14 ⇒ k = 2/3. Sides = 26/3, 28/3, 10. Check R = abc/(4Δ). Δ = √[14*(14-26/3)*(14-28/3)*(14-10)] = 112/3. Then R = (26/3*28/3*10)/(4*112/3) = 65/8. Matches given R. So sides are 26/3, 28/3, 10.
5. In a triangle, b = 4, c = 6, B = 30°. How many triangles are possible?
- 0
- 1
- 3
- 2
Answer: 2
Using the sine rule, sin C = c sin B / b = 6 * 0.5 / 4 = 0.75. Since c sin B = 3, and 3 < b < c, there are two possible values for C: C = sin⁻¹(0.75) ≈ 48.59° and C = 180° - 48.59° = 131.41°, both giving positive A. Hence two triangles.
6. A student solves a triangle with A=40°, B=70°, a=10 and gets b=14.62, c=15.32. Using Mollweide (a+b)/c = cos((A−B)/2)/sin(C/2), which check result indicates an error?
- LHS ≈ 1.607, RHS ≈ 1.607
- LHS ≈ 1.683, RHS ≈ 1.607
- LHS ≈ 1.683, RHS ≈ 1.683
- LHS ≈ 1.607, RHS ≈ 1.683
Answer: LHS ≈ 1.607, RHS ≈ 1.683
C = 180° - 40° - 70° = 70°. LHS = (10+14.62)/15.32 ≈ 1.607. RHS = cos((40°-70°)/2)/sin(70°/2) = cos(-15°)/sin35° = cos15°/sin35° ≈ 0.9659/0.5736 ≈ 1.684. The mismatch (LHS=1.607, RHS≈1.684) indicates an error, matching option d.
7. For a triangle with sides 13, 14, 15, what is the value of r1 + r2 + r3 - r?
- 32.5
- 26
- 65/4
- 4R
Answer: 32.5
Using identity r1 + r2 + r3 - r = 4R. For sides 13,14,15, semiperimeter s=21, area Δ=84 (Heron's formula). Circumradius R = abc/(4Δ) = (13×14×15)/(4×84) = 2730/336 = 65/8. So 4R = 4×(65/8) = 65/2 = 32.5. Hence answer is 32.5.
8. For a triangle with sides 13, 14, 15, which of the following equals its area?
- 42
- 84
- 168
- √7056
Answer: 84
Using Heron's formula: semi-perimeter s = (13+14+15)/2 = 21. Area Δ = √[21(21-13)(21-14)(21-15)] = √[21×8×7×6] = √7056 = 84. The same result is obtained from Δ = abc/(4R) = (13×14×15)/(4×65/8) = 84 and Δ = rs = 4×21 = 84.
9. In a triangle ABC, if a^2 + b^2 + c^2 = 8R^2, then the triangle is
- right-angled
- equilateral
- isosceles
- obtuse-angled
Answer: right-angled
Using sine rule a = 2R sin A etc., condition becomes sin²A + sin²B + sin²C = 2. Identity sin²A + sin²B + sin²C = 2 + 2 cos A cos B cos C gives cos A cos B cos C = 0, so one angle is 90°. Hence triangle is right-angled.
10. In a triangle, if cos A + cos B + cos C = 1 + r/R, what is the value of cos A + cos B + cos C for the (13,14,15) triangle?
- 32/65
- 1
- 97/65
- 1.492
Answer: 97/65
For the (13,14,15) triangle, r = 4 and R = 65/8. So r/R = 4/(65/8) = 32/65. Then 1 + r/R = 1 + 32/65 = 97/65. This matches the sum of cosines computed directly: cos A = 0.6, cos B = 33/65, cos C = 5/13, sum = 97/65.
11. In a triangle ABC, which of the following equals a cos A + b cos B + c cos C?
- 2R cos A cos B cos C
- 2R sin A sin B sin C
- 4R cos A cos B cos C
- 4R sin A sin B sin C
Answer: 4R sin A sin B sin C
Using sine rule a = 2R sin A, etc., we get a cos A = R sin 2A. Summing gives R(sin 2A + sin 2B + sin 2C). In a triangle, sin 2A + sin 2B + sin 2C = 4 sin A sin B sin C. Hence the sum equals 4R sin A sin B sin C.
12. For a triangle with sides 3, 4, 5, what is the exradius r1 opposite the side of length 5?
- 4
- 3
- 2
- 6
Answer: 6
For a triangle, exradius opposite side a is r1 = Δ/(s-a). Here sides 3,4,5 form a right triangle. Semi-perimeter s = (3+4+5)/2 = 6. Area Δ = (1/2)*3*4 = 6. Side opposite r1 is a=5, so s-a=1. Thus r1 = 6/1 = 6.