IB Mathematics: Applications and Interpretation Quiz: Triangle Trigonometry
20 questions · exam conditions
0:00
Triangle TrigonometryQuestion 1 of 20

Two forces, one of 100 N and one of 150 N, act on an object. The angle between the directions of the forces is 45°. Find the magnitude of the resultant force.

106.2 N
231.8 N
180.3 N
250.0 N
← Back to quizzes

IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Triangle Trigonometry

Practice Triangle Trigonometry in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Triangle Trigonometry, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Two forces, one of 100 N and one of 150 N, act on an object. The angle between the directions of the forces is 45°. Find the magnitude of the resultant force.

  1. 106.2 N
  2. 231.8 N (correct answer)
  3. 180.3 N
  4. 250.0 N
Explanation: When you encounter forces acting at angles, you need vector addition to find the resultant force. Since these forces aren't acting in the same direction, you can't simply add their magnitudes—you must account for the angle between them. Use the law of cosines to find the magnitude of the resultant force. When two forces F1F_1 and F2F_2 act at angle θ\theta, the resultant magnitude is: R=F12+F22+2F1F2cos⁡θR = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} Substituting the given values: F1=100F_1 = 100 N, F2=150F_2 = 150 N, and θ=45°\theta = 45°: R=1002+1502+2(100)(150)cos⁡(45°)R = \sqrt{100^2 + 150^2 + 2(100)(150)\cos(45°)} R=10000+22500+30000(22)R = \sqrt{10000 + 22500 + 30000(\frac{\sqrt{2}}{2})} R=32500+21213.2=53713.2=231.8R = \sqrt{32500 + 21213.2} = \sqrt{53713.2} = 231.8 N This confirms answer B is correct. Answer A (106.2 N) likely comes from incorrectly subtracting the forces or using sine instead of cosine. Answer C (180.3 N) suggests someone might have used the wrong angle or made an arithmetic error in the calculation. Answer D (250.0 N) is what you'd get if you simply added the magnitudes (100 + 150), ignoring the angle entirely—this would only be correct if the forces acted in exactly the same direction. Remember: whenever forces act at angles, visualize them as vectors and use the law of cosines. The resultant will always be less than the simple sum of magnitudes unless the angle is 0°.

Question 2

Two forces, one of 100 N and one of 150 N, act on an object. The angle between the directions of the forces is 45°. Find the magnitude of the resultant force.

  1. 106.2 N
  2. 231.8 N (correct answer)
  3. 180.3 N
  4. 250.0 N
Explanation: When you encounter forces acting at angles, you need vector addition to find the resultant force. Since these forces aren't acting in the same direction, you can't simply add their magnitudes—you must account for the angle between them. Use the law of cosines to find the magnitude of the resultant force. When two forces F1F_1 and F2F_2 act at angle θ\theta, the resultant magnitude is: R=F12+F22+2F1F2cos⁡θR = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\theta} Substituting the given values: F1=100F_1 = 100 N, F2=150F_2 = 150 N, and θ=45°\theta = 45°: R=1002+1502+2(100)(150)cos⁡(45°)R = \sqrt{100^2 + 150^2 + 2(100)(150)\cos(45°)} R=10000+22500+30000(22)R = \sqrt{10000 + 22500 + 30000(\frac{\sqrt{2}}{2})} R=32500+21213.2=53713.2=231.8R = \sqrt{32500 + 21213.2} = \sqrt{53713.2} = 231.8 N This confirms answer B is correct. Answer A (106.2 N) likely comes from incorrectly subtracting the forces or using sine instead of cosine. Answer C (180.3 N) suggests someone might have used the wrong angle or made an arithmetic error in the calculation. Answer D (250.0 N) is what you'd get if you simply added the magnitudes (100 + 150), ignoring the angle entirely—this would only be correct if the forces acted in exactly the same direction. Remember: whenever forces act at angles, visualize them as vectors and use the law of cosines. The resultant will always be less than the simple sum of magnitudes unless the angle is 0°.

Question 3

The area of a triangular field is 12,000 m². Two of its sides measure 150 m and 210 m. The angle between these two sides is known to be obtuse. What is the length of the third side of the field, correct to the nearest metre?

  1. 161 m
  2. 258 m
  3. 328 m (correct answer)
  4. 354 m
Explanation: Let the sides be a = 150 m and b = 210 m, and the angle between them be C. The area is given by Area = (1/2)ab sin(C). So, 12,000 = (1/2)(150)(210)sin(C), which simplifies to 12,000 = 15,750 sin(C). Therefore, sin(C) = 12,000 / 15,750 ≈ 0.7619. Since the angle C is obtuse, C = 180° - arcsin(0.7619) = 180° - 49.63° = 130.37°. Now, use the Cosine Rule to find the third side, c: c² = a² + b² - 2ab cos(C). c² = 150² + 210² - 2(150)(210)cos(130.37°). c² = 22,500 + 44,100 - 63,000(-0.6478) = 66,600 + 40,811.4 = 107,411.4. So, c = √107,411.4 ≈ 327.7 m. To the nearest metre, the length is 328 m.

Question 4

A landscape architect is designing a triangular garden ABC. She is given the measurements AB = 35 m, AC = 28 m, and the angle ABC = 40°. Which of the following statements about the length of the third side, BC, is true?

  1. No such triangle can be formed with these measurements.
  2. There is only one possible length for the side BC.
  3. There are exactly two possible lengths for the side BC. (correct answer)
  4. The length of BC must be greater than the length of AB.
Explanation: This is the ambiguous case (SSA) for the Sine Rule. Let side c = AB = 35, side b = AC = 28, and angle B = 40°. We can find angle C using the Sine Rule: sin(C)/c = sin(B)/b, so sin(C)/35 = sin(40°)/28. This gives sin(C) = 35 * sin(40°) / 28 ≈ 0.8035. Since 0 < 0.8035 < 1, there are two possible values for angle C. C₁ = arcsin(0.8035) ≈ 53.5° and C₂ = 180° - 53.5° = 126.5°. We check if both are valid. Case 1: Angle A₁ = 180° - 40° - 53.5° = 86.5°. This is a valid triangle. Case 2: Angle A₂ = 180° - 40° - 126.5° = 13.5°. This is also a valid triangle. Since two distinct triangles can be formed, there are two possible lengths for the side BC.

Question 5

From the top of a 120 m tall cliff, the angle of depression to a buoy is 18°. A boat is located 200 m from the buoy on the straight line connecting the cliff base and the buoy. If the boat is further from the cliff than the buoy, what is the angle of depression from the top of the cliff to the boat?

  1. 11.9° (correct answer)
  2. 15.5°
  3. 28.5°
  4. 35.3°
Explanation: Let T be the top of the cliff and B be the base. The height is TB = 120 m. Let U be the buoy. The angle of depression to U is 18°, so the angle of elevation from U to T is also 18° (angle TUB = 18°). The horizontal distance from the base of the cliff to the buoy is BU. In the right-angled triangle TBU, tan(18°) = TB/BU = 120/BU. So, BU = 120 / tan(18°) ≈ 369.32 m. The boat, A, is 200 m from the buoy and further from the cliff. So, the distance from the cliff base to the boat is BA = BU + 200 = 369.32 + 200 = 569.32 m. The angle of depression, θ, to the boat is found from the right-angled triangle TBA. tan(θ) = TB/BA = 120 / 569.32 ≈ 0.2108. θ = arctan(0.2108) ≈ 11.9°.

Question 6

A farmer wants to find the area of a triangular field ABC. From point A, he walks 400 m to point B. He turns and walks 500 m to point C. The final distance from C back to A is 600 m. Find the total area of the field.

  1. 150,000 m²
  2. 100,000 m²
  3. 120,000 m²
  4. 99,200 m² (correct answer)
Explanation: When you encounter a triangle problem with all three side lengths given, you should immediately think of Heron's formula for finding area. This is the most direct method when you know all sides but no angles or heights. First, calculate the semi-perimeter: s=a+b+c2=400+500+6002=750 ms = \frac{a + b + c}{2} = \frac{400 + 500 + 600}{2} = 750 \text{ m} Then apply Heron's formula: Area=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} Substituting the values: Area=750(750−400)(750−500)(750−600)\text{Area} = \sqrt{750(750-400)(750-500)(750-600)} =750×350×250×150= \sqrt{750 \times 350 \times 250 \times 150} =9,843,750,000= \sqrt{9,843,750,000} =99,216.75 m2= 99,216.75 \text{ m}^2 This rounds to approximately 99,200 m², confirming answer D. Answer A (150,000 m²) likely comes from incorrectly using the formula 12×500×600=150,000\frac{1}{2} \times 500 \times 600 = 150,000, treating two sides as if they were base and height at a right angle. Answer B (100,000 m²) might result from using 12×400×500=100,000\frac{1}{2} \times 400 \times 500 = 100,000 with the same misconception. Answer C (120,000 m²) could come from 12×400×600=120,000\frac{1}{2} \times 400 \times 600 = 120,000, again incorrectly assuming perpendicularity. Remember: when you have three sides of any triangle, Heron's formula is your go-to method. Don't assume right angles unless explicitly stated or unless the sides satisfy the Pythagorean theorem.

Question 7

The area of a triangular field is 12,000 m². Two of its sides measure 150 m and 210 m. The angle between these two sides is known to be obtuse. What is the length of the third side of the field, correct to the nearest metre?

  1. 161 m
  2. 258 m
  3. 328 m (correct answer)
  4. 354 m
Explanation: Let the sides be a = 150 m and b = 210 m, and the angle between them be C. The area is given by Area = (1/2)ab sin(C). So, 12,000 = (1/2)(150)(210)sin(C), which simplifies to 12,000 = 15,750 sin(C). Therefore, sin(C) = 12,000 / 15,750 ≈ 0.7619. Since the angle C is obtuse, C = 180° - arcsin(0.7619) = 180° - 49.63° = 130.37°. Now, use the Cosine Rule to find the third side, c: c² = a² + b² - 2ab cos(C). c² = 150² + 210² - 2(150)(210)cos(130.37°). c² = 22,500 + 44,100 - 63,000(-0.6478) = 66,600 + 40,811.4 = 107,411.4. So, c = √107,411.4 ≈ 327.7 m. To the nearest metre, the length is 328 m.

Question 8

A landscape architect is designing a triangular garden ABC. She is given the measurements AB = 35 m, AC = 28 m, and the angle ABC = 40°. Which of the following statements about the length of the third side, BC, is true?

  1. No such triangle can be formed with these measurements.
  2. There is only one possible length for the side BC.
  3. There are exactly two possible lengths for the side BC. (correct answer)
  4. The length of BC must be greater than the length of AB.
Explanation: This is the ambiguous case (SSA) for the Sine Rule. Let side c = AB = 35, side b = AC = 28, and angle B = 40°. We can find angle C using the Sine Rule: sin(C)/c = sin(B)/b, so sin(C)/35 = sin(40°)/28. This gives sin(C) = 35 * sin(40°) / 28 ≈ 0.8035. Since 0 < 0.8035 < 1, there are two possible values for angle C. C₁ = arcsin(0.8035) ≈ 53.5° and C₂ = 180° - 53.5° = 126.5°. We check if both are valid. Case 1: Angle A₁ = 180° - 40° - 53.5° = 86.5°. This is a valid triangle. Case 2: Angle A₂ = 180° - 40° - 126.5° = 13.5°. This is also a valid triangle. Since two distinct triangles can be formed, there are two possible lengths for the side BC.

Question 9

From the top of a 120 m tall cliff, the angle of depression to a buoy is 18°. A boat is located 200 m from the buoy on the straight line connecting the cliff base and the buoy. If the boat is further from the cliff than the buoy, what is the angle of depression from the top of the cliff to the boat?

  1. 11.9° (correct answer)
  2. 15.5°
  3. 28.5°
  4. 35.3°
Explanation: Let T be the top of the cliff and B be the base. The height is TB = 120 m. Let U be the buoy. The angle of depression to U is 18°, so the angle of elevation from U to T is also 18° (angle TUB = 18°). The horizontal distance from the base of the cliff to the buoy is BU. In the right-angled triangle TBU, tan(18°) = TB/BU = 120/BU. So, BU = 120 / tan(18°) ≈ 369.32 m. The boat, A, is 200 m from the buoy and further from the cliff. So, the distance from the cliff base to the boat is BA = BU + 200 = 369.32 + 200 = 569.32 m. The angle of depression, θ, to the boat is found from the right-angled triangle TBA. tan(θ) = TB/BA = 120 / 569.32 ≈ 0.2108. θ = arctan(0.2108) ≈ 11.9°.

Question 10

A rectangular room has a length of 8 m, a width of 5 m, and a height of 3 m. A thin wire is stretched from one corner of the ceiling to the diametrically opposite corner on the floor. Calculate the angle this wire makes with the floor.

  1. 17.7° (correct answer)
  2. 20.6°
  3. 32.0°
  4. 72.3°
Explanation: Let the angle be θ. This problem can be modeled with a right-angled triangle. The three sides of this triangle are: the height of the room (h = 3 m), the diagonal of the floor (d), and the wire itself (the space diagonal, which is the hypotenuse). The angle θ is between the wire and the floor diagonal. The length of the floor diagonal is found using Pythagoras' theorem on the floor dimensions: d = √(8² + 5²) = √(64 + 25) = √89 ≈ 9.434 m. Now, consider the vertical right-angled triangle with the height as the side opposite to angle θ, and the floor diagonal as the adjacent side. We have tan(θ) = opposite/adjacent = h/d = 3/√89. So, θ = arctan(3/√89) ≈ 17.65°. This rounds to 17.7°.

Question 11

A rectangular room has a length of 8 m, a width of 5 m, and a height of 3 m. A thin wire is stretched from one corner of the ceiling to the diametrically opposite corner on the floor. Calculate the angle this wire makes with the floor.

  1. 17.7° (correct answer)
  2. 20.6°
  3. 32.0°
  4. 72.3°
Explanation: Let the angle be θ. This problem can be modeled with a right-angled triangle. The three sides of this triangle are: the height of the room (h = 3 m), the diagonal of the floor (d), and the wire itself (the space diagonal, which is the hypotenuse). The angle θ is between the wire and the floor diagonal. The length of the floor diagonal is found using Pythagoras' theorem on the floor dimensions: d = √(8² + 5²) = √(64 + 25) = √89 ≈ 9.434 m. Now, consider the vertical right-angled triangle with the height as the side opposite to angle θ, and the floor diagonal as the adjacent side. We have tan(θ) = opposite/adjacent = h/d = 3/√89. So, θ = arctan(3/√89) ≈ 17.65°. This rounds to 17.7°.

Question 12

A farmer wants to find the area of a triangular field ABC. From point A, he walks 400 m to point B. He turns and walks 500 m to point C. The final distance from C back to A is 600 m. Find the total area of the field.

  1. 150,000 m²
  2. 100,000 m²
  3. 120,000 m²
  4. 99,200 m² (correct answer)
Explanation: When you encounter a triangle problem with all three side lengths given, you should immediately think of Heron's formula for finding area. This is the most direct method when you know all sides but no angles or heights. First, calculate the semi-perimeter: s=a+b+c2=400+500+6002=750 ms = \frac{a + b + c}{2} = \frac{400 + 500 + 600}{2} = 750 \text{ m} Then apply Heron's formula: Area=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} Substituting the values: Area=750(750−400)(750−500)(750−600)\text{Area} = \sqrt{750(750-400)(750-500)(750-600)} =750×350×250×150= \sqrt{750 \times 350 \times 250 \times 150} =9,843,750,000= \sqrt{9,843,750,000} =99,216.75 m2= 99,216.75 \text{ m}^2 This rounds to approximately 99,200 m², confirming answer D. Answer A (150,000 m²) likely comes from incorrectly using the formula 12×500×600=150,000\frac{1}{2} \times 500 \times 600 = 150,000, treating two sides as if they were base and height at a right angle. Answer B (100,000 m²) might result from using 12×400×500=100,000\frac{1}{2} \times 400 \times 500 = 100,000 with the same misconception. Answer C (120,000 m²) could come from 12×400×600=120,000\frac{1}{2} \times 400 \times 600 = 120,000, again incorrectly assuming perpendicularity. Remember: when you have three sides of any triangle, Heron's formula is your go-to method. Don't assume right angles unless explicitly stated or unless the sides satisfy the Pythagorean theorem.

Question 13

A triangular paddock has a perimeter of 800 m. Two of the sides measure 200 m and 300 m. What is the area of the paddock, correct to three significant figures?

  1. 35,000 m²
  2. 29,500 m²
  3. 30,000 m²
  4. 28,300 m² (correct answer)
Explanation: When you encounter a triangle problem where you know all three sides, think Heron's formula - it's specifically designed to find the area when you have the perimeter information but no height. First, find the third side using the perimeter: 800−200−300=300800 - 200 - 300 = 300 m. So you have sides of 200 m, 300 m, and 300 m. Heron's formula states that for a triangle with sides aa, bb, cc, the area is A=s(s−a)(s−b)(s−c)A = \sqrt{s(s-a)(s-b)(s-c)}, where ss is the semi-perimeter (half the perimeter). Calculate the semi-perimeter: s=8002=400s = \frac{800}{2} = 400 m Now apply the formula: A=400(400−200)(400−300)(400−300)A = \sqrt{400(400-200)(400-300)(400-300)} A=400×200×100×100A = \sqrt{400 \times 200 \times 100 \times 100} A=800,000,000A = \sqrt{800,000,000} A=28,284.3A = 28,284.3 m² Rounded to three significant figures, this gives 28,300 m². Option A (35,000 m²) likely comes from incorrectly using the formula 12×base×height\frac{1}{2} \times base \times height with estimated values. Option B (29,500 m²) might result from calculation errors in the square root. Option C (30,000 m²) appears to be a rough estimation trap, possibly from using 12×300×200\frac{1}{2} \times 300 \times 200. Study tip: Memorize Heron's formula for the IB exam. When you see a triangle problem with three known sides (or perimeter plus two sides), Heron's formula is almost always the intended method. Practice the arithmetic carefully - these problems often involve large numbers under square roots.

Question 14

A triangular park ABC has side AB = 180 m, angle CAB = 55°, and angle ABC = 70°. The cost of laying turf is $25 per square metre. Calculate the total cost to lay turf on the entire park.

  1. $138,500
  2. $361,400
  3. $380,600 (correct answer)
  4. $761,200
Explanation: First, find the third angle of the triangle: angle ACB = 180° - 55° - 70° = 55°. Since angle CAB = angle ACB = 55°, the triangle is isosceles with sides BC = AB = 180 m. Now we can find the area of the park using the formula Area = (1/2)ab sin(C). Using sides AB, BC and the included angle ABC: Area = (1/2) * (AB) * (BC) * sin(ABC) = (1/2) * 180 * 180 * sin(70°). Area = 16200 * sin(70°) ≈ 16200 * 0.9397 ≈ 15223.1 m². The total cost is the area multiplied by the cost per square metre: Cost = 15223.1 * $25 ≈ $380,577. The closest answer is $380,600.

Question 15

In triangle PQR, PQ = 12 cm, QR = 15 cm and angle QPR = 40°. Which of the following is a possible value for angle PRQ?

  1. Cannot be determined
  2. 48.6°
  3. 51.4°
  4. 31.4° (correct answer)
Explanation: When you encounter a triangle with two sides and one angle given, you're dealing with the ambiguous case of the sine rule, which can potentially yield two different triangles. Given: PQ = 12 cm, QR = 15 cm, and angle QPR = 40°. Using the sine rule: sin⁡(∠PRQ)PQ=sin⁡(∠QPR)QR\frac{\sin(\angle PRQ)}{PQ} = \frac{\sin(\angle QPR)}{QR} Substituting the values: sin⁡(∠PRQ)12=sin⁡(40°)15\frac{\sin(\angle PRQ)}{12} = \frac{\sin(40°)}{15} Solving for angle PRQ: sin⁡(∠PRQ)=12×sin⁡(40°)15=12×0.64315=0.514\sin(\angle PRQ) = \frac{12 \times \sin(40°)}{15} = \frac{12 \times 0.643}{15} = 0.514 This gives us ∠PRQ=arcsin⁡(0.514)=31.0°\angle PRQ = \arcsin(0.514) = 31.0° (approximately 31.4°). However, since sine is positive in both the first and second quadrants, there's potentially a second solution: 180°−31.4°=148.6°180° - 31.4° = 148.6°. But this would make the sum of angles exceed 180°, so it's invalid. Looking at the options: (A) is wrong because we can determine the angle using the sine rule. (B) 48.6° and (C) 51.4° likely represent common calculation errors or confusion with the complementary relationships in the triangle. (D) 31.4° matches our calculation. Study tip: In ambiguous triangle problems, always check if both potential solutions create valid triangles by ensuring the sum of all angles equals 180°. The sine rule is your primary tool, but geometric constraints eliminate impossible solutions.

Question 16

Three communication towers are located at coordinates A(2, 5), B(8, 1), and C(9, 8). What is the measure of the smallest angle in the triangle formed by these three towers?

  1. 56.9° (correct answer)
  2. 58.7°
  3. 64.4°
  4. 68.2°
Explanation: First, calculate the square of the lengths of the sides using the distance formula d²=(x₂-x₁)²+(y₂-y₁)²: a² = BC² = (9-8)² + (8-1)² = 1² + 7² = 50. b² = AC² = (9-2)² + (8-5)² = 7² + 3² = 58. c² = AB² = (8-2)² + (1-5)² = 6² + (-4)² = 52. The shortest side is 'a' (BC), so the smallest angle is A, which is opposite side BC. Use the Cosine Rule to find angle A: a² = b² + c² - 2bc cos(A). It is easier to use the formula in the form cos(A) = (b² + c² - a²)/(2bc). cos(A) = (58 + 52 - 50) / (2 * √58 * √52) = 60 / (2 * √3016) ≈ 60 / 109.836 ≈ 0.5462. A = arccos(0.5462) ≈ 56.9°.

Question 17

A triangular paddock has a perimeter of 800 m. Two of the sides measure 200 m and 300 m. What is the area of the paddock, correct to three significant figures?

  1. 35,000 m²
  2. 29,500 m²
  3. 30,000 m²
  4. 28,300 m² (correct answer)
Explanation: When you encounter a triangle problem where you know all three sides, think Heron's formula - it's specifically designed to find the area when you have the perimeter information but no height. First, find the third side using the perimeter: 800−200−300=300800 - 200 - 300 = 300 m. So you have sides of 200 m, 300 m, and 300 m. Heron's formula states that for a triangle with sides aa, bb, cc, the area is A=s(s−a)(s−b)(s−c)A = \sqrt{s(s-a)(s-b)(s-c)}, where ss is the semi-perimeter (half the perimeter). Calculate the semi-perimeter: s=8002=400s = \frac{800}{2} = 400 m Now apply the formula: A=400(400−200)(400−300)(400−300)A = \sqrt{400(400-200)(400-300)(400-300)} A=400×200×100×100A = \sqrt{400 \times 200 \times 100 \times 100} A=800,000,000A = \sqrt{800,000,000} A=28,284.3A = 28,284.3 m² Rounded to three significant figures, this gives 28,300 m². Option A (35,000 m²) likely comes from incorrectly using the formula 12×base×height\frac{1}{2} \times base \times height with estimated values. Option B (29,500 m²) might result from calculation errors in the square root. Option C (30,000 m²) appears to be a rough estimation trap, possibly from using 12×300×200\frac{1}{2} \times 300 \times 200. Study tip: Memorize Heron's formula for the IB exam. When you see a triangle problem with three known sides (or perimeter plus two sides), Heron's formula is almost always the intended method. Practice the arithmetic carefully - these problems often involve large numbers under square roots.

Question 18

In triangle PQR, PQ = 12 cm, QR = 15 cm and angle QPR = 40°. Which of the following is a possible value for angle PRQ?

  1. Cannot be determined
  2. 48.6°
  3. 51.4°
  4. 31.4° (correct answer)
Explanation: When you encounter a triangle with two sides and one angle given, you're dealing with the ambiguous case of the sine rule, which can potentially yield two different triangles. Given: PQ = 12 cm, QR = 15 cm, and angle QPR = 40°. Using the sine rule: sin⁡(∠PRQ)PQ=sin⁡(∠QPR)QR\frac{\sin(\angle PRQ)}{PQ} = \frac{\sin(\angle QPR)}{QR} Substituting the values: sin⁡(∠PRQ)12=sin⁡(40°)15\frac{\sin(\angle PRQ)}{12} = \frac{\sin(40°)}{15} Solving for angle PRQ: sin⁡(∠PRQ)=12×sin⁡(40°)15=12×0.64315=0.514\sin(\angle PRQ) = \frac{12 \times \sin(40°)}{15} = \frac{12 \times 0.643}{15} = 0.514 This gives us ∠PRQ=arcsin⁡(0.514)=31.0°\angle PRQ = \arcsin(0.514) = 31.0° (approximately 31.4°). However, since sine is positive in both the first and second quadrants, there's potentially a second solution: 180°−31.4°=148.6°180° - 31.4° = 148.6°. But this would make the sum of angles exceed 180°, so it's invalid. Looking at the options: (A) is wrong because we can determine the angle using the sine rule. (B) 48.6° and (C) 51.4° likely represent common calculation errors or confusion with the complementary relationships in the triangle. (D) 31.4° matches our calculation. Study tip: In ambiguous triangle problems, always check if both potential solutions create valid triangles by ensuring the sum of all angles equals 180°. The sine rule is your primary tool, but geometric constraints eliminate impossible solutions.

Question 19

A triangular field has an area of 1500 m². Two of the sides adjacent to one corner measure 60 m and 80 m. What is the smallest possible perimeter of the field, correct to one decimal place?

  1. 50.1 m
  2. 190.1 m (correct answer)
  3. 240.0 m
  4. 272.3 m
Explanation: Let the sides be a = 60 m, b = 80 m, and the included angle be C. Area = (1/2)ab sin(C). 1500 = (1/2)(60)(80)sin(C) => 1500 = 2400 sin(C) => sin(C) = 1500/2400 = 0.625. There are two possible angles: C₁ = arcsin(0.625) ≈ 38.68° (acute) and C₂ = 180° - 38.68° = 141.32° (obtuse). The perimeter is P = a + b + c = 140 + c. To find the smallest perimeter, we need the smallest possible third side, c. The length of c depends on angle C via the Cosine Rule: c² = a² + b² - 2ab cos(C). For the smallest c, we need the largest cos(C), which corresponds to the smallest positive angle C. Using C₁ = 38.68°: c² = 60² + 80² - 2(60)(80)cos(38.68°) = 10000 - 9600(0.7806) ≈ 2506.24. So, c ≈ 50.06 m. The smallest perimeter is P₁ = 140 + 50.06 = 190.06 m, or 190.1 m.

Question 20

Three communication towers are located at coordinates A(2, 5), B(8, 1), and C(9, 8). What is the measure of the smallest angle in the triangle formed by these three towers?

  1. 56.9° (correct answer)
  2. 58.7°
  3. 64.4°
  4. 68.2°
Explanation: First, calculate the square of the lengths of the sides using the distance formula d²=(x₂-x₁)²+(y₂-y₁)²: a² = BC² = (9-8)² + (8-1)² = 1² + 7² = 50. b² = AC² = (9-2)² + (8-5)² = 7² + 3² = 58. c² = AB² = (8-2)² + (1-5)² = 6² + (-4)² = 52. The shortest side is 'a' (BC), so the smallest angle is A, which is opposite side BC. Use the Cosine Rule to find angle A: a² = b² + c² - 2bc cos(A). It is easier to use the formula in the form cos(A) = (b² + c² - a²)/(2bc). cos(A) = (58 + 52 - 50) / (2 * √58 * √52) = 60 / (2 * √3016) ≈ 60 / 109.836 ≈ 0.5462. A = arccos(0.5462) ≈ 56.9°.