Solve the following trigonometry problems: 1. In triangle ABC, ABC = 80, CAB = 40, AC = 50. Find BC. 2. In triangle PQR, RPQ = 60, QRP = 50, RP = 8. Find PQ.

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Asked on April 22, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-04-22T12:51:24.701Z

Hey scoddyevarau87@gmail.com, good to see you again. Here are the solutions to your trigonometry problems: 1. In the triangle ABC, ABC = 80, CAB = 40 and AC = 50. Find BC. Step 1: Find the third angle of the triangle. The sum of angles in a triangle is 180^. ACB = 180^ - ABC - CAB ACB = 180^ - 80^ - 40^ ACB = 60^ Step 2: Apply the Sine Rule. The Sine Rule states (a)/( A) = (b)/( B) = (c)/( C). We want to find BC (side a). (BC)/(( CAB)) = (AC)/(( ABC)) (BC)/((40^)) = (50)/((80^)) Step 3: Solve for BC. BC = (50 × (40^))/((80^)) BC = (50 × 0.6428)/(0.9848) BC = (32.14)/(0.9848) BC ≈ 32.636 The length of BC is approximately 32.64 units. 2. In the triangle PQR, RPQ = 60, QRP = 50 and RP = 8. Find PQ. Step 1: Find the third angle of the triangle. PQR = 180^ - RPQ - QRP PQR = 180^ - 60^ - 50^ PQR = 70^ Step 2: Apply the Sine Rule. We want to find PQ (side r). (PQ)/(( QRP)) = (RP)/(( PQR)) (PQ)/((50^)) = (8)/((70^)) Step 3: Solve for PQ. PQ = (8 × (50^))/((70^)) PQ = (8 × 0.7660)/(0.9397) PQ = (6.128)/(0.9397) PQ ≈ 6.521 The length of PQ is approximately 6.52 units. 3. Find the height of a flagpole which casts a shadow of 6.35m when the sun makes an angle of 65 to the horizontal. Step 1: Identify the trigonometric relationship. Let h be the height of the flagpole and s be the length of the shadow. The angle of elevation is = 65^. This forms a right-angled triangle where h is the opposite side and s is the adjacent side. () = oppositeadjacent = (h)/(s) Step 2: Substitute the given values. (65^) = (h)/(6.35 m) Step 3: Solve for h. h = 6.35 m × (65^) h = 6.35 m × 2.1445 h ≈ 13.618 m The height of the flagpole is approximately 13.62 m. 4. A rectangle is 9.2 m by 3.8 m. What angle does its diagonal make with its longer side? Step 1: Identify the sides and the angle. The longer side (adjacent) is L = 9.2 m. The shorter side (opposite) is W = 3.8 m. Let be the angle the diagonal makes with the longer side. () = oppositeadjacent = (W)/(L) Step 2: Substitute the values. () = 3.8 m9.2 m () ≈ 0.41304 Step 3: Solve for . = (0.41304) ≈ 22.43^ The angle is approximately 22.43^. 5. The diagonal and the longer side of a rectangle make an angle of 43.2. If the longer side is 12.6 cm, find the length of the shorter side. Step 1: Identify the sides and the angle. The longer side (adjacent) is L = 12.6 cm. Let W be the shorter side (opposite). The angle = 43.2^. () = oppositeadjacent = (W)/(L) Step 2: Substitute the given values. (43.2^) = (W)/(12.6 cm) Step 3: Solve for W. W = 12.6 cm × (43.2^) W = 12.6 cm × 0.9391 W ≈ 11.833 cm The length of the shorter side is approximately 11.83 cm. 6. A rhombus has diagonals of length 12 cm and 7 cm respectively. Find the larger angle of the rhombus. Step 1: Understand the properties of a rhombus. The diagonals of a rhombus bisect each other at right angles. This forms four congruent right-angled triangles. The half-diagonals are the legs of these right-angled triangles. Half of diagonal 1: d_1/2 = 12 cm / 2 = 6 cm. Half of diagonal 2: d_2/2 = 7 cm / 2 = 3.5 cm. Step 2: Find the angles within one of the right-angled triangles. Let be one of the angles at a vertex of the rhombus, bisected by a diagonal. Consider the right-angled triangle formed by the half-diagonals. Let be the angle opposite the 3.5 cm side and adjacent to the 6 cm side. () = oppositeadjacent = (3.5)/(6) \

Accepted Answer · 2026-04-22T12:51:24.701Z

Hey scoddyevarau87@gmail.com, good to see you again. Here are the solutions to your trigonometry problems: 1. In the triangle ABC, ABC = 80, CAB = 40 and AC = 50. Find BC. Step 1: Find the third angle of the triangle. The sum of angles in a triangle is 180^. ACB = 180^ - ABC - CAB ACB = 180^ - 80^ - 40^ ACB = 60^ Step 2: Apply the Sine Rule. The Sine Rule states (a)/( A) = (b)/( B) = (c)/( C). We want to find BC (side a). (BC)/(( CAB)) = (AC)/(( ABC)) (BC)/((40^)) = (50)/((80^)) Step 3: Solve for BC. BC = (50 × (40^))/((80^)) BC = (50 × 0.6428)/(0.9848) BC = (32.14)/(0.9848) BC ≈ 32.636 The length of BC is approximately 32.64 units. 2. In the triangle PQR, RPQ = 60, QRP = 50 and RP = 8. Find PQ. Step 1: Find the third angle of the triangle. PQR = 180^ - RPQ - QRP PQR = 180^ - 60^ - 50^ PQR = 70^ Step 2: Apply the Sine Rule. We want to find PQ (side r). (PQ)/(( QRP)) = (RP)/(( PQR)) (PQ)/((50^)) = (8)/((70^)) Step 3: Solve for PQ. PQ = (8 × (50^))/((70^)) PQ = (8 × 0.7660)/(0.9397) PQ = (6.128)/(0.9397) PQ ≈ 6.521 The length of PQ is approximately 6.52 units. 3. Find the height of a flagpole which casts a shadow of 6.35m when the sun makes an angle of 65 to the horizontal. Step 1: Identify the trigonometric relationship. Let h be the height of the flagpole and s be the length of the shadow. The angle of elevation is = 65^. This forms a right-angled triangle where h is the opposite side and s is the adjacent side. () = oppositeadjacent = (h)/(s) Step 2: Substitute the given values. (65^) = (h)/(6.35 m) Step 3: Solve for h. h = 6.35 m × (65^) h = 6.35 m × 2.1445 h ≈ 13.618 m The height of the flagpole is approximately 13.62 m. 4. A rectangle is 9.2 m by 3.8 m. What angle does its diagonal make with its longer side? Step 1: Identify the sides and the angle. The longer side (adjacent) is L = 9.2 m. The shorter side (opposite) is W = 3.8 m. Let be the angle the diagonal makes with the longer side. () = oppositeadjacent = (W)/(L) Step 2: Substitute the values. () = 3.8 m9.2 m () ≈ 0.41304 Step 3: Solve for . = (0.41304) ≈ 22.43^ The angle is approximately 22.43^. 5. The diagonal and the longer side of a rectangle make an angle of 43.2. If the longer side is 12.6 cm, find the length of the shorter side. Step 1: Identify the sides and the angle. The longer side (adjacent) is L = 12.6 cm. Let W be the shorter side (opposite). The angle = 43.2^. () = oppositeadjacent = (W)/(L) Step 2: Substitute the given values. (43.2^) = (W)/(12.6 cm) Step 3: Solve for W. W = 12.6 cm × (43.2^) W = 12.6 cm × 0.9391 W ≈ 11.833 cm The length of the shorter side is approximately 11.83 cm. 6. A rhombus has diagonals of length 12 cm and 7 cm respectively. Find the larger angle of the rhombus. Step 1: Understand the properties of a rhombus. The diagonals of a rhombus bisect each other at right angles. This forms four congruent right-angled triangles. The half-diagonals are the legs of these right-angled triangles. Half of diagonal 1: d_1/2 = 12 cm / 2 = 6 cm. Half of diagonal 2: d_2/2 = 7 cm / 2 = 3.5 cm. Step 2: Find the angles within one of the right-angled triangles. Let be one of the angles at a vertex of the rhombus, bisected by a diagonal. Consider the right-angled triangle formed by the half-diagonals. Let be the angle opposite the 3.5 cm side and adjacent to the 6 cm side. () = oppositeadjacent = (3.5)/(6) \

Solve the following trigonometry problems: 1. In triangle ABC, ABC = 80, CAB = 40, AC = 50. Find BC. 2. In triangle PQR, RPQ = 60, QRP = 50, RP = 8. Find PQ.

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Solve the following trigonometry problems: 1. In triangle ABC, ABC = 80, CAB = 40, AC = 50. Find BC. 2. In triangle PQR, RPQ = 60, QRP = 50, RP = 8. Find PQ.

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