Solve the following trigonometry problems: 1. In triangle ABC, angle ABC = 80, angle CAB = 45, AC = 50. Find BC. 2. In triangle PQR, angle RPQ = 60, angle QRP = 50, RP = 6.5. Find PQ.

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Asked on April 13, 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 · 2026-04-13T08:40:57.337Z

Hey scoddyevarau87@gmail.com — ready when you are. Here are the solutions to your trigonometry problems. Problem 1: In the triangle ABC, ABC = 80^, CAB = 45^ and AC = 50. Find BC. Step 1: Find the third angle, BCA. The sum of angles in a triangle is 180^. BCA = 180^ - ABC - CAB BCA = 180^ - 80^ - 45^ BCA = 180^ - 125^ BCA = 55^ Step 2: Use the Sine Rule to find BC. The Sine Rule states (a)/( A) = (b)/( B) = (c)/( C). We want to find side BC (let's call it a), which is opposite CAB. We are given side AC (let's call it b), which is opposite ABC. (BC)/(( CAB)) = (AC)/(( ABC)) (BC)/((45^)) = (50)/((80^)) Step 3: Solve for BC. BC = (50 × (45^))/((80^)) BC = (50 × 0.7071)/(0.9848) BC = (35.355)/(0.9848) BC ≈ 35.90 Problem 2: In the triangle PQR, RPQ = 60^, QRP = 50^ and RP = 6.5. Find PQ. Step 1: Find the third angle, PQR. PQR = 180^ - RPQ - QRP PQR = 180^ - 60^ - 50^ PQR = 180^ - 110^ PQR = 70^ Step 2: Use the Sine Rule to find PQ. We want to find side PQ (opposite QRP). We are given side RP (opposite PQR). (PQ)/(( QRP)) = (RP)/(( PQR)) (PQ)/((50^)) = (6.5)/((70^)) Step 3: Solve for PQ. PQ = (6.5 × (50^))/((70^)) PQ = (6.5 × 0.7660)/(0.9397) PQ = (4.980)/(0.9397) PQ ≈ 5.30 Problem 3: Find the height of a flagpole which casts a shadow of 6.35 m 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 to the angle . = oppositeadjacent = (h)/(s) Step 2: Substitute the given values. Given s = 6.35 m and = 65^. (65^) = (h)/(6.35 m) Step 3: Solve for h. h = 6.35 m × (65^) h = 6.35 m × 2.1445 h ≈ 13.61 m Problem 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 of the right-angled triangle. The longer side is 9.2 m and the shorter side is 3.8 m. The diagonal forms a right-angled triangle with these two sides. Let be the angle the diagonal makes with the longer side. In this triangle, the shorter side is opposite to , and the longer side is adjacent to . = oppositeadjacent = shorter sidelonger side Step 2: Substitute the given values. = 3.8 m9.2 m = 0.41304 Step 3: Solve for . = (0.41304) ≈ 22.43^ Problem 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 trigonometric relationship. Let the angle be = 43.2^. The longer side is adjacent to this angle, and the shorter side is opposite to this angle in the right-angled triangle formed by the sides and the diagonal. = oppositeadjacent = shorter sidelonger side Step 2: Substitute the given values. Given longer side = 12.6 cm and = 43.2^. Let the shorter side be x. (43.2^) = (x)/(12.6 cm) Step 3: Solve for x. x = 12.6 cm × (43.2^) x = 12.6 cm × 0.9391 x ≈ 11.83 cm Problem 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 creates four congruent right-angled triangles. The half-diagonals are the legs of these triangles. Half of the longer diagonal = 12 cm2 = 6 cm. Half of the shorter diagonal = 7 cm2 = 3.5 cm. Step 2: Find the angles within one of the right-angled triangles. Let be the angle opposite the 3.5 cm side and adjacent to the 6 cm side. = 3.5 cm6 cm = 0.5833 = (0.5833) ≈ 30.25^ Let be the angle opposite the 6 cm side and adjacent to the 3.5 cm side. = 6 cm3.5 cm = 1.7143 = (1.7143) ≈ 59.75^ Note that + = 30.25^ + 59.75^ = 90^, which is correct for a right-angled triangle. Step 3: Find the angles of the rhombus. The angles of the rhombus are formed by two of these angles from the right-angled triangles. The smaller angle of the rhombus is 2. 2 = 2 × 30.25^ = 60.5^ The larger angle of the rhombus is 2. 2 = 2 × 59.75^ = 119.5^ The sum of adjacent angles in a rhombus is 180^, so 60.5^ + 119.5^ = 180^. The larger angle of the rhombus is 119.5^. Problem 7: Find the angle of elevation to the top of a 62 m high building from point A, which is at ground level 155 m from its base. What is the angle of depression from the top of the building to A? Step 1: Find the angle of elevation. Let h be the height of the building (62 m) and d be the distance from the base (155 m). Let be the angle of elevation. = oppositeadjacent = (h)/(d) = 62 m155 m = 0.4 = (0.4) ≈ 21.80^ Step 2: Find the angle of depression. The angle of depression from the top of the building to point A is equal to the angle of elevation from point A to the top of the building, due to alternate interior angles being equal when parallel lines (ground and horizontal line from top of building) are intersected by a transversal (line of sight). Therefore, the angle of depression is also 21.80^. Problem 8: The angle of depression from the top of a 180 m high vertical cliff to a boat B is 26^. Find how far the boat is from the base of the cliff. Step 1: Identify the trigonometric relationship. Let h be the height of the cliff (180 m). Let x be the distance of the boat from the base of the cliff. The angle of depression from the top of the cliff to the boat is 26^. This means the angle of elevation from the boat to the top of the cliff is also 26^. In the right-angled triangle formed, h is the opposite side and x is the adjacent side to the angle of elevation. (angle of elevation) = oppositeadjacent = (h)/(x) Step 2: Substitute the given values. (26^) = 180 mx Step 3: Solve for x. x = 180 m(26^) x = 180 m0.4877 x ≈ 369.08 m Problem 9: Sarah measures the angle of elevation to the top of a tree as 25^ from where she is standing at a point 150 m from its base. Her eye level, where the angle measurement was taken, is 2.5 m above the ground. Assuming the ground is horizontal, find the height of the tree. Step 1: Calculate the height of the tree above Sarah's eye level. Let h' be the height of the tree above Sarah's eye level. The distance from the base of the tree to Sarah is 150 m. The angle of elevation is 25^. (25^) = (h')/(150 m) h' = 150 m × (25^) h' = 150 m × 0.4663 h' ≈ 69.945 m Step 2: Calculate the total height of the tree. The total height of the tree (H) is the height above eye level plus Sarah's eye level height. H = h' + eye level height H = 69.945 m + 2.5 m H = 72.445 m Send me the next one 📸

Accepted Answer · 2026-04-13T08:40:57.337Z

Hey scoddyevarau87@gmail.com — ready when you are. Here are the solutions to your trigonometry problems. Problem 1: In the triangle ABC, ABC = 80^, CAB = 45^ and AC = 50. Find BC. Step 1: Find the third angle, BCA. The sum of angles in a triangle is 180^. BCA = 180^ - ABC - CAB BCA = 180^ - 80^ - 45^ BCA = 180^ - 125^ BCA = 55^ Step 2: Use the Sine Rule to find BC. The Sine Rule states (a)/( A) = (b)/( B) = (c)/( C). We want to find side BC (let's call it a), which is opposite CAB. We are given side AC (let's call it b), which is opposite ABC. (BC)/(( CAB)) = (AC)/(( ABC)) (BC)/((45^)) = (50)/((80^)) Step 3: Solve for BC. BC = (50 × (45^))/((80^)) BC = (50 × 0.7071)/(0.9848) BC = (35.355)/(0.9848) BC ≈ 35.90 Problem 2: In the triangle PQR, RPQ = 60^, QRP = 50^ and RP = 6.5. Find PQ. Step 1: Find the third angle, PQR. PQR = 180^ - RPQ - QRP PQR = 180^ - 60^ - 50^ PQR = 180^ - 110^ PQR = 70^ Step 2: Use the Sine Rule to find PQ. We want to find side PQ (opposite QRP). We are given side RP (opposite PQR). (PQ)/(( QRP)) = (RP)/(( PQR)) (PQ)/((50^)) = (6.5)/((70^)) Step 3: Solve for PQ. PQ = (6.5 × (50^))/((70^)) PQ = (6.5 × 0.7660)/(0.9397) PQ = (4.980)/(0.9397) PQ ≈ 5.30 Problem 3: Find the height of a flagpole which casts a shadow of 6.35 m 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 to the angle . = oppositeadjacent = (h)/(s) Step 2: Substitute the given values. Given s = 6.35 m and = 65^. (65^) = (h)/(6.35 m) Step 3: Solve for h. h = 6.35 m × (65^) h = 6.35 m × 2.1445 h ≈ 13.61 m Problem 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 of the right-angled triangle. The longer side is 9.2 m and the shorter side is 3.8 m. The diagonal forms a right-angled triangle with these two sides. Let be the angle the diagonal makes with the longer side. In this triangle, the shorter side is opposite to , and the longer side is adjacent to . = oppositeadjacent = shorter sidelonger side Step 2: Substitute the given values. = 3.8 m9.2 m = 0.41304 Step 3: Solve for . = (0.41304) ≈ 22.43^ Problem 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 trigonometric relationship. Let the angle be = 43.2^. The longer side is adjacent to this angle, and the shorter side is opposite to this angle in the right-angled triangle formed by the sides and the diagonal. = oppositeadjacent = shorter sidelonger side Step 2: Substitute the given values. Given longer side = 12.6 cm and = 43.2^. Let the shorter side be x. (43.2^) = (x)/(12.6 cm) Step 3: Solve for x. x = 12.6 cm × (43.2^) x = 12.6 cm × 0.9391 x ≈ 11.83 cm Problem 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 creates four congruent right-angled triangles. The half-diagonals are the legs of these triangles. Half of the longer diagonal = 12 cm2 = 6 cm. Half of the shorter diagonal = 7 cm2 = 3.5 cm. Step 2: Find the angles within one of the right-angled triangles. Let be the angle opposite the 3.5 cm side and adjacent to the 6 cm side. = 3.5 cm6 cm = 0.5833 = (0.5833) ≈ 30.25^ Let be the angle opposite the 6 cm side and adjacent to the 3.5 cm side. = 6 cm3.5 cm = 1.7143 = (1.7143) ≈ 59.75^ Note that + = 30.25^ + 59.75^ = 90^, which is correct for a right-angled triangle. Step 3: Find the angles of the rhombus. The angles of the rhombus are formed by two of these angles from the right-angled triangles. The smaller angle of the rhombus is 2. 2 = 2 × 30.25^ = 60.5^ The larger angle of the rhombus is 2. 2 = 2 × 59.75^ = 119.5^ The sum of adjacent angles in a rhombus is 180^, so 60.5^ + 119.5^ = 180^. The larger angle of the rhombus is 119.5^. Problem 7: Find the angle of elevation to the top of a 62 m high building from point A, which is at ground level 155 m from its base. What is the angle of depression from the top of the building to A? Step 1: Find the angle of elevation. Let h be the height of the building (62 m) and d be the distance from the base (155 m). Let be the angle of elevation. = oppositeadjacent = (h)/(d) = 62 m155 m = 0.4 = (0.4) ≈ 21.80^ Step 2: Find the angle of depression. The angle of depression from the top of the building to point A is equal to the angle of elevation from point A to the top of the building, due to alternate interior angles being equal when parallel lines (ground and horizontal line from top of building) are intersected by a transversal (line of sight). Therefore, the angle of depression is also 21.80^. Problem 8: The angle of depression from the top of a 180 m high vertical cliff to a boat B is 26^. Find how far the boat is from the base of the cliff. Step 1: Identify the trigonometric relationship. Let h be the height of the cliff (180 m). Let x be the distance of the boat from the base of the cliff. The angle of depression from the top of the cliff to the boat is 26^. This means the angle of elevation from the boat to the top of the cliff is also 26^. In the right-angled triangle formed, h is the opposite side and x is the adjacent side to the angle of elevation. (angle of elevation) = oppositeadjacent = (h)/(x) Step 2: Substitute the given values. (26^) = 180 mx Step 3: Solve for x. x = 180 m(26^) x = 180 m0.4877 x ≈ 369.08 m Problem 9: Sarah measures the angle of elevation to the top of a tree as 25^ from where she is standing at a point 150 m from its base. Her eye level, where the angle measurement was taken, is 2.5 m above the ground. Assuming the ground is horizontal, find the height of the tree. Step 1: Calculate the height of the tree above Sarah's eye level. Let h' be the height of the tree above Sarah's eye level. The distance from the base of the tree to Sarah is 150 m. The angle of elevation is 25^. (25^) = (h')/(150 m) h' = 150 m × (25^) h' = 150 m × 0.4663 h' ≈ 69.945 m Step 2: Calculate the total height of the tree. The total height of the tree (H) is the height above eye level plus Sarah's eye level height. H = h' + eye level height H = 69.945 m + 2.5 m H = 72.445 m Send me the next one 📸

Solve the following trigonometry problems: 1. In triangle ABC, angle ABC = 80, angle CAB = 45, AC = 50. Find BC. 2. In triangle PQR, angle RPQ = 60, angle QRP = 50, RP = 6.5. Find PQ.

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Solve the following trigonometry problems: 1. In triangle ABC, angle ABC = 80, angle CAB = 45, AC = 50. Find BC. 2. In triangle PQR, angle RPQ = 60, angle QRP = 50, RP = 6.5. Find PQ.

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