In △RST ~ △XYZ, RS = 6, ST = 9, RT = 12, and XY = 4. Find YZ and XZ. — YZ = 6, XZ = 8.Scale factor: RS/XY = 6/4 = 3/2. So YZ = ST × (2/3) = 9 × 2/3 = 6 and XZ = RT × (2/3) = 12 × 2/3 = 8.
A 6-foot person casts a 4-foot shadow. At the same time, a tree casts a 20-foot shadow. How tall is the tree? — 30 ft.Set up proportion: person height/shadow = tree height/shadow → 6/4 = h/20 → h = 30 feet.
In △ABC, DE ∥ BC where D is on AB and E is on AC. If AD = 4, DB = 6, and AE = 5, find EC. — 7.5.By the Triangle Proportionality Theorem: AD/DB = AE/EC → 4/6 = 5/EC → EC = 5 × 6/4 = 7.5.
If △ABC ~ △DEF with areas 36 and 16 respectively, what is the scale factor of △ABC to △DEF? — 3/2.Areas scale as the square of the scale factor: (k)² = 36/16 = 9/4, so k = 3/2.
SSS similarity requires all three pairs of corresponding sides to be: — Proportional.SSS similarity requires all three ratios of corresponding sides to be equal (proportional) — not necessarily equal in length.
In right △ABC (right angle at C), altitude CD is drawn to the hypotenuse AB. Which triangles are similar? — △ABC ~ △ACD ~ △CBD.When the altitude is drawn to the hypotenuse of a right triangle, three similar triangles are formed: △ABC ~ △ACD ~ △CBD (all share angle A or angle B plus the right angle).
△PQR ~ △STU with PQ = 5, QR = 7, PR = 9, and ST = 10. What is TU? — 14.Scale factor: ST/PQ = 10/5 = 2. TU = QR × 2 = 7 × 2 = 14.
Two triangles have sides 3, 4, 5 and 6, 8, 10 respectively. Are they similar? — Yes, by SSS similarity (ratio 1:2).3/6 = 4/8 = 5/10 = 1/2. All three ratios are equal, so SSS similarity applies. The triangles are similar with scale factor 1:2.
In △ABC, A = (0,0), B = (6,0), C = (4,8). Find the midpoint M of AB and midpoint N of AC. — M = (3,0), N = (2,4).M = midpoint of AB = ((0+6)/2, (0+0)/2) = (3, 0). N = midpoint of AC = ((0+4)/2, (0+8)/2) = (2, 4).
Continuing from the previous question: What is the length of midsegment MN? — √17.MN: from (3,0) to (2,4). Distance = √((3−2)² + (0−4)²) = √(1 + 16) = √17.
What is the length of BC in △ABC with A = (0,0), B = (6,0), C = (4,8)? — Both A and B are correct.BC: from (6,0) to (4,8). Distance = √((6−4)² + (0−8)²) = √(4 + 64) = √68 = 2√17. So both A and B are equivalent correct answers.
A midsegment has length 4x − 1 and the parallel base has length 6x + 4. Set up and solve for x. — x = 2.Midsegment = ½ base → 4x − 1 = ½(6x + 4) → 4x − 1 = 3x + 2 → x = 3. Check: midsegment = 11, base = 22, 11 = 22/2 ✓. Actually x = 3 is correct.
If the midsegment of a triangle is 15 and the base is 3y − 6, find y. — y = 12.Midsegment = ½ base → 15 = ½(3y − 6) → 30 = 3y − 6 → 3y = 36 → y = 12.
The medial triangle (formed by all three midsegments) is similar to the original triangle with what scale factor? — 1:2.The medial triangle has sides equal to the midsegments, which are each half of the original sides. Scale factor = 1:2.
Can the Midsegment Theorem be proven using similar triangles? Which criterion? — Yes, by AA similarity.Since M and N are midpoints, AM/AB = AN/AC = 1/2, and ∠A is shared. By SAS similarity, △AMN ~ △ABC. Then MN/BC = 1/2 and MN ∥ BC follow from the similar triangles.
In trapezoid ABCD with AB ∥ CD, the midsegment connects the midpoints of the non-parallel sides. Its length is: — (AB + CD)/2.The midsegment of a trapezoid equals the average of the two parallel sides (bases): midsegment = (AB + CD)/2.
In a trapezoid with parallel sides 8 and 14, what is the midsegment length? — 11.Trapezoid midsegment = (8 + 14)/2 = 22/2 = 11.
In right △ABC with ∠B = 90°, AB = 5 (adjacent to ∠A), BC = 12 (opposite to ∠A). Find tan A. — 12/5.tan A = opposite/adjacent = BC/AB = 12/5.
Given cos θ = 8/17, find sin θ using the Pythagorean Identity. — 15/17.sin²θ = 1 − cos²θ = 1 − 64/289 = 225/289. sin θ = 15/17.
A ladder leans against a wall, making a 65° angle with the ground. If the ladder is 12 feet long, how high on the wall does it reach? — 12 sin 65°.The wall height is opposite the 65° angle; the ladder is the hypotenuse. sin 65° = height/12 → height = 12 sin 65° ≈ 10.88 ft.
tan 45° equals: — 1.In a 45-45-90 triangle, opposite and adjacent sides are equal, so tan 45° = 1.
If sin θ = 5/13 and θ is an acute angle, find cos θ. — 12/13.cos²θ = 1 − sin²θ = 1 − 25/169 = 144/169. cos θ = 12/13.
A surveyor is 50 m from the base of a building and measures the angle of elevation to the top as 40°. What is the height of the building? — 50 tan 40°.tan 40° = height/50 → height = 50 tan 40° ≈ 50 × 0.839 ≈ 41.95 m.
In right △ABC with legs 6 and 8, find angle A opposite the side of length 8. — All of the above.Hypotenuse = √(36+64) = 10. sin A = 8/10, cos A = 6/10, tan A = 8/6. All three expressions give angle A — they are all correct.
sin 30° = 1/2 and cos 60° = 1/2. What does this say about sin and cos? — sin θ = cos(90° − θ) (cofunctions).sin and cos are cofunctions: sin θ = cos(90° − θ). So sin 30° = cos 60° = 1/2 because 30° + 60° = 90°.
From the top of a 30-foot cliff, the angle of depression to a boat is 25°. How far is the boat from the base of the cliff? — 30/tan 25°.Angle of depression = angle of elevation from boat = 25°. tan 25° = 30/d → d = 30/tan 25° ≈ 64.3 ft.
A 45-45-90 triangle has a hypotenuse of 8√2. What is the length of each leg? — 8.hypotenuse = leg × √2 → 8√2 = leg × √2 → leg = 8.
A 30-60-90 triangle has a hypotenuse of 14. Find the short leg and the long leg. — Short = 7, Long = 7√3.Hypotenuse = 2x → 2x = 14 → x = 7. Short leg = 7, long leg = 7√3.
A 30-60-90 triangle has a long leg of 9. Find the short leg and hypotenuse. — Short = 9/√3 = 3√3, Hyp = 6√3.Long leg = x√3 = 9 → x = 9/√3 = 3√3. Hypotenuse = 2x = 6√3.
What is the exact value of tan 30°? — 1/√3 = √3/3.tan 30° = opposite/adjacent = x/(x√3) = 1/√3 = √3/3.
An equilateral triangle has side length 10. What is its height? — 5√3.An altitude of an equilateral triangle creates two 30-60-90 triangles. The altitude (long leg) = short leg × √3 = 5√3.
A square has diagonal 12. What is its side length? — 12/√2 and 6√2 are equal.The diagonal of a square creates a 45-45-90 triangle. Diagonal = side × √2 → 12 = s√2 → s = 12/√2 = 6√2. Both "6√2" and "12/√2" are equal.
In a 30-60-90 triangle, the 60° angle is opposite which side? — Long leg.In a 30-60-90 triangle: the 30° angle is opposite the short leg (x), the 60° angle is opposite the long leg (x√3), and the 90° angle is opposite the hypotenuse (2x).
An isosceles right triangle has legs of length 5. What is the area? — 12.5.Area = ½ × base × height = ½ × 5 × 5 = 12.5. In a 45-45-90 triangle, both legs serve as base and height.
In △ABC, A = 30°, B = 45°, a = 8. Find the value of b using Law of Sines. — b ≈ 11.3.b/sinB = a/sinA → b = 8·sin45°/sin30° = 8·(√2/2)/(1/2) = 8√2 ≈ 11.31.
In △ABC with a = 7, b = 9, C = 120°. Which formula finds c? — c² = a² + b² − 2ab·cosC.SAS case (two sides + included angle) → use Law of Cosines.
Compute c² if a = 6, b = 8, C = 90°. Use Law of Cosines. — 100.c² = 36 + 64 − 2(6)(8)cos90° = 100 − 0 = 100. c = 10 (reduces to Pythagorean Theorem!).
In △ABC with a = 8, b = 5, C = 60°, what is c? — c = √49 = 7.c² = 64 + 25 − 2(8)(5)cos60° = 89 − 40 = 49. c = 7.
Law of Sines: a = 12, sinA = 0.6. What is the common ratio a/sinA? — 20.a/sinA = 12/0.6 = 20. All three ratios b/sinB = c/sinC = 20 in this triangle.
In a triangle, which case DOES NOT require the Law of Sines or Cosines? — Right triangle with two sides given.A right triangle can be solved with basic trigonometric ratios (SOH-CAH-TOA) and the Pythagorean Theorem.
The formula c² = a² + b² − 2ab·cosC becomes c² = a² + b² when C = ? — 90°.cos90° = 0, so the term −2ab·cosC disappears, giving c² = a² + b² — the Pythagorean Theorem.
In △ABC, sinA/a = sinB/b = sinC/c = 1/10. If a = 8, what is sinA? — 0.8.sinA/a = 1/10 → sinA = a/10 = 8/10 = 0.8.
In △ABC, a = 5, b = 5, c = 5 (equilateral). Use Law of Cosines to find angle A. — A = 60°.cosA = (b² + c² − a²)/(2bc) = (25 + 25 − 25)/(50) = 25/50 = 0.5. A = cos⁻¹(0.5) = 60°.
In △ABC, B = 90°, b = 13, a = 5. Using Law of Sines: sinA = ? — sinA = 5/13.a/sinA = b/sinB → 5/sinA = 13/sin90° = 13. sinA = 5/13.
Two sides and the included angle are known: a = 4, c = 6, B = 110°. Which law should you use? — Law of Cosines.SAS case (two sides + the angle BETWEEN them) → use Law of Cosines to find the third side.