Trigonometry — Flashcards

Answer

Step 1: tan A = opposite/adjacent = 3/4, so take opposite = 3k and adjacent = 4k. Step 2: Use Pythagoras theorem. AC^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2 AC = 5k Step 3: Find the ratios. sin A =…

Answer

Step 1: For angle R, opposite side = PQ = 3 cm and hypotenuse = PR = 6 cm. Step 2: Find sine of angle R. sin R = PQ/PR = 3/6 = 1/2 Step 3: sin 30° = 1/2, so angle PRQ = 30°. Step 4: The two acute angl…

Answer

Step 1: At angle C, AB is opposite and BC is adjacent. Step 2: Use tan C. tan 30° = AB/BC = 5/BC Step 3: tan 30° = 1/√3. So, 5/BC = 1/√3 BC = 5√3 cm Step 4: Use sin 30°. sin 30° = AB/AC = 5/AC = 1/2 A…

Answer

Step 1: Use the identity sin^2 A + cos^2 A = 1. Step 2: Substitute sin A = 15/17. (15/17)^2 + cos^2 A = 1 225/289 + cos^2 A = 1 cos^2 A = 64/289 Step 3: Since A is acute, cos A = 8/17. Step 4: Use tan…

Answer

Use it when one trigonometric ratio is known and another ratio is needed. Example: If sin A = 3/5, then sin^2 A + cos^2 A = 1 9/25 + cos^2 A = 1 cos^2 A = 16/25 cos A = 4/5 for an acute angle. This id…

Answer

Step 1: Use sin^2 A + cos^2 A = 1. Step 2: Substitute cos A = 12/13. sin^2 A + (12/13)^2 = 1 sin^2 A + 144/169 = 1 sin^2 A = 25/169 Step 3: Since A is acute, sin A = 5/13. Step 4: Find tan A. tan A = …

Answer

Step 1: 3 tan A = 4 means tan A = 4/3. Step 2: Take opposite = 4k and adjacent = 3k. Step 3: Hypotenuse = sqrt((4k)^2 + (3k)^2) = sqrt(16k^2 + 9k^2) = 5k. Step 4: Find ratios. sin A = 4k/5k = 4/5 cos …

Answer

Step 1: Use Pythagoras theorem. PR^2 = PQ^2 - QR^2 = 29^2 - 21^2 = 841 - 441 = 400 PR = 20 units Step 2: For angle θ at Q, opposite side = PR and adjacent side = QR. sin θ = PR/PQ = 20/29 cos θ = QR/P…