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Use when you know: 2 angles + 1 side (AAS/ASA) or 2 sides + non-included angle (SSA)
In △ABC: A=35°, B=65°, a=8 cm. Find b and C.
C = 180°−35°−65° = 80°
b/sin65° = 8/sin35° → b = 8×sin65°/sin35° = 12.65 cm
a=7, b=10, A=30°. Find B.
sinB = 10×sin30°/7 = 5/7 → B = 45.6° or B = 134.4°
Both solutions are valid (ambiguous case) → Two triangles possible
Use when: 3 sides known (SSS) or 2 sides + included angle (SAS)
b=5, c=7, A=60°. Find a.
a² = 25+49−2(5)(7)cos60° = 74−35 = 39
a = √39 ≈ 6.24 cm
a=6, b=8, c=10. Find angle A.
cosA = (64+100−36)/(2×8×10) = 128/160 = 0.8
A = cos⁻¹(0.8) ≈ 36.87°
where a, b are two sides and C is the included angle
a=8 cm, b=10 cm, C=45°. Find area.
Area = ½×8×10×sin45° = 40×0.7071 = 28.28 cm²
a=5, b=6, c=7. s = 9. Area = √[9×4×3×2] = √216 ≈ 14.7 cm²
Bearings are measured clockwise from North (000° to 360°).
Ship sails from A on bearing 060° for 20 km to B. Then on bearing 150° for 15 km to C. Find AC and bearing of C from A.
Angle ABC = 150°−60° = 90° (or use geometry)
AC = √(20²+15²) = √625 = 25 km
From a point 50m from base of a tower, angle of elevation of top is 40°. Find height.
h = 50×tan40° = 50×0.839 = 41.95 m
Reference angle = sin⁻¹(0.5) = 30°
Sin positive in Q1 and Q2: x = 30° or x = 180°−30° = 30° or 150°
cos x = 0.5 → reference = 60°
Cos positive in Q1 and Q4: x = 60° or 300°
| Identity | Formula |
|---|---|
| Pythagorean | sin²θ + cos²θ = 1 |
| Quotient | tan θ = sin θ / cos θ |
| From Pythag | 1 + tan²θ = sec²θ |
| From Pythag | 1 + cot²θ = cosec²θ |
LHS = sin²θ + 2sinθcosθ + cos²θ = (sin²θ+cos²θ) + 2sinθcosθ = 1 + 2sinθcosθ = RHS ✓
Cuboid 6m × 4m × 3m. Find length of space diagonal AG.
Face diagonal AC = √(6²+4²) = √52 = 2√13
Space diagonal AG = √(AC²+3²) = √(52+9) = √61 ≈ 7.81 m
The angle of elevation of point P from point A on the ground where AP = 10 m and the height of P = 6 m:
tan θ = 6/10 → θ = tan⁻¹(0.6) ≈ 30.96°
In △PQR: PQ=8, QR=10, angle Q=50°. Find PR and area of △PQR.
PR² = 8²+10²−2(8)(10)cos50° = 64+100−102.8 = 61.2 → PR = 7.82
Area = ½×8×10×sin50° = 30.64 cm²
Solve: tan x = −1 for 0° ≤ x ≤ 360°
Reference = 45°. Tan negative in Q2 and Q4.
x = 180°−45° = 135° or x = 360°−45° = 135° or 315°
1. The rule a/sinA = b/sinB is called the .
2. The rule a² = b²+c²−2bc·cosA is called the .
3. The angles of a triangle sum to degrees.
4. The area formula ½ab·sinC uses sides and their included angle.
5. The formula ½ab·sinC is used to find the of a triangle.
6. (4 marks) In △ABC: a=9, b=12, C=60°. Find c and the area of the triangle.
7. (3 marks) Solve: 2sin x = 1 for 0° ≤ x ≤ 360°.
8. (3 marks) From the top of a cliff 80m high, the angle of depression of a boat is 25°. Find the distance of the boat from the base of the cliff.