Cosine rule

Using the cosine rule, x² = 13² + 10² - 2 × 13 × 10 × cos(100°), so x = 17.7 cm (1 d.p.).

Using the cosine rule, x² = 11² + 14² - 2 × 11 × 14 × cos(75°), so x = 15.4 cm (1 d.p.).

Using the cosine rule, x² = 9² + 12² - 2 × 9 × 12 × cos(50°), so x = 9.3 cm (1 d.p.).

Using the cosine rule, x² = 8.5² + 8.5² - 2 × 8.5 × 8.5 × cos(130°), so x = 15.4 cm (1 d.p.).

Using the cosine rule, x² = 7² + 11² - 2 × 7 × 11 × cos(115°), so x = 15.3 m (1 d.p.).

Using the cosine rule, x² = 14² + 18² - 2 × 14 × 18 × cos(38°), so x = 11.1 mm (1 d.p.).

Using the cosine rule to find cos x° and then x = cos⁻¹(...), x = 80.3° (3 s.f.).

Using the cosine rule to find cos x° and then x = cos⁻¹(...), x = 54.6° (3 s.f.).

Using the cosine rule to find cos x° and then x = cos⁻¹(...), x = 32.2° (3 s.f.).

Using the cosine rule to find cos x° and then x = cos⁻¹(...), x = 89.8° (3 s.f.).

Using the cosine rule to find cos x° and then x = cos⁻¹(...), x = 67.4° (3 s.f.).

Using the cosine rule to find cos x° and then x = cos⁻¹(...), x = 38.7° (3 s.f.).

Shared vertical h from the cosine rule: h² = 10² + 13² - 2 × 10 × 13 × cos(40°), so h = 8.4 cm. Then tan x° = h ÷ 5, so x = 59.1° (1 d.p.).

Shared vertical h from the cosine rule: h² = 9² + 11² - 2 × 9 × 11 × cos(55°), so h = 9.4 cm. Then tan x° = h ÷ 6, so x = 57.5° (1 d.p.).

Shared vertical h from the cosine rule: h² = 8² + 12² - 2 × 8 × 12 × cos(48°), so h = 8.9 cm. Then tan x° = h ÷ 4.5, so x = 63.2° (1 d.p.).

Shared vertical h from the cosine rule: h² = 11² + 14² - 2 × 11 × 14 × cos(35°), so h = 8.0 cm. Then tan x° = h ÷ 7, so x = 49.0° (1 d.p.).

Shared vertical h from the cosine rule: h² = 12² + 15² - 2 × 12 × 15 × cos(50°), so h = 11.7 cm. Then tan x° = h ÷ 5.5, so x = 64.9° (1 d.p.).

Shared vertical h from the cosine rule: h² = 7² + 10² - 2 × 7 × 10 × cos(65°), so h = 9.5 cm. Then tan x° = h ÷ 4, so x = 67.1° (1 d.p.).

Let the sides adjacent to the 60° angle be the two algebraic lengths and apply the cosine rule with cos(60°). This gives a quadratic equation in x. Solving gives x = 6 (reject any root that would make a side length non-positive).

Let the sides adjacent to the 60° angle be the two algebraic lengths and apply the cosine rule with cos(60°). This gives a quadratic equation in x. Solving gives x = 5 (reject any root that would make a side length non-positive).

Let the sides adjacent to the 120° angle be the two algebraic lengths and apply the cosine rule with cos(120°). This gives a quadratic equation in x. Solving gives x = 4 (reject any root that would make a side length non-positive).

Let the sides adjacent to the 60° angle be the two algebraic lengths and apply the cosine rule with cos(60°). This gives a quadratic equation in x. Solving gives x = 7 (reject any root that would make a side length non-positive).

Let the sides adjacent to the 60° angle be the two algebraic lengths and apply the cosine rule with cos(60°). This gives a quadratic equation in x. Solving gives x = 5 (reject any root that would make a side length non-positive).

Let the sides adjacent to the 120° angle be the two algebraic lengths and apply the cosine rule with cos(120°). This gives a quadratic equation in x. Solving gives x = 3 (reject any root that would make a side length non-positive).

sin BAC = 2 × area ÷ (AB × AC). Since angle BAC is obtuse, BAC = 180° − sin⁻¹(...). Then BC from the cosine rule and perimeter AB + AC + BC = 68.0 m (3 s.f.).

sin BAC = 2 × area ÷ (AB × AC). With BAC acute, BAC = sin⁻¹(...). Cosine rule gives BC = 13.5 cm (3 s.f.).

sin BAC = 2 × area ÷ (AB × AC). Obtuse angle: BAC = 180° − sin⁻¹(...) = 125.3° (1 d.p.).

Find BAC using the area formula, then BC using the cosine rule. Perimeter = 40.6 cm (3 s.f.).

(a) BC = 37.7 km (3 s.f.).
(b) Perimeter = 78.7 km (3 s.f.).

sin BAC = 2 × area ÷ (AB × AC). With BAC acute, use the cosine rule to get BC = 13.0 m. Perimeter = 36.0 m (3 s.f.).