Ответ:
а) Обозначим BC = a, AC = b = 7,5 см, AB = c = 5 см.
По теореме косинусов:
\[a^2=b^2+c^2-2bc\cos A\]
\[a^2=7.5^2+5^2-2\cdot7.5\cdot5\cos135^\circ=56.25+25+37.5\sqrt2\]
\[BC=\sqrt{81.25+37.5\sqrt2}\approx11.59\text{ см}.\]
Далее:
\[\cos B=\frac{a^2+c^2-b^2}{2ac}\approx0.889,\qquad B\approx27.25^\circ.\]
\[C=180^\circ-A-B=180^\circ-135^\circ-27.25^\circ\approx17.75^\circ.\]
Ответ: BC ≈ 11,59 см, ∠B ≈ 27,25°, ∠C ≈ 17,75°.
б) Обозначим AC = b, AB = c = 2√2 дм, BC = a = 3 дм.
\[b^2=a^2+c^2-2ac\cos B\]
\[b^2=3^2+(2\sqrt2)^2-2\cdot3\cdot2\sqrt2\cdot\cos45^\circ=9+8-12=5.\]
\[AC=b=\sqrt5\text{ дм}.\]
\[\cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{5+8-9}{2\cdot\sqrt5\cdot2\sqrt2}=\frac1{\sqrt{10}},\]
\[A\approx71.57^\circ.\]
\[C=180^\circ-45^\circ-71.57^\circ\approx63.43^\circ.\]
Ответ: AC = √5 дм, ∠A ≈ 71,57°, ∠C ≈ 63,43°.
в) Обозначим AB = c, AC = b = 0,6 м, BC = a = √3/2 м.
\[c^2=a^2+b^2-2ab\cos C\]
\[c^2=\left(\frac{\sqrt3}{2}\right)^2+0.6^2-2\cdot\frac{\sqrt3}{2}\cdot0.6\cos150^\circ\]
\[c^2=\frac34+\frac9{25}+\frac9{10}=\frac{201}{100}.\]
\[AB=c=\frac{\sqrt{201}}{10}\text{ м}\approx1.42\text{ м}.\]
\[\sin A=\frac{a\sin C}{c}=\frac{5\sqrt3}{2\sqrt{201}},\qquad A\approx17.77^\circ.\]
\[B=180^\circ-150^\circ-17.77^\circ\approx12.23^\circ.\]
Ответ: AB = √201/10 м ≈ 1,42 м, ∠A ≈ 17,77°, ∠B ≈ 12,23°.
