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МатематикаВопрос:
Заполните в таблице как можно больше клеток, делая чертёж к каждой задаче. В треугольнике ABC угол A = 90°. 1) AB = 3, AC = 4, BC = 5; 2) AB = 1, sin C = 1/2; 3) BC = 4, cos C = 1/8; 4) AB = 10, cos B = 2/5; 5) AC = 4, tg B = 5; 6) AB = 6, AC = 8, BC = 10.
Ответ:
Так как \(\angle A=90^\circ\), сторона \(BC\) — гипотенуза. Для угла \(B\): \(\sin B=\frac{AC}{BC}\), \(\cos B=\frac{AB}{BC}\), \( g B=\frac{AC}{AB}\). Для угла \(C\): \(\sin C=\frac{AB}{BC}\), \(\cos C=\frac{AC}{BC}\), \( g C=\frac{AB}{AC}\).
| № | AB | AC | BC | sin B | sin C | cos B | cos C | tg B | tg C |
|---|---|---|---|---|---|---|---|---|---|
| 1 | 3 | 4 | 5 | \(\frac45\) | \(\frac35\) | \(\frac35\) | \(\frac45\) | \(\frac43\) | \(\frac34\) |
| 2 | 1 | \(\sqrt3\) | 2 | \(\frac{\sqrt3}{2}\) | \(\frac12\) | \(\frac12\) | \(\frac{\sqrt3}{2}\) | \(\sqrt3\) | \(\frac1{\sqrt3}=\frac{\sqrt3}{3}\) |
| 3 | \(\frac12\) | \(\frac{\sqrt{63}}2=\frac{3\sqrt7}2\) | 4 | \(\frac{\sqrt{63}}8=\frac{3\sqrt7}{8}\) | \(\frac18\) | \(\frac{\sqrt{63}}8=\frac{3\sqrt7}{8}\) | \(\frac18\) | \(\frac1{\sqrt{63}}=\frac{\sqrt7}{21}\) | \(\frac{\sqrt{63}}1=3\sqrt7\) |
| 4 | 10 | \(\frac{10\sqrt{21}}5=2\sqrt{21}\) | \(\frac{10\sqrt{21}}{\sqrt5}=2\sqrt{105}\) | \(\frac{\sqrt{21}}{\sqrt{105}}=\frac1{\sqrt5}=\frac{\sqrt5}5\) | \(\frac1{\sqrt5}=\frac{\sqrt5}5\) | \(\frac25\) | \(\frac{\sqrt{21}}{\sqrt{105}}=\frac1{\sqrt5}=\frac{\sqrt5}5\) | \(\frac{\sqrt{21}}5\) | \(\sqrt5\) |
| 5 | \(\frac45\) | 4 | \(\frac{4\sqrt{26}}5\) | \(\frac5{\sqrt{26}}=\frac{5\sqrt{26}}{26}\) | \(\frac1{\sqrt{26}}=\frac{\sqrt{26}}{26}\) | \(\frac1{\sqrt{26}}=\frac{\sqrt{26}}{26}\) | \(\frac5{\sqrt{26}}=\frac{5\sqrt{26}}{26}\) | 5 | \(\frac15\) |
| 6 | 6 | 8 | 10 | \(\frac45\) | \(\frac35\) | \(\frac35\) | \(\frac45\) | \(\frac43\) | \(\frac34\) |
В строке 4 использовано \(\sin B=\sqrt{1-\cos^2B}=\sqrt{1-\frac4{25}}=\frac{\sqrt{21}}5\). В строке 5 использовано \(AB=\frac{AC}{ g B}=\frac45\) и \(BC=\sqrt{AB^2+AC^2}=\frac{4\sqrt{26}}5\).
