В ∆ABC, &ang;С = 90°, ВC = 18, &ang;A = 60°. Найдите АВ, АС, &ang;В.. Created by 4ebyrekmdaoyfsfg. geometriya-ru

В ∆ABC, ∠С = 90°, ВC = 18, ∠A = 60°. Найдите АВ, АС, ∠В...

4ebyrekmdaoyfsfg @4ebyrekmdaoyfsfg

July 2022 1 5 Report

В ∆ABC, ∠С = 90°, ВC = 18, ∠A = 60°.

Найдите АВ, АС, ∠В.

Answers & Comments

Vopoxov

Ответ:

$AB = 12\sqrt{3} \\ AC = 6\sqrt{3} \\ \angle {B} = 30\degree$

Объяснение:

∠В = 180° - ∠С -∠A

∠В = 180° - 90° - 60° = 30°

АС и ВС - катеты; АВ - гипотенуза.

ВС = 18,

Следовательно:

$\sin (\angle A)= \frac{BC}{AB} \: = > AB = \frac{BC}{ \sin (\angle A)} \\ \angle A = 60 \degree = > \sin (\angle A) = \frac{ \sqrt{3} }{2} \\ AB = \frac{BC}{ \sin (\angle A)} = \frac{2 \times 18}{ \sqrt{3} } = \\ = \frac{36}{ \sqrt{3} } = \frac{12 \times 3}{ \sqrt{3} } \\ AB = 12 \sqrt{3} \\ \\ \small{AC^{2} + {BC}^{2} = {AB}^{2} = > AC^{2} = {AB}^{2} - {BC}^{2} } \\ AC = \sqrt{ {AB}^{2} - {BC}^{2}} \\ AC = \sqrt{(12 \sqrt{3})^{2} - 18^{2} } = \\ = \sqrt{432 -324 } = \sqrt{108} = \sqrt{36 \cdot3} \\ AC = 6 \sqrt{3}$

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