Trang 99 —
Trang 99 — Chương 4: Hệ thức lượng trong tam giác. Vectơ
Bài tập 1. Nếu hai điểm M , N M, N M , N M N → ⋅ N M → = − 4 \overrightarrow{MN} \cdot \overrightarrow{NM} = -4 M N ⋅ N M = − 4 M N MN M N
Lời giải:
M N → ⋅ N M → = − 4 \overrightarrow{MN} \cdot \overrightarrow{NM} = -4 M N ⋅ N M = − 4
⇔ ∣ M N → ∣ . ∣ N M → ∣ . cos 180 ∘ = − 4 \Leftrightarrow \left| \overrightarrow{MN} \right| . \left| \overrightarrow{NM} \right| . \cos 180^\circ = -4 ⇔ M N . N M . cos 18 0 ∘ = − 4
⇔ ∣ M N → ∣ . ∣ N M → ∣ . ( − 1 ) = − 4 \Leftrightarrow \left| \overrightarrow{MN} \right| . \left| \overrightarrow{NM} \right| . (-1) = -4 ⇔ M N . N M . ( − 1 ) = − 4
⇔ ∣ M N → ∣ . ∣ N M → ∣ = 4 \Leftrightarrow \left| \overrightarrow{MN} \right| . \left| \overrightarrow{NM} \right| = 4 ⇔ M N . N M = 4
⇔ ∣ M N → ∣ 2 = 4 \Leftrightarrow \left| \overrightarrow{MN} \right| ^2 = 4 ⇔ M N 2 = 4
⇔ ∣ M N → ∣ = 2 \Leftrightarrow \left| \overrightarrow{MN} \right| = 2 ⇔ M N = 2
Vậy ∣ M N → ∣ = 2 \left| \overrightarrow{MN} \right| = 2 M N = 2
Kết quả: B. M N = 2 MN=2 M N = 2
Trang 100 —
Bài 2. Phát biểu nào sau đây là đúng?
A. Nếu a → \overrightarrow{a} a b → \overrightarrow{b} b 0 → \overrightarrow{0} 0 ( a → , b → ) < 90 ∘ (\overrightarrow{a}, \overrightarrow{b}) < 90^\circ ( a , b ) < 9 0 ∘ a → ⋅ b → < 0 \overrightarrow{a} \cdot \overrightarrow{b} < 0 a ⋅ b < 0
B. Nếu a → \overrightarrow{a} a b → \overrightarrow{b} b 0 → \overrightarrow{0} 0 ( a → , b → ) > 90 ∘ (\overrightarrow{a}, \overrightarrow{b}) > 90^\circ ( a , b ) > 9 0 ∘ a → ⋅ b → > 0 \overrightarrow{a} \cdot \overrightarrow{b} > 0 a ⋅ b > 0
C. Nếu a → \overrightarrow{a} a b → \overrightarrow{b} b 0 → \overrightarrow{0} 0 ( a → , b → ) < 90 ∘ (\overrightarrow{a}, \overrightarrow{b}) < 90^\circ ( a , b ) < 9 0 ∘ a → ⋅ b → > 0 \overrightarrow{a} \cdot \overrightarrow{b} > 0 a ⋅ b > 0
D. Nếu a → \overrightarrow{a} a b → \overrightarrow{b} b 0 → \overrightarrow{0} 0 ( a → , b → ) > 90 ∘ (\overrightarrow{a}, \overrightarrow{b}) > 90^\circ ( a , b ) > 9 0 ∘ a → ⋅ b → < 0 \overrightarrow{a} \cdot \overrightarrow{b} < 0 a ⋅ b < 0
Lời giải:
Ta có
$$
\begin{aligned}
&\overrightarrow{a} \cdot \overrightarrow{b} = \left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{b} \right| \cdot \cos(\overrightarrow{a}, \overrightarrow{b})\
&\quad \cdot \text{Nếu } (\overrightarrow{a}, \overrightarrow{b}) < 90^\circ \Rightarrow \cos(\overrightarrow{a}, \overrightarrow{b}) > 0 \Rightarrow \overrightarrow{a} \cdot \overrightarrow{b} > 0.\
&\quad \cdot \text{Nếu } (\overrightarrow{a}, \overrightarrow{b}) > 90^\circ \Rightarrow \cos(\overrightarrow{a}, \overrightarrow{b}) < 0 \Rightarrow \overrightarrow{a} \cdot \overrightarrow{b} < 0.\
\end{aligned}
$$
Kết quả: C
Bài 3. Tính ∣ a → ∣ \left| \overrightarrow{a} \right| a
a) ∣ a → ∣ = 3 \left| \overrightarrow{a} \right| = 3 a = 3 ∣ b → ∣ = 4 \left| \overrightarrow{b} \right| = 4 b = 4 ( a → , b → ) = 30 ∘ (\overrightarrow{a}, \overrightarrow{b}) = 30^\circ ( a , b ) = 3 0 ∘
b) ∣ a → ∣ = 5 \left| \overrightarrow{a} \right| = 5 a = 5 ∣ b → ∣ = 6 \left| \overrightarrow{b} \right| = 6 b = 6 ( a → , b → ) = 120 ∘ (\overrightarrow{a}, \overrightarrow{b}) = 120^\circ ( a , b ) = 12 0 ∘
c) ∣ a → ∣ = 2 \left| \overrightarrow{a} \right| = 2 a = 2 ∣ b → ∣ = 3 \left| \overrightarrow{b} \right| = 3 b = 3 a → \overrightarrow{a} a b → \overrightarrow{b} b
d) ∣ a → ∣ = 2 \left| \overrightarrow{a} \right| = 2 a = 2 ∣ b → ∣ = 3 \left| \overrightarrow{b} \right| = 3 b = 3 a → \overrightarrow{a} a b → \overrightarrow{b} b
Lời giải:
a)
$$
\begin{aligned}
&\overrightarrow{a} \cdot \overrightarrow{b} = \left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{b} \right| \cdot \cos(\overrightarrow{a}, \overrightarrow{b})\
&= 3 \cdot 4 \cdot \cos 30^\circ\
&= 12 \cdot \frac{\sqrt{3}}{2} = 6\sqrt{3}.
\end{aligned}
$$
b)
$$
\begin{aligned}
&\overrightarrow{a} \cdot \overrightarrow{b} = \left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{b} \right| \cdot \cos(\overrightarrow{a}, \overrightarrow{b})\
&= 5 \cdot 6 \cdot \cos 120^\circ\
&= 30 \cdot \left( -\frac{1}{2} \right) = -15.
\end{aligned}
$$
c)
$$
\begin{aligned}
&\overrightarrow{a} \cdot \overrightarrow{b} = \left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{b} \right| \cdot \cos(\overrightarrow{a}, \overrightarrow{b})\
&= 2 \cdot 3 \cdot \cos 0^\circ\
&= 6 \cdot 1 = 6.
\end{aligned}
$$
d)
$$
\begin{aligned}
&\overrightarrow{a} \cdot \overrightarrow{b} = \left| \overrightarrow{a} \right| \cdot \left| \overrightarrow{b} \right| \cdot \cos(\overrightarrow{a}, \overrightarrow{b})\
&= 2 \cdot 3 \cdot \cos 180^\circ\
&= 6 \cdot (-1) = -6.
\end{aligned}
$$
Bài 4. Cho hình vuông A B C D ABCD A B C D a a a
a) A B → ⋅ A C → \overrightarrow{AB} \cdot \overrightarrow{AC} A B ⋅ A C
b) A C → ⋅ A D → \overrightarrow{AC} \cdot \overrightarrow{AD} A C ⋅ A D
Lời giải:
Ta có
∣ A B → ∣ = a ; ∣ A C → ∣ = a 2 + a 2 = a 2 ; ( A B → , A C → ) = 45 ∘ . \left| \overrightarrow{AB} \right| = a; \left| \overrightarrow{AC} \right| = \sqrt{a^2 + a^2} = a\sqrt{2}; (\overrightarrow{AB}, \overrightarrow{AC}) = 45^\circ. A B = a ; A C = a 2 + a 2 = a 2 ; ( A B , A C ) = 4 5 ∘ .
a)
$$
\begin{aligned}
&\overrightarrow{AB} \cdot \overrightarrow{AC} = \left| \overrightarrow{AB} \right| \cdot \left| \overrightarrow{AC} \right| \cdot \cos(\overrightarrow{AB}, \overrightarrow{AC})\
&= a \cdot a\sqrt{2} \cdot \cos 45^\circ\
&= a^2 \sqrt{2} \cdot \frac{\sqrt{2}}{2} = a^2.
\end{aligned}
$$
b)
$$
\begin{aligned}
&\overrightarrow{AC} \cdot \overrightarrow{AD} = \left| \overrightarrow{AC} \right| \cdot \left| \overrightarrow{AD} \right| \cdot \cos(\overrightarrow{AC}, \overrightarrow{AD})\
&= a\sqrt{2} \cdot a \cdot \cos 90^\circ\
&= a^2 \sqrt{2} \cdot 0 = 0.
\end{aligned}
$$
Bài 5. Cho tam giác A B C ABC A B C
A B → + B C → + C A → = 0 → . \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{0}. A B + B C + C A = 0 . Lời giải:
Ta có
$$
\begin{aligned}
&\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA}\
&= \overrightarrow{AB} + \overrightarrow{BC} + (-\overrightarrow{AC})\
&= \overrightarrow{AB} + \overrightarrow{BC} - \overrightarrow{AC} \
&= \overrightarrow{AC} - \overrightarrow{AC} = \overrightarrow{0}.
\end{aligned}
$$
Bài 6. Cho tam giác nhọn A B C ABC A B C A H AH A H
a) A B → ⋅ A H → = A C → ⋅ A H → \overrightarrow{AB} \cdot \overrightarrow{AH} = \overrightarrow{AC} \cdot \overrightarrow{AH} A B ⋅ A H = A C ⋅ A H
b) A B → ⋅ B C → = H B → ⋅ B C → \overrightarrow{AB} \cdot \overrightarrow{BC} = \overrightarrow{HB} \cdot \overrightarrow{BC} A B ⋅ B C = H B ⋅ B C
Lời giải:
a)
$$
\begin{aligned}
&\overrightarrow{AB} \cdot \overrightarrow{AH} - \overrightarrow{AC} \cdot \overrightarrow{AH} = (\overrightarrow{AB} - \overrightarrow{AC}) \cdot \overrightarrow{AH}\
&= \overrightarrow{CB} \cdot \overrightarrow{AH} = 0 \quad \text{vì } \overrightarrow{CB} \perp \overrightarrow{AH}.
\end{aligned}
$$
b)
$$
\begin{aligned}
&\overrightarrow{AB} \cdot \overrightarrow{BC} - \overrightarrow{HB} \cdot \overrightarrow{BC} = (\overrightarrow{AB} - \overrightarrow{HB}) \cdot \overrightarrow{BC}\
&= \overrightarrow{AH} \cdot \overrightarrow{BC} = 0 \quad \text{vì } \overrightarrow{AH} \perp \overrightarrow{BC}.
\end{aligned}
$$
Bài 7. Một máy bay đang bay từ hướng đông sang hướng tây với tốc độ bay 700 700 700 40 40 40
Lời giải:
Gọi v 1 → \overrightarrow{v_1} v 1
v 2 → \overrightarrow{v_2} v 2
Khi đó, ∣ v 1 → ∣ = 700 \left| \overrightarrow{v_1} \right| = 700 v 1 = 700 ∣ v 2 → ∣ = 40 \left| \overrightarrow{v_2} \right| = 40 v 2 = 40
Tốc độ của máy bay khi có gió là:
∣ v → ∣ = ∣ v 1 → − v 2 → ∣ = ∣ v 1 → ∣ 2 + ∣ v 2 → ∣ 2 = 700 2 + 40 2 ≈ 700.06 km/h . \begin{aligned}
&\left| \overrightarrow{v} \right| = \left| \overrightarrow{v_1} - \overrightarrow{v_2} \right|\\
&= \sqrt{\left| \overrightarrow{v_1} \right|^2 + \left| \overrightarrow{v_2} \right|^2}\\
&= \sqrt{700^2 + 40^2}\\
& \approx 700.06 \text{ km/h}.
\end{aligned} v = v 1 − v 2 = v 1 2 + v 2 2 = 70 0 2 + 4 0 2 ≈ 700.06 km/h . Bài 8. Cho tam giác A B C ABC A B C A B = 2 AB = 2 A B = 2 A C = 3 AC = 3 A C = 3 R A C = 60 ∘ RAC = 60^\circ R A C = 6 0 ∘ M M M B C BC B C D D D A D → = 7 12 A C → \overrightarrow{AD} = \frac{7}{12} \overrightarrow{AC} A D = 12 7 A C
a) Tính A B → ⋅ A C → \overrightarrow{AB} \cdot \overrightarrow{AC} A B ⋅ A C
b) Biểu diễn A M → \overrightarrow{AM} A M B D → \overrightarrow{BD} B D A B → \overrightarrow{AB} A B A C → \overrightarrow{AC} A C
c) Chứng minh A M ⊥ B D AM \perp BD A M ⊥ B D
Lời giải:
a)
$$
\begin{aligned}
&\overrightarrow{AB} \cdot \overrightarrow{AC} = \left| \overrightarrow{AB} \right| \cdot \left| \overrightarrow{AC} \right| \cdot \cos(\overrightarrow{AB}, \overrightarrow{AC})\
&= 2 \cdot 3 \cdot \cos 60^\circ\
&= 6 \cdot \frac{1}{2} = 3.
\end{aligned}
$$
b)
A M → = 1 2 ( A B → + A C → ) . \overrightarrow{AM} = \frac{1}{2} (\overrightarrow{AB} + \overrightarrow{AC}). A M = 2 1 ( A B + A C ) .
B D → = A D → − A B → = 7 12 A C → − A B → . \overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB} = \frac{7}{12} \overrightarrow{AC} - \overrightarrow{AB}. B D = A D − A B = 12 7 A C − A B .
c)
$$
\begin{aligned}
&\overrightarrow{AM} \cdot \overrightarrow{BD} = \frac{1}{2} (\overrightarrow{AB} + \overrightarrow{AC}) \cdot (\frac{7}{12} \overrightarrow{AC} - \overrightarrow{AB})\
&= \frac{1}{2} \left[ \frac{7}{12} (\overrightarrow{AB} \cdot \overrightarrow{AC}) - \left| \overrightarrow{AB} \right|^2 + \frac{7}{12} \left| \overrightarrow{AC} \right|^2 - (\overrightarrow{AB} \cdot \overrightarrow{AC}) \right]\
&= \frac{1}{2} \left[ \frac{7}{12} \cdot 3 - 4 + \frac{7}{12} \cdot 9 - 3 \right]\
&= 0.
\end{aligned}
$$
Trang 101 — Bài tập cuối chương IV
Bài 1. Cho tam giác A B C ABC A B C A B = 3 AB = 3 A B = 3 A C = 4 AC = 4 A C = 4 B A C ^ = 120 ∘ \widehat{BAC} = 120^\circ B A C = 12 0 ∘
a) Độ dài cạnh B C BC B C B B B
b) Bán kính đường tròn ngoại tiếp;
c) Diện tích của tam giác;
d) A B → ⋅ A C → \overrightarrow{AB} \cdot \overrightarrow{AC} A B ⋅ A C A M → ⋅ B C → \overrightarrow{AM} \cdot \overrightarrow{BC} A M ⋅ B C M M M B C BC B C
Lời giải:
a) Áp dụng định lý Cosin vào tam giác A B C ABC A B C
B C 2 = A B 2 + A C 2 − 2 A B ⋅ A C ⋅ cos B A C ^ = 3 2 + 4 2 − 2 ⋅ 3 ⋅ 4 ⋅ cos 120 ∘ = 9 + 16 − 24 ⋅ ( − 1 2 ) = 9 + 16 + 12 = 37 BC^2 = AB^2 + AC^2 - 2AB \cdot AC \cdot \cos \widehat{BAC} = 3^2 + 4^2 - 2 \cdot 3 \cdot 4 \cdot \cos 120^\circ = 9 + 16 - 24 \cdot \left( -\frac{1}{2} \right) = 9 + 16 + 12 = 37 B C 2 = A B 2 + A C 2 − 2 A B ⋅ A C ⋅ cos B A C = 3 2 + 4 2 − 2 ⋅ 3 ⋅ 4 ⋅ cos 12 0 ∘ = 9 + 16 − 24 ⋅ ( − 2 1 ) = 9 + 16 + 12 = 37
B C = 37 ≈ 6 BC = \sqrt{37} \approx 6 B C = 37 ≈ 6
Áp dụng định lý Sin vào tam giác A B C ABC A B C
B C sin B A C ^ = A B sin C ⇒ sin C = A B ⋅ sin B A C ^ B C = 3 ⋅ sin 120 ∘ 37 ≈ 3 ⋅ 3 2 6 ≈ 0 , 433 \frac{BC}{\sin \widehat{BAC}} = \frac{AB}{\sin C} \Rightarrow \sin C = \frac{AB \cdot \sin \widehat{BAC}}{BC} = \frac{3 \cdot \sin 120^\circ}{\sqrt{37}} \approx \frac{3 \cdot \frac{\sqrt{3}}{2}}{6} \approx 0,433 sin B A C B C = sin C A B ⇒ sin C = B C A B ⋅ sin B A C = 37 3 ⋅ sin 12 0 ∘ ≈ 6 3 ⋅ 2 3 ≈ 0 , 433
⇒ C ^ ≈ arcsin 0 , 433 ≈ 26 ∘ \Rightarrow \widehat{C} \approx \arcsin 0,433 \approx 26^\circ ⇒ C ≈ arcsin 0 , 433 ≈ 2 6 ∘
B ^ = 180 ∘ − A ^ − C ^ ≈ 180 ∘ − 120 ∘ − 26 ∘ = 34 ∘ \widehat{B} = 180^\circ - \widehat{A} - \widehat{C} \approx 180^\circ - 120^\circ - 26^\circ = 34^\circ B = 18 0 ∘ − A − C ≈ 18 0 ∘ − 12 0 ∘ − 2 6 ∘ = 3 4 ∘
Kết quả: B C ≈ 6 BC \approx 6 B C ≈ 6 B ^ ≈ 34 ∘ \widehat{B} \approx 34^\circ B ≈ 3 4 ∘
b) Bán kính đường tròn ngoại tiếp R = B C 2 sin B A C ^ = 37 2 sin 120 ∘ = 37 2 ⋅ 3 2 = 37 3 ≈ 6 1 , 73 ≈ 3 R = \frac{BC}{2 \sin \widehat{BAC}} = \frac{\sqrt{37}}{2 \sin 120^\circ} = \frac{\sqrt{37}}{2 \cdot \frac{\sqrt{3}}{2}} = \frac{\sqrt{37}}{\sqrt{3}} \approx \frac{6}{1,73} \approx 3 R = 2 s i n B A C B C = 2 s i n 12 0 ∘ 37 = 2 ⋅ 2 3 37 = 3 37 ≈ 1 , 73 6 ≈ 3
Kết quả: R ≈ 3 R \approx 3 R ≈ 3
c) Diện tích của tam giác S = 1 2 A B ⋅ A C ⋅ sin B A C ^ = 1 2 ⋅ 3 ⋅ 4 ⋅ sin 120 ∘ = 6 ⋅ 3 2 = 3 3 ≈ 5 S = \frac{1}{2} AB \cdot AC \cdot \sin \widehat{BAC} = \frac{1}{2} \cdot 3 \cdot 4 \cdot \sin 120^\circ = 6 \cdot \frac{\sqrt{3}}{2} = 3\sqrt{3} \approx 5 S = 2 1 A B ⋅ A C ⋅ sin B A C = 2 1 ⋅ 3 ⋅ 4 ⋅ sin 12 0 ∘ = 6 ⋅ 2 3 = 3 3 ≈ 5
Kết quả: S ≈ 5 S \approx 5 S ≈ 5
d) A B → ⋅ A C → = A B ⋅ A C ⋅ cos B A C ^ = 3 ⋅ 4 ⋅ cos 120 ∘ = 12 ⋅ ( − 1 2 ) = − 6 \overrightarrow{AB} \cdot \overrightarrow{AC} = AB \cdot AC \cdot \cos \widehat{BAC} = 3 \cdot 4 \cdot \cos 120^\circ = 12 \cdot \left( -\frac{1}{2} \right) = -6 A B ⋅ A C = A B ⋅ A C ⋅ cos B A C = 3 ⋅ 4 ⋅ cos 12 0 ∘ = 12 ⋅ ( − 2 1 ) = − 6
M M M B C BC B C A M → = 1 2 ( A B → + A C → ) \overrightarrow{AM} = \frac{1}{2} \left( \overrightarrow{AB} + \overrightarrow{AC} \right) A M = 2 1 ( A B + A C )
A M → ⋅ B C → = 1 2 ( A B → + A C → ) ⋅ ( A C → − A B → ) = 1 2 ( A C → 2 − A B → 2 ) = 1 2 ( A C 2 − A B 2 ) = 1 2 ( 4 2 − 3 2 ) = 7 2 \overrightarrow{AM} \cdot \overrightarrow{BC} = \frac{1}{2} \left( \overrightarrow{AB} + \overrightarrow{AC} \right) \cdot \left( \overrightarrow{AC} - \overrightarrow{AB} \right) = \frac{1}{2} \left( \overrightarrow{AC}^2 - \overrightarrow{AB}^2 \right) = \frac{1}{2} \left( AC^2 - AB^2 \right) = \frac{1}{2} \left( 4^2 - 3^2 \right) = \frac{7}{2} A M ⋅ B C = 2 1 ( A B + A C ) ⋅ ( A C − A B ) = 2 1 ( A C 2 − A B 2 ) = 2 1 ( A C 2 − A B 2 ) = 2 1 ( 4 2 − 3 2 ) = 2 7
Kết quả: A B → ⋅ A C → = − 6 \overrightarrow{AB} \cdot \overrightarrow{AC} = -6 A B ⋅ A C = − 6 A M → ⋅ B C → = 7 2 \overrightarrow{AM} \cdot \overrightarrow{BC} = \frac{7}{2} A M ⋅ B C = 2 7
Bài 2. Không dùng máy tính cầm tay, hãy tính giá trị của các biểu thức sau:
A = ( sin 20 ∘ + sin 70 ∘ ) + ( cos 20 ∘ + cos 110 ∘ ) A = \left( \sin 20^\circ + \sin 70^\circ \right) + \left( \cos 20^\circ + \cos 110^\circ \right) A = ( sin 2 0 ∘ + sin 7 0 ∘ ) + ( cos 2 0 ∘ + cos 11 0 ∘ )
B = tan 20 ∘ + tan 40 ∘ + tan 20 ∘ tan 40 ∘ B = \tan 20^\circ + \tan 40^\circ + \tan 20^\circ \tan 40^\circ B = tan 2 0 ∘ + tan 4 0 ∘ + tan 2 0 ∘ tan 4 0 ∘
Lời giải:
A = ( sin 20 ∘ + sin 70 ∘ ) + ( cos 20 ∘ + cos 110 ∘ ) A = \left( \sin 20^\circ + \sin 70^\circ \right) + \left( \cos 20^\circ + \cos 110^\circ \right) A = ( sin 2 0 ∘ + sin 7 0 ∘ ) + ( cos 2 0 ∘ + cos 11 0 ∘ )
= 2 sin 20 ∘ + 70 ∘ 2 cos 20 ∘ − 70 ∘ 2 + 2 cos 20 ∘ + 110 ∘ 2 cos 20 ∘ − 110 ∘ 2 = 2 \sin \frac{20^\circ + 70^\circ}{2} \cos \frac{20^\circ - 70^\circ}{2} + 2 \cos \frac{20^\circ + 110^\circ}{2} \cos \frac{20^\circ - 110^\circ}{2} = 2 sin 2 2 0 ∘ + 7 0 ∘ cos 2 2 0 ∘ − 7 0 ∘ + 2 cos 2 2 0 ∘ + 11 0 ∘ cos 2 2 0 ∘ − 11 0 ∘
= 2 sin 45 ∘ cos ( − 25 ∘ ) + 2 cos 65 ∘ cos ( − 45 ∘ ) = 2 \sin 45^\circ \cos \left( -25^\circ \right) + 2 \cos 65^\circ \cos \left( -45^\circ \right) = 2 sin 4 5 ∘ cos ( − 2 5 ∘ ) + 2 cos 6 5 ∘ cos ( − 4 5 ∘ )
= 2 ⋅ 2 2 ⋅ cos 25 ∘ + 2 cos 65 ∘ ⋅ 2 2 = 2 \cdot \frac{\sqrt{2}}{2} \cdot \cos 25^\circ + 2 \cos 65^\circ \cdot \frac{\sqrt{2}}{2} = 2 ⋅ 2 2 ⋅ cos 2 5 ∘ + 2 cos 6 5 ∘ ⋅ 2 2
= 2 cos 25 ∘ + 2 cos 65 ∘ = \sqrt{2} \cos 25^\circ + \sqrt{2} \cos 65^\circ = 2 cos 2 5 ∘ + 2 cos 6 5 ∘
= 2 ( cos 25 ∘ + cos 65 ∘ ) = \sqrt{2} \left( \cos 25^\circ + \cos 65^\circ \right) = 2 ( cos 2 5 ∘ + cos 6 5 ∘ )
= 2 ⋅ 2 cos 25 ∘ + 65 ∘ 2 cos 25 ∘ − 65 ∘ 2 = \sqrt{2} \cdot 2 \cos \frac{25^\circ + 65^\circ}{2} \cos \frac{25^\circ - 65^\circ}{2} = 2 ⋅ 2 cos 2 2 5 ∘ + 6 5 ∘ cos 2 2 5 ∘ − 6 5 ∘
= 2 2 cos 45 ∘ cos ( − 20 ∘ ) = 2 2 ⋅ 2 2 cos 20 ∘ = 2 cos 20 ∘ = 2 \sqrt{2} \cos 45^\circ \cos \left( -20^\circ \right) = 2 \sqrt{2} \cdot \frac{\sqrt{2}}{2} \cos 20^\circ = 2 \cos 20^\circ = 2 2 cos 4 5 ∘ cos ( − 2 0 ∘ ) = 2 2 ⋅ 2 2 cos 2 0 ∘ = 2 cos 2 0 ∘
B = tan 20 ∘ + tan 40 ∘ + tan 20 ∘ tan 40 ∘ B = \tan 20^\circ + \tan 40^\circ + \tan 20^\circ \tan 40^\circ B = tan 2 0 ∘ + tan 4 0 ∘ + tan 2 0 ∘ tan 4 0 ∘
= sin 20 ∘ cos 20 ∘ + sin 40 ∘ cos 40 ∘ + sin 20 ∘ sin 40 ∘ cos 20 ∘ cos 40 ∘ = \frac{\sin 20^\circ}{\cos 20^\circ} + \frac{\sin 40^\circ}{\cos 40^\circ} + \frac{\sin 20^\circ \sin 40^\circ}{\cos 20^\circ \cos 40^\circ} = cos 2 0 ∘ sin 2 0 ∘ + cos 4 0 ∘ sin 4 0 ∘ + cos 2 0 ∘ cos 4 0 ∘ sin 2 0 ∘ sin 4 0 ∘
= sin 20 ∘ cos 40 ∘ + sin 40 ∘ cos 20 ∘ + sin 20 ∘ sin 40 ∘ cos 20 ∘ cos 40 ∘ = \frac{\sin 20^\circ \cos 40^\circ + \sin 40^\circ \cos 20^\circ + \sin 20^\circ \sin 40^\circ}{\cos 20^\circ \cos 40^\circ} = cos 2 0 ∘ cos 4 0 ∘ sin 2 0 ∘ cos 4 0 ∘ + sin 4 0 ∘ cos 2 0 ∘ + sin 2 0 ∘ sin 4 0 ∘
= sin ( 20 ∘ + 40 ∘ ) + sin 20 ∘ sin 40 ∘ cos 20 ∘ cos 40 ∘ = \frac{\sin \left( 20^\circ + 40^\circ \right) + \sin 20^\circ \sin 40^\circ}{\cos 20^\circ \cos 40^\circ} = cos 2 0 ∘ cos 4 0 ∘ sin ( 2 0 ∘ + 4 0 ∘ ) + sin 2 0 ∘ sin 4 0 ∘
= sin 60 ∘ + sin 20 ∘ sin 40 ∘ cos 20 ∘ cos 40 ∘ = \frac{\sin 60^\circ + \sin 20^\circ \sin 40^\circ}{\cos 20^\circ \cos 40^\circ} = cos 2 0 ∘ cos 4 0 ∘ sin 6 0 ∘ + sin 2 0 ∘ sin 4 0 ∘
= 3 2 + sin 20 ∘ sin 40 ∘ cos 20 ∘ cos 40 ∘ = \frac{\frac{\sqrt{3}}{2} + \sin 20^\circ \sin 40^\circ}{\cos 20^\circ \cos 40^\circ} = cos 2 0 ∘ cos 4 0 ∘ 2 3 + sin 2 0 ∘ sin 4 0 ∘
Sử dụng công thức sin α sin β = 1 2 [ cos ( α − β ) − cos ( α + β ) ] \sin \alpha \sin \beta = \frac{1}{2} \left[ \cos \left( \alpha - \beta \right) - \cos \left( \alpha + \beta \right) \right] sin α sin β = 2 1 [ cos ( α − β ) − cos ( α + β ) ]
= 3 2 + 1 2 [ cos 20 ∘ − cos 60 ∘ ] cos 20 ∘ cos 40 ∘ = \frac{\frac{\sqrt{3}}{2} + \frac{1}{2} \left[ \cos 20^\circ - \cos 60^\circ \right]}{\cos 20^\circ \cos 40^\circ} = cos 2 0 ∘ cos 4 0 ∘ 2 3 + 2 1 [ cos 2 0 ∘ − cos 6 0 ∘ ]
= 3 2 + 1 2 cos 20 ∘ − 1 4 cos 20 ∘ cos 40 ∘ = \frac{\frac{\sqrt{3}}{2} + \frac{1}{2} \cos 20^\circ - \frac{1}{4}}{\cos 20^\circ \cos 40^\circ} = cos 2 0 ∘ cos 4 0 ∘ 2 3 + 2 1 cos 2 0 ∘ − 4 1
= 2 3 + 2 cos 20 ∘ − 1 2 cos 20 ∘ cos 40 ∘ = \frac{\frac{2 \sqrt{3} + 2 \cos 20^\circ - 1}{2}}{\cos 20^\circ \cos 40^\circ} = cos 2 0 ∘ cos 4 0 ∘ 2 2 3 + 2 c o s 2 0 ∘ − 1
= 2 3 + 2 cos 20 ∘ − 1 2 cos 20 ∘ cos 40 ∘ = \frac{2 \sqrt{3} + 2 \cos 20^\circ - 1}{2 \cos 20^\circ \cos 40^\circ} = 2 cos 2 0 ∘ cos 4 0 ∘ 2 3 + 2 cos 2 0 ∘ − 1
Sử dụng công thức cos α cos β = 1 2 [ cos ( α + β ) + cos ( α − β ) ] \cos \alpha \cos \beta = \frac{1}{2} \left[ \cos \left( \alpha + \beta \right) + \cos \left( \alpha - \beta \right) \right] cos α cos β = 2 1 [ cos ( α + β ) + cos ( α − β ) ]
= 2 3 + 2 cos 20 ∘ − 1 cos 60 ∘ + cos 20 ∘ = \frac{2 \sqrt{3} + 2 \cos 20^\circ - 1}{\cos 60^\circ + \cos 20^\circ} = cos 6 0 ∘ + cos 2 0 ∘ 2 3 + 2 cos 2 0 ∘ − 1
= 2 3 + 2 cos 20 ∘ − 1 1 2 + cos 20 ∘ = \frac{2 \sqrt{3} + 2 \cos 20^\circ - 1}{\frac{1}{2} + \cos 20^\circ} = 2 1 + cos 2 0 ∘ 2 3 + 2 cos 2 0 ∘ − 1
= 4 3 + 4 cos 20 ∘ − 2 1 + 2 cos 20 ∘ = \frac{4 \sqrt{3} + 4 \cos 20^\circ - 2}{1 + 2 \cos 20^\circ} = 1 + 2 cos 2 0 ∘ 4 3 + 4 cos 2 0 ∘ − 2
= 4 ( 3 + cos 20 ∘ ) − 2 2 cos 20 ∘ + 1 = \frac{4 \left( \sqrt{3} + \cos 20^\circ \right) - 2}{2 \cos 20^\circ + 1} = 2 cos 2 0 ∘ + 1 4 ( 3 + cos 2 0 ∘ ) − 2
Kết quả: A = 2 cos 20 ∘ A = 2 \cos 20^\circ A = 2 cos 2 0 ∘ B = 3 B = \sqrt{3} B = 3
Bài 3. Không dùng thước góc, làm thế nào để biết được góc của một nhọn hay tù? Bạn Hoài giải như sau:
Chọn hai điểm A A A B B B O x Ox O x O y Oy O y O A = O B = 2 OA = OB = 2 O A = O B = 2
Dùng dây đo A B AB A B A B = 3 , 1 AB = 3,1 A B = 3 , 1 x O y xOy x O y
Bạn làm như vậy có đúng không? Nếu không đúng, hãy trình bày cách làm đúng.
Lời giải:
Cách làm của bạn Hoài không đúng.
Để tính góc x O y xOy x O y
Chọn hai điểm A A A B B B O x Ox O x O y Oy O y O A = O B = 2 OA = OB = 2 O A = O B = 2
Dùng dây đo A B AB A B A B = 3 , 1 AB = 3,1 A B = 3 , 1
Áp dụng định lý Cosin vào tam giác O A B OAB O A B
A B 2 = O A 2 + O B 2 − 2 O A ⋅ O B ⋅ cos x O y ^ AB^2 = OA^2 + OB^2 - 2OA \cdot OB \cdot \cos \widehat{xOy} A B 2 = O A 2 + O B 2 − 2 O A ⋅ O B ⋅ cos x O y
3 , 1 2 = 2 2 + 2 2 − 2 ⋅ 2 ⋅ 2 ⋅ cos x O y ^ 3,1^2 = 2^2 + 2^2 - 2 \cdot 2 \cdot 2 \cdot \cos \widehat{xOy} 3 , 1 2 = 2 2 + 2 2 − 2 ⋅ 2 ⋅ 2 ⋅ cos x O y
cos x O y ^ = 2 2 + 2 2 − 3 , 1 2 2 ⋅ 2 ⋅ 2 = 4 + 4 − 9 , 61 8 = − 1 , 61 8 = − 0 , 20125 \cos \widehat{xOy} = \frac{2^2 + 2^2 - 3,1^2}{2 \cdot 2 \cdot 2} = \frac{4 + 4 - 9,61}{8} = \frac{-1,61}{8} = -0,20125 cos x O y = 2 ⋅ 2 ⋅ 2 2 2 + 2 2 − 3 , 1 2 = 8 4 + 4 − 9 , 61 = 8 − 1 , 61 = − 0 , 20125
x O y ^ = arccos ( − 0 , 20125 ) ≈ 101 , 5 ∘ > 90 ∘ \widehat{xOy} = \arccos \left( -0,20125 \right) \approx 101,5^\circ > 90^\circ x O y = arccos ( − 0 , 20125 ) ≈ 101 , 5 ∘ > 9 0 ∘
Vậy góc x O y xOy x O y
Bài 4. Có hai trạm quan sát A A A B B B C C C A B AB A B
Khoảng cách giữa hai trạm A A A B B B
Lời giải:
A C B ^ = 180 ∘ − B A C ^ − A B C ^ = 180 ∘ − 60 ∘ − 45 ∘ = 75 ∘ \widehat{ACB} = 180^\circ - \widehat{BAC} - \widehat{ABC} = 180^\circ - 60^\circ - 45^\circ = 75^\circ A C B = 18 0 ∘ − B A C − A B C = 18 0 ∘ − 6 0 ∘ − 4 5 ∘ = 7 5 ∘
Áp dụng định lý Sin vào tam giác A B C ABC A B C
A B sin A C B ^ = A C sin A B C ^ \frac{AB}{\sin \widehat{ACB}} = \frac{AC}{\sin \widehat{ABC}} sin A C B A B = sin A B C A C
A B = A C ⋅ sin A C B ^ sin A B C ^ = 1200 ⋅ sin 75 ∘ sin 45 ∘ ≈ 1200 ⋅ 0 , 9659 0 , 7071 ≈ 1639 , 9 ≈ 1640 AB = \frac{AC \cdot \sin \widehat{ACB}}{\sin \widehat{ABC}} = \frac{1200 \cdot \sin 75^\circ}{\sin 45^\circ} \approx \frac{1200 \cdot 0,9659}{0,7071} \approx 1639,9 \approx 1640 A B = sin A B C A C ⋅ sin A C B = sin 4 5 ∘ 1200 ⋅ sin 7 5 ∘ ≈ 0 , 7071 1200 ⋅ 0 , 9659 ≈ 1639 , 9 ≈ 1640
Kết quả: A B ≈ 1640 AB \approx 1640 A B ≈ 1640
Bài 5. Một người đứng ở vị trí A A A 35 ∘ 35^\circ 3 5 ∘ C D = 50 CD = 50 C D = 50
Từ vị trí A A A B B B a = 75 ∘ a = 75^\circ a = 7 5 ∘ A B = 70 AB = 70 A B = 70
Lời giải:
Đặt h = C D = C E = 50 h = CD = CE = 50 h = C D = C E = 50
Trong tam giác A C E ACE A C E
tan C A E ^ = C E A E ⇒ A E = C E tan 35 ∘ = 50 tan 35 ∘ ≈ 50 0 , 70 ≈ 71 , 4 \tan \widehat{CAE} = \frac{CE}{AE} \Rightarrow AE = \frac{CE}{\tan 35^\circ} = \frac{50}{\tan 35^\circ} \approx \frac{50}{0,70} \approx 71,4 tan C A E = A E C E ⇒ A E = tan 3 5 ∘ C E = tan 3 5 ∘ 50 ≈ 0 , 70 50 ≈ 71 , 4
Trong tam giác B C E BCE B C E
tan C B E ^ = C E B E ⇒ B E = C E tan 75 ∘ = 50 tan 75 ∘ ≈ 50 3 , 73 ≈ 13 , 4 \tan \widehat{CBE} = \frac{CE}{BE} \Rightarrow BE = \frac{CE}{\tan 75^\circ} = \frac{50}{\tan 75^\circ} \approx \frac{50}{3,73} \approx 13,4 tan C B E = B E C E ⇒ B E = tan 7 5 ∘ C E = tan 7 5 ∘ 50 ≈ 3 , 73 50 ≈ 13 , 4
A B = A E − B E ≈ 71 , 4 − 13 , 4 = 58 AB = AE - BE \approx 71,4 - 13,4 = 58 A B = A E − B E ≈ 71 , 4 − 13 , 4 = 58
Ta lại có:
tan 35 ∘ = h A C ⇒ A C = h tan 35 ∘ \tan 35^\circ = \frac{h}{AC} \Rightarrow AC = \frac{h}{\tan 35^\circ} tan 3 5 ∘ = A C h ⇒ A C = tan 3 5 ∘ h
tan 75 ∘ = h B C ⇒ B C = h tan 75 ∘ \tan 75^\circ = \frac{h}{BC} \Rightarrow BC = \frac{h}{\tan 75^\circ} tan 7 5 ∘ = B C h ⇒ B C = tan 7 5 ∘ h
A B = A C − B C = h ( 1 tan 35 ∘ − 1 tan 75 ∘ ) = 50 ( 1 0 , 7 − 1 3 , 73 ) ≈ 58 AB = AC - BC = h \left( \frac{1}{\tan 35^\circ} - \frac{1}{\tan 75^\circ} \right) = 50 \left( \frac{1}{0,7} - \frac{1}{3,73} \right) \approx 58 A B = A C − B C = h ( tan 3 5 ∘ 1 − tan 7 5 ∘ 1 ) = 50 ( 0 , 7 1 − 3 , 73 1 ) ≈ 58
Kết quả: A B ≈ 58 AB \approx 58 A B ≈ 58
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Bài 6. Để đo khoảng cách giữa hai điểm A A A N N N O O O A N AN A N O M OM O M O N ON O N M O N MON M O N O M = 200 m OM = 200\,\text{m} O M = 200 m O N = 500 m ON = 500\,\text{m} O N = 500 m M O N ^ = 135 ∘ \widehat{MON} = 135^{\circ} M O N = 13 5 ∘ A A A N N N
Lời giải:
Trước hết, ta cần xác định vị trí điểm A A A B B B O A ⊥ O M OA \perp OM O A ⊥ O M O B ⊥ O N OB \perp ON O B ⊥ O N
Với các dữ kiện đã cho:
O M = 200 m OM = 200\,\text{m} O M = 200 m O N = 500 m ON = 500\,\text{m} O N = 500 m M O N ^ = 135 ∘ \widehat{MON} = 135^{\circ} M O N = 13 5 ∘
Ta có thể tính khoảng cách A N AN A N
Sử dụng công thức tính độ dài vectơ A N → \overrightarrow{AN} A N
∣ A N → ∣ = ∣ O M → ∣ 2 + ∣ O N → ∣ 2 − 2 ∣ O M → ∣ ⋅ ∣ O N → ∣ ⋅ cos ( M O N ^ ) = 200 2 + 500 2 − 2 ⋅ 200 ⋅ 500 ⋅ cos 135 ∘ = 200 2 + 500 2 − 2 ⋅ 200 ⋅ 500 ⋅ ( − 2 2 ) = 40 000 + 250 000 + 100 000 2 ≈ 40 000 + 250 000 + 141 421.36 ≈ 431 421.36 ≈ 656.85 m . \begin{aligned}
|\overrightarrow{AN}| &= \sqrt{|\overrightarrow{OM}|^2 + |\overrightarrow{ON}|^2 - 2|\overrightarrow{OM}|\cdot |\overrightarrow{ON}|\cdot \cos(\widehat{MON})} \\
&= \sqrt{200^2 + 500^2 - 2 \cdot 200 \cdot 500 \cdot \cos 135^{\circ}}\\
&= \sqrt{200^2 + 500^2 - 2 \cdot 200 \cdot 500 \cdot \left(-\frac{\sqrt{2}}{2}\right)}\\
&= \sqrt{40\,000 + 250\,000 + 100\,000 \sqrt{2}}\\
&\approx \sqrt{40\,000 + 250\,000 + 141\,421.36}\\
&\approx \sqrt{431\,421.36}\\
&\approx 656.85\,\text{m}.
\end{aligned} ∣ A N ∣ = ∣ O M ∣ 2 + ∣ O N ∣ 2 − 2∣ O M ∣ ⋅ ∣ O N ∣ ⋅ cos ( M O N ) = 20 0 2 + 50 0 2 − 2 ⋅ 200 ⋅ 500 ⋅ cos 13 5 ∘ = 20 0 2 + 50 0 2 − 2 ⋅ 200 ⋅ 500 ⋅ ( − 2 2 ) = 40 000 + 250 000 + 100 000 2 ≈ 40 000 + 250 000 + 141 421.36 ≈ 431 421.36 ≈ 656.85 m .
Kết quả: 657 m 657\,\text{m} 657 m
Bài 7. Chứng minh:
a) Nếu A B → + A D → + C D → = A E → − D E → \overrightarrow{AB} + \overrightarrow{AD} + \overrightarrow{CD} = \overrightarrow{AE} - \overrightarrow{DE} A B + A D + C D = A E − D E A B → = C E → \overrightarrow{AB} = \overrightarrow{CE} A B = C E
b) Nếu M A → + M B → + 2 M N → = 0 → \overrightarrow{MA} + \overrightarrow{MB} + 2\overrightarrow{MN} = \overrightarrow{0} M A + M B + 2 M N = 0 M M M N N N
c) Nếu G G G A B C ABC A B C M A → + M B → + M C → − 3 M N → = 3 N G → \overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC} - 3\overrightarrow{MN} = 3\overrightarrow{NG} M A + M B + M C − 3 M N = 3 N G M M M N N N
Lời giải:
a) Ta có:
A B → + A D → + C D → = A E → − D E → ⇔ A B → + A D → + C D → + D E → = A E → ⇔ A B → + A E → = A E → ⇔ A B → = C E → . \begin{aligned}
\overrightarrow{AB} + \overrightarrow{AD} + \overrightarrow{CD} &= \overrightarrow{AE} - \overrightarrow{DE}\\
\Leftrightarrow \overrightarrow{AB} + \overrightarrow{AD} + \overrightarrow{CD} + \overrightarrow{DE} &= \overrightarrow{AE}\\
\Leftrightarrow \overrightarrow{AB} + \overrightarrow{AE} &= \overrightarrow{AE}\\
\Leftrightarrow \overrightarrow{AB} &= \overrightarrow{CE}.
\end{aligned} A B + A D + C D ⇔ A B + A D + C D + D E ⇔ A B + A E ⇔ A B = A E − D E = A E = A E = C E . b) Ta có:
M A → + M B → + 2 M N → = 0 → ⇔ M A → + M B → = − 2 M N → ⇔ M A → + M B → = 2 N M → . \begin{aligned}
\overrightarrow{MA} + \overrightarrow{MB} + 2\overrightarrow{MN} &= \overrightarrow{0}\\
\Leftrightarrow \overrightarrow{MA} + \overrightarrow{MB} &= -2\overrightarrow{MN}\\
\Leftrightarrow \overrightarrow{MA} + \overrightarrow{MB} &= 2\overrightarrow{NM}.
\end{aligned} M A + M B + 2 M N ⇔ M A + M B ⇔ M A + M B = 0 = − 2 M N = 2 N M . Giả sử M A → + M B → = 2 M N → = 0 → \overrightarrow{MA} + \overrightarrow{MB} = 2\overrightarrow{MN} = \overrightarrow{0} M A + M B = 2 M N = 0 M M M A B AB A B N M → = 0 → \overrightarrow{NM} = \overrightarrow{0} N M = 0 N N N M M M M M M N N N
c) Ta có:
M A → + M B → + M C → − 3 M N → = 3 N G → ⇔ M A → + M B → + M C → = 3 M N → + 3 N G → = 3 ( M N → + N G → ) = 3 ( M G → ) . \begin{aligned}
\overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC} - 3\overrightarrow{MN} &= 3\overrightarrow{NG}\\
\Leftrightarrow \overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC} &= 3\overrightarrow{MN} + 3\overrightarrow{NG}\\
&= 3(\overrightarrow{MN} + \overrightarrow{NG})\\
&= 3(\overrightarrow{MG}).
\end{aligned} M A + M B + M C − 3 M N ⇔ M A + M B + M C = 3 N G = 3 M N + 3 N G = 3 ( M N + N G ) = 3 ( M G ) . Vì G G G △ A B C \triangle ABC △ A B C
M G → = 1 3 ( M A → + M B → + M C → ) . \overrightarrow{MG} = \frac{1}{3}(\overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC}). M G = 3 1 ( M A + M B + M C ) . Do đó:
M A → + M B → + M C → = 3 M G → = 3 N G → . \overrightarrow{MA} + \overrightarrow{MB} + \overrightarrow{MC} = 3\overrightarrow{MG} = 3\overrightarrow{NG}. M A + M B + M C = 3 M G = 3 N G . Bài 8. Cho hình bình hành A B C D ABCD A B C D A B = 4 AB = 4 A B = 4 A D = 6 AD = 6 A D = 6 B A D ^ = 60 ∘ \widehat{BAD} = 60^{\circ} B A D = 6 0 ∘
a) Biểu diễn các vectơ B D → , A C → \overrightarrow{BD}, \overrightarrow{AC} B D , A C A B → , A D → \overrightarrow{AB}, \overrightarrow{AD} A B , A D
b) Tính các tích vô hướng A B → ⋅ A D → \overrightarrow{AB} \cdot \overrightarrow{AD} A B ⋅ A D A B → ⋅ A C → \overrightarrow{AB} \cdot \overrightarrow{AC} A B ⋅ A C B D → ⋅ A C → \overrightarrow{BD} \cdot \overrightarrow{AC} B D ⋅ A C
c) Tính độ dài các đường chéo B D BD B D A C AC A C
Lời giải:
a) Ta có:
B D → = A D → − A B → \overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB} B D = A D − A B A C → = A B → + A D → \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD} A C = A B + A D
b) Ta có:
A B → ⋅ A D → = ∣ A B → ∣ ⋅ ∣ A D → ∣ ⋅ cos ( B A D ^ ) = 4 ⋅ 6 ⋅ cos 60 ∘ = 4 ⋅ 6 ⋅ 1 2 = 12 \overrightarrow{AB} \cdot \overrightarrow{AD} = |\overrightarrow{AB}|\cdot |\overrightarrow{AD}|\cdot \cos(\widehat{BAD}) = 4 \cdot 6 \cdot \cos 60^{\circ} = 4 \cdot 6 \cdot \frac{1}{2} = 12 A B ⋅ A D = ∣ A B ∣ ⋅ ∣ A D ∣ ⋅ cos ( B A D ) = 4 ⋅ 6 ⋅ cos 6 0 ∘ = 4 ⋅ 6 ⋅ 2 1 = 12 A B → ⋅ A C → = A B → ⋅ ( A B → + A D → ) = A B → 2 + A B → ⋅ A D → = 4 2 + 12 = 28 \overrightarrow{AB} \cdot \overrightarrow{AC} = \overrightarrow{AB} \cdot (\overrightarrow{AB} + \overrightarrow{AD}) = \overrightarrow{AB}^2 + \overrightarrow{AB} \cdot \overrightarrow{AD} = 4^2 + 12 = 28 A B ⋅ A C = A B ⋅ ( A B + A D ) = A B 2 + A B ⋅ A D = 4 2 + 12 = 28 Ta lại có:
B D → ⋅ A C → = ( A D → − A B → ) ⋅ ( A B → + A D → ) = A D → 2 − A B → 2 = 6 2 − 4 2 = 20 \overrightarrow{BD} \cdot \overrightarrow{AC} = (\overrightarrow{AD} - \overrightarrow{AB}) \cdot (\overrightarrow{AB} + \overrightarrow{AD}) = \overrightarrow{AD}^2 - \overrightarrow{AB}^2 = 6^2 - 4^2 = 20 B D ⋅ A C = ( A D − A B ) ⋅ ( A B + A D ) = A D 2 − A B 2 = 6 2 − 4 2 = 20
c) Ta có:
B D = ∣ B D → ∣ = ∣ A D → − A B → ∣ = A D → 2 + A B → 2 − 2 A D → ⋅ A B → BD = \sqrt{|\overrightarrow{BD}|} = \sqrt{|\overrightarrow{AD} - \overrightarrow{AB}|} = \sqrt{\overrightarrow{AD}^2 + \overrightarrow{AB}^2 - 2\overrightarrow{AD} \cdot \overrightarrow{AB}} B D = ∣ B D ∣ = ∣ A D − A B ∣ = A D 2 + A B 2 − 2 A D ⋅ A B = 6 2 + 4 2 − 2 ⋅ 12 = 52 = 2 13 = \sqrt{6^2 + 4^2 - 2 \cdot 12} = \sqrt{52} = 2\sqrt{13} = 6 2 + 4 2 − 2 ⋅ 12 = 52 = 2 13 A C = ∣ A C → ∣ = ∣ A B → + A D → ∣ = A B → 2 + A D → 2 + 2 A B → ⋅ A D → AC = \sqrt{|\overrightarrow{AC}|} = \sqrt{|\overrightarrow{AB} + \overrightarrow{AD}|} = \sqrt{\overrightarrow{AB}^2 + \overrightarrow{AD}^2 + 2\overrightarrow{AB} \cdot \overrightarrow{AD}} A C = ∣ A C ∣ = ∣ A B + A D ∣ = A B 2 + A D 2 + 2 A B ⋅ A D = 4 2 + 6 2 + 2 ⋅ 12 = 76 = 2 19 = \sqrt{4^2 + 6^2 + 2 \cdot 12} = \sqrt{76} = 2\sqrt{19} = 4 2 + 6 2 + 2 ⋅ 12 = 76 = 2 19
Bài 9. Hai lực F 1 → \overrightarrow{F_1} F 1 F 2 → \overrightarrow{F_2} F 2 O O O α = 60 ∘ \alpha = 60^{\circ} α = 6 0 ∘ F → \overrightarrow{F} F O O O C C C F 1 → \overrightarrow{F_1} F 1 F 2 → \overrightarrow{F_2} F 2 30 N 30\,\text{N} 30 N
Lời giải:
Ta có:
∣ F 1 → ∣ = 30 N |\overrightarrow{F_1}| = 30\,\text{N} ∣ F 1 ∣ = 30 N ∣ F 2 → ∣ = 30 N |\overrightarrow{F_2}| = 30\,\text{N} ∣ F 2 ∣ = 30 N α = ( F 1 → , F 2 → ) = 60 ∘ \alpha = (\overrightarrow{F_1}, \overrightarrow{F_2}) = 60^{\circ} α = ( F 1 , F 2 ) = 6 0 ∘
Độ lớn của hợp lực F → \overrightarrow{F} F
∣ F → ∣ = ∣ F 1 → ∣ 2 + ∣ F 2 → ∣ 2 + 2 ∣ F 1 → ∣ ⋅ ∣ F 2 → ∣ ⋅ cos ( α ) = 30 2 + 30 2 + 2 ⋅ 30 ⋅ 30 ⋅ cos 60 ∘ = 900 + 900 + 1800 ⋅ 1 2 = 900 + 900 + 900 = 2700 = 30 3 N . \begin{aligned}
|\overrightarrow{F}| &= \sqrt{|\overrightarrow{F_1}|^2 + |\overrightarrow{F_2}|^2 + 2|\overrightarrow{F_1}|\cdot |\overrightarrow{F_2}|\cdot \cos(\alpha)}\\
&= \sqrt{30^2 + 30^2 + 2 \cdot 30 \cdot 30 \cdot \cos 60^{\circ}}\\
&= \sqrt{900 + 900 + 1800 \cdot \frac{1}{2}}\\
&= \sqrt{900 + 900 + 900}\\
&= \sqrt{2700}\\
&= 30\sqrt{3}\,\text{N}.
\end{aligned} ∣ F ∣ = ∣ F 1 ∣ 2 + ∣ F 2 ∣ 2 + 2∣ F 1 ∣ ⋅ ∣ F 2 ∣ ⋅ cos ( α ) = 3 0 2 + 3 0 2 + 2 ⋅ 30 ⋅ 30 ⋅ cos 6 0 ∘ = 900 + 900 + 1800 ⋅ 2 1 = 900 + 900 + 900 = 2700 = 30 3 N . Vậy độ lớn của hợp lực F → \overrightarrow{F} F 30 3 N 30\sqrt{3}\,\text{N} 30 3 N