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Do tam giác ABC vuông tại A và \(\widehat{B}=30^o\) \(\Rightarrow C=60^o\)
\(\Rightarrow\left(\overrightarrow{AB},\overrightarrow{BC}\right)=150^o;\)\(\left(\overrightarrow{BA},\overrightarrow{BC}\right)=30^o;\left(\overrightarrow{AC},\overrightarrow{CB}\right)=120^o\)
\(\left(\overrightarrow{AB},\overrightarrow{AC}\right)=90^o;\left(\overrightarrow{BC},\overrightarrow{BA}\right)=30^o\).Do vậy:
a) \(\cos\left(\overrightarrow{AB},\overrightarrow{BC}\right)+\sin\left(\overrightarrow{BA},\overrightarrow{BC}\right)+\tan\frac{\left(\overrightarrow{AC},\overrightarrow{CB}\right)}{2}\)
\(=\cos150^o+\sin30^o+\tan60^o\)
\(=-\frac{\sqrt{3}}{2}+\frac{1}{2}+\sqrt{3}\)
\(=\frac{\sqrt{3}+1}{2}\)
b) \(\sin\left(\overrightarrow{AB},\overrightarrow{AC}\right)+\cos\left(\overrightarrow{BC},\overrightarrow{AB}\right)+\cos\left(\overrightarrow{CA},\overrightarrow{BA}\right)\)
\(=\sin90^o+\cos30^o+\cos0^o\)
\(=1+\frac{\sqrt{3}}{2}\)
\(=\frac{2+\sqrt{3}}{2}\)
Tính \(\overrightarrow{a}.\overrightarrow{b}\) hả bạn?
\(\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|.\left|\overrightarrow{b}\right|cos\left(\overrightarrow{a};\overrightarrow{b}\right)=2.\sqrt{3}.cos30^0=3\)
Tính \(\left|\overrightarrow{a}+\overrightarrow{b}\right|\)
\(\overrightarrow{a}\perp\overrightarrow{b}\Rightarrow\overrightarrow{a}.\overrightarrow{b}=0\)
\(\left(2\overrightarrow{a}-\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)=2a^2+2\overrightarrow{a}.\overrightarrow{b}-\overrightarrow{a}.\overrightarrow{b}-b^2\)
\(=2a^2-b^2+\overrightarrow{a}.\overrightarrow{b}\)
\(=2.1-2+0=0\)
\(\Rightarrow\left(2\overrightarrow{a}-\overrightarrow{b}\right)\perp\left(\overrightarrow{a}+\overrightarrow{b}\right)\)
\(\left(\overrightarrow{a}+\overrightarrow{b}\right)^2=\left(\overrightarrow{a}+\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)\)\(=\left|\overrightarrow{a}\right|^2+\left|\overrightarrow{b}\right|^2+2\overrightarrow{a}\overrightarrow{b}\).
\(\left(\overrightarrow{a}-\overrightarrow{b}\right)^2=\left(\overrightarrow{a}-\overrightarrow{b}\right)\left(\overrightarrow{a}-\overrightarrow{b}\right)\)\(=\left|\overrightarrow{a}\right|^2+\left|\overrightarrow{b}\right|^2-2\overrightarrow{a}\overrightarrow{b}\).
\(\left(\overrightarrow{a}-\overrightarrow{b}\right)\left(\overrightarrow{a}+\overrightarrow{b}\right)=\left|\overrightarrow{a}\right|^2+\overrightarrow{a}\overrightarrow{b}-\overrightarrow{a}\overrightarrow{b}+\left|\overrightarrow{b}\right|^2\)\(=\left|\overrightarrow{a}\right|^2-\left|\overrightarrow{b}\right|^2\).
a) cos(;
) =
= 0
=> (;
) = 900
b) cos(;
) =
=
=> (;
) = 450
c) cos(;
) =
=
=> (;
) = 1500
Đăng những câu khác đi em mỏi tay rồi
Vì AH=(BC.1/2)tan60 ct lương giác
=BC.tan60.1/2=\(\sqrt{3}\)/2
họk tốt!
