Trắc nghiệm Ôn tập Toán 10 Chương 2 Hình học có đáp án (Nhận biết)

A. $\sin {150}^0=-\frac{\sqrt{3}}{2}$

B. $\cos {150}^0=\frac{\sqrt{3}}{2}$

C. $\tan {150}^0=-\frac{1}{\sqrt{3}}$

D. $\cot {150}^0=\sqrt{3}$

A. $\stackrel{\to }{MC}-\stackrel{\to }{MA}=\stackrel{\to }{MB}-\stackrel{\to }{MD}$

B. $\stackrel{\to }{MC}-\stackrel{\to }{MA}=\stackrel{\to }{DA}-\stackrel{\to }{DC}$

C. $\stackrel{\to }{MC}-\stackrel{\to }{MA}=\stackrel{\to }{AB}+\stackrel{\to }{AD}$

D. $\stackrel{\to }{MC}-\stackrel{\to }{MA}=\stackrel{\to }{BA}-\stackrel{\to }{BC}$

A. ABCD là hình bình hành

B. DA = BC

C. $\stackrel{\to }{AC}=\stackrel{\to }{BD}$

D. $\stackrel{\to }{AB}=\stackrel{\to }{DC}$

A. $\exists k<0:\stackrel{\to }{AB}=k\stackrel{\to }{AC}$

B. $\exists k\ne 0:\stackrel{\to }{AB}=k\stackrel{\to }{AC}$

C. $\exists k>0:\stackrel{\to }{AB}=k\stackrel{\to }{AC}$

D. $\exists k=0:\stackrel{\to }{AB}=k\stackrel{\to }{AC}$

A. Nếu $b^2+c^2-a^2>0$ thì góc A nhọn

B. Nếu $b^2+c^2-a^2>0$ thì góc A tù

C. Nếu $b^2+c^2-a^2<0$ thì góc A nhọn

D. Nếu $b^2+c^2-a^2>0$ thì góc A vuông

A. ${90}^{\circ }$

B. ${60}^{\circ }$

C. ${45}^{\circ }$

D. ${30}^{\circ }$

A. $\stackrel{\to }{a}.\stackrel{\to }{b}=a_1.b_1+a_2.b_2$

B. $\stackrel{\to }{a}.\stackrel{\to }{b}=\left|\stackrel{\to }{a}\right|.\left|\stackrel{\to }{b}\right|.\cos \left(\stackrel{\to }{a},\stackrel{\to }{b}\right)$

C. $\stackrel{\to }{a}.\stackrel{\to }{b}=\frac{1}{2}\left[\stackrel{\to }{a^2}+\stackrel{\to }{b^2}-{\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)}^2\right]$

D. $\stackrel{\to }{a}.\stackrel{\to }{b}=\frac{1}{2}\left[{\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)}^2-\stackrel{\to }{a^2}-\stackrel{\to }{b^2}\right]$

A. $\frac{1}{5}$

B. 1

C. $\frac{6}{5}$

D. $\frac{7}{5}$

A. $\stackrel{\to }{a}.\stackrel{\to }{b}=-30$

B. $\stackrel{\to }{a}.\stackrel{\to }{b}=3$

C. $\stackrel{\to }{a}.\stackrel{\to }{b}=30$

D. $\stackrel{\to }{a}.\stackrel{\to }{b}=43$

A. $\stackrel{\to }{CA}.\stackrel{\to }{CB}=a^2$

B. $\stackrel{\to }{CA}.\stackrel{\to }{CB}=a$

C. $\stackrel{\to }{CA}.\stackrel{\to }{CB}=\frac{a\sqrt{2}}{2}$

D. $\stackrel{\to }{CA}.\stackrel{\to }{CB}=a\sqrt{2}$

A. 1,5a

B. $a\sqrt{2}$

C. $a\sqrt{3}$

D. $\frac{a\sqrt{5}}{2}$

A. ABCD là hình bình hành

B. DA = BC

C. $\stackrel{\to }{AC}=\stackrel{\to }{BD}$

D. $\stackrel{\to }{AB}=\stackrel{\to }{DC}$

A. ${90}^{\circ }$

B. ${60}^{\circ }$

C. ${45}^{\circ }$

D. ${30}^{\circ }$

A. $\stackrel{\to }{a}.\stackrel{\to }{b}=a_1.b_1+a_2.b_2$

B. $\stackrel{\to }{a}.\stackrel{\to }{b}=\left(\stackrel{\to }{a}\right).\left(\stackrel{\to }{b}\right).\cos \left(\stackrel{\to }{a},\stackrel{\to }{b}\right)$

C. $\stackrel{\to }{a}.\stackrel{\to }{b}=\frac{1}{2}\left(\stackrel{\to }{a^2}+\stackrel{\to }{b^2}-{\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)}^2\right)$

D. $\stackrel{\to }{a}.\stackrel{\to }{b}=\frac{1}{2}\left({\left(\stackrel{\to }{a}+\stackrel{\to }{b}\right)}^2-\stackrel{\to }{a^2}-\stackrel{\to }{b^2}\right)$

A. $\frac{1}{5}$

B. 1

C. $\frac{6}{5}$

D. $\frac{7}{5}$

A. $\stackrel{\to }{a}.\stackrel{\to }{b}=-30$

B. $\stackrel{\to }{a}.\stackrel{\to }{b}=3$

C. $\stackrel{\to }{a}.\stackrel{\to }{b}=30$

D. $\stackrel{\to }{a}.\stackrel{\to }{b}=43$

A. $\stackrel{\to }{CA}.\stackrel{\to }{CB}=a^2$

B. $\stackrel{\to }{CA}.\stackrel{\to }{CB}=a$

C. $\stackrel{\to }{CA}.\stackrel{\to }{CB}=\frac{a\sqrt{2}}{2}$

D. $\stackrel{\to }{CA}.\stackrel{\to }{CB}=a\sqrt{2}$