Trắc nghiệm Toán 9 Bài 4 (có đáp án): Một số hệ thức về cạnh và góc trong tam giác...

A. MN = MP. sin P

B. MN = MP. cos P

C. MN = MP. tan P

D. MN = MP. cot P

A. NP = MP. sin P

B. NP = MN. cot P

C. NP = MN. tan P

D. NP = MP. cot P

A. b = c.sin ${50}^0$.

B. b = a.tan ${50}^0$.

C. b = c.cot ${50}^0$.

D. c = b.cot ${50}^0$.

A. AB = $\frac{5\sqrt{3}}{3}$ ; BC = $\frac{20\sqrt{3}}{3}$

B. AB = $\frac{10\sqrt{3}}{3}$; BC = $\frac{14\sqrt{3}}{3}$

C. AB = $\frac{10\sqrt{3}}{3}$; BC = 20$\sqrt{3}$

D. AB = $\frac{10\sqrt{3}}{3}$ ; BC = $\frac{20\sqrt{3}}{3}$

A. AB = 20$\sqrt{3}$; BC = 40

B. AB = 20$\sqrt{3}$; BC = 40$\sqrt{3}$

C. AB = 20; BC = 40

D. AB = 20; BC = 20$\sqrt{3}$

A. $AC\approx 7,71; \hat{C}={40}^0$

B. $AC\approx 7,72; \hat{C}={50}^0$

C. $AC\approx 7,71; \hat{C}={50}^0$

D. $AC\approx 7,73; \hat{C}={50}^0$

A. AC = 8 (cm); $\hat{B}\approx {36}^{0}52\text{'}$

B. AC = 9 (cm); $\hat{B}\approx {36}^{0}52\text{'}$

C. AC = 9 (cm); $\hat{B}\approx {37}^{0}52\text{'}$

D. AC = 9 (cm); $\hat{B}\approx {36}^{0}55\text{'}$

A. $BC=\sqrt{74}\left(cm\right); \hat{C}\approx {35}^{0}32\text{'}$

B. $BC=\sqrt{74}\left(cm\right); \hat{C}\approx {36}^{0}32\text{'}$

C. $BC=\sqrt{74}\left(cm\right); \hat{C}\approx {35}^{0}33\text{'}$

D. $BC=\sqrt{74}\left(cm\right); \hat{B}\approx {35}^{0}32\text{'}$

A. AC = 22; $\hat{B}\approx {67}^0$

B. AC = 24; $\hat{B}\approx {66}^0$

C. AC = 24; $\hat{B}\approx {67}^0$

D. AC = 24; $\hat{B}\approx {68}^0$

A. BC = 10

B. BC = 11

C. BC = 9

D. BC = 12

A. BC = $3\sqrt{3}+6$

B. BC =$3\sqrt{13}+6$

C. BC = 9

D. BC = 6

A. 4

B. 5

C. 7

D.8

A. 4

B. 5

C. 6

D. 8

A. 60 $c{m}^2$.

B. 80 $c{m}^2$.

C. 40 $c{m}^2$.

D. 160 $c{m}^2$.

A. $\hat{A}={50}^0$; $\hat{C}={65}^0$; AB = AC = 5,6; BC = 8,52

B. $\hat{A}={50}^0$; $\hat{C}={65}^0$; AB = AC = 5,6; BC = 4,42

C. $\hat{A}={50}^0$; $\hat{C}={65}^0$; AB = AC = 4,7; BC = 4,24

D. $\hat{A}={50}^0$; $\hat{C}={65}^0$; AB = AC = 4,7; BC = 3,97

A. $\hat{B}={53}^{0}8\text{'}$; $\hat{C}={36}^{0}52\text{'}$

B. $\hat{B}={36}^{0}52\text{'}$; $\hat{C}={53}^{0}8\text{'}$

C. $\hat{B}={48}^{0}35\text{'}$; $\hat{C}={41}^{0}25\text{'}$

D. $\hat{B}={41}^{0}25\text{'}$; $\hat{C}={48}^{0}35\text{'}$

A. $\hat{A}={45}^0$; $\hat{B}=\hat{C}={67}^{0}30\text{'}$

B. $\hat{A}={30}^0$; $\hat{B}=\hat{C}={75}^0$

C. $\hat{A}={48}^{0}6\text{'}$; $\hat{B}=\hat{C}={65}^{0}57\text{'}$

D. $\hat{A}={53}^{0}8\text{'}$; $\hat{B}=\hat{C}={63}^{0}26\text{'}$

A. $AD=a.\cos 22,5^0$; $DC=a-a.\cos 22,5^0$.

B. $AD=a.\sin 22,5^0$; $DC=a-a.\sin 22,5^0$.

C. $AD=a.\tan 22,5^0$; $DC=a-a.\tan 22,5^0$.

D. $AD=a.\mathrm{co}t 22,5^0$; $DC=a-a.cot 22,5^0$.

A. $\left\{\begin{array}{l}AB=2\sqrt{3}a\\ AC=2a\end{array}\right.$

B. $\left\{\begin{array}{l}AB=\sqrt{3}a\\ AC=\frac{1}{2}a\end{array}\right.$

C. $\left\{\begin{array}{l}AB=a\\ AC=3a\end{array}\right.$

D. $\left\{\begin{array}{l}AB=\sqrt{3}a\\ AC=a\end{array}\right.$

A. $S_{ABCD}$ = 2 (đvdt)

B. $S_{ABCD}$ = 3 (đvdt)

C. $S_{ABCD}$ = 4 (đvdt)

D. $S_{ABCD}$ = $\frac{5}{2}$ (đvdt)

A. AH = 2,8 cm; $cos\widehat{ACB}=\frac{3}{5}$

B. AH = 2,4 cm; $cos\widehat{ACB}=\frac{4}{5}$

C. AH = 2,5 cm; $cos\widehat{ACB}=\frac{3}{4}$

D. AH = 1,8 cm; $cos\widehat{ACB}=\frac{2}{3}$

A. $\frac{1}{2}{a}^2\sin \alpha .\cos \alpha$

B. $a^2\sin \alpha .\cos \alpha$

C. $2a^2\sin \alpha .\cos \alpha$

D. $3a^2\sin \alpha .\cos \alpha$

A. $\alpha ={45}^0$; $ma{x}_{S_{ABC}}=\frac{1}{2}{a}^2$.

B. $\alpha ={30}^0$; $ma{x}_{S_{ABC}}=\frac{\sqrt{3}}{4}{a}^2$.

C. $\alpha ={60}^0$; $ma{x}_{S_{ABC}}=\frac{\sqrt{3}}{4}{a}^2$.

D. $\alpha ={45}^0$; $ma{x}_{S_{ABC}}=\frac{1}{4}{a}^2$.

A. EI = 4,5cm

B. EI = 5,4cm

C. EI = 5,9cm

D. EI = 5,6cm