10 câu Trắc nghiệm Toán 10 Kết nối tri thức Bài ôn tập cuối chương 3 (Nhận biết) có đáp án

a² = b² + c² + 2bcsinA;

a² = b² + c² – 2bccosA;

a² = b² + c² – 2acsinA;

a² = b² + c² + 2abcosA.

sin45° = $\frac{–\sqrt{2}}{2}$;

cos45° =1 ;

tan45° = 1;

cot45° = $\frac{\sqrt{2}}{2}$.

S = $\frac{1}{2}$bcsinA;

S = $\frac{1}{2}$absinB;

S = 2acsinB;

S = 2bcsinA.

$\frac{\text{a}}{\text{sinA}}\text{ = }\frac{\text{b}}{\text{sinB}}\text{ = }\frac{\text{c}}{\text{sinC}}$

a² = b² + c² – 2bccosA;

S = $\frac{1}{2}$bcsinA = $\frac{1}{2}$acsinB = $\frac{1}{2}$absinC;

b² = a² + c² – 2accosB .

$co{s}^{2}x+si{n}^{2}x=1$;

$co{s}^{2}x+si{n}^{2}x=0;$

$co{s}^{2}x+si{n}^{2}x=2$

$co{s}^{2}x+si{n}^{2}x=\frac{1}{4}$

$\frac{\text{b}}{\text{sinB}}=2\text{R}$

$\frac{\text{a}}{\text{2sinA}}\text{ = R}$

$\frac{\text{c}}{\text{sinC}}\text{ = 2R}$

$\frac{\text{2c}}{\text{sinC}}\text{ = R}$

S = $\sqrt{\text{2p(p + a)(p + b)(p + c)}}$

S = $2\sqrt{\text{p(p }-\text{ a)(p }-\text{ b)(p }-\text{ c)}}$

S = $2\sqrt{\text{p(p }-\text{ a)(p }-\text{ b)(p }-\text{ c)}}$

S = $\sqrt{\text{p(p }-\text{ a)(p }-\text{ b)(p }-\text{ c)}}$

sin( 180° – α ) = –sinα;

cos( 180° – α ) = cosα;

sin( 90° – α ) = –cosα;

cos( 90° – α ) = sinα.

$\frac{\text{a}}{\text{sinA}}\text{ = }\frac{\text{b}}{\text{sinB}}\text{ = }\frac{\text{c}}{\text{sinC}}$

a² = b² + c² – 2bccosA;

S =$\frac{1}{2}$ bcsinA =$\frac{1}{2}$ acsinB = $\frac{1}{2}$absinC;

b² = a² + c² – 2bccosB .