Chọn gốc tọa độ trùng với điểm A(0;0;0);

Tia AB nằm trên trục Ox.

Tia AD nằm trên trục Oy.

Tia Az (chứa SA) nằm trên trục Oz.

SA=2

=>S(0;0;2)

AB=1

=>B(1;0;0)

AD=3

=>D(0;3;0)

ABCD là hình bình hành

=>BC=AD=3

=>C(1;3;0)

Tọa độ trung điểm M của BC là:

\(\begin{cases}x=\frac{1+1}{2}=\frac22=1\\ y=\frac{0+3}{2}=\frac32=1,5\\ z=\frac{0+0}{2}=0\end{cases}\)

=>M(1;1,5;0)

S(0;0;2); A(0;0;0); M(1;1,5;0)

\(\overrightarrow{SA}=\left(0-0;0-0;0-2\right)=\left(0;0;-2\right)\)

\(\overrightarrow{AM}=\left(1-0;1,5-0;0-0\right)=\left(1;1,5;0\right)\)

\(\overrightarrow{SA}\cdot\overrightarrow{AM}=0\cdot1+0\cdot1,5+2\cdot0=0\)

=>\(cos\left(SA;AM\right)=0\)

b: S(0;0;2); C(1;3;0); A(0;0;0); D(0;3;0)

\(\overrightarrow{SC}=\left(1-0;3-0;0-2\right)=\left(1;3;-2\right)\) ; \(\overrightarrow{AD}=\left(0-0;3-0;0-0\right)=\left(0;3;0\right)\)

\(SC=\sqrt{1^2+3^2+\left(-2\right)^2}=\sqrt{1+4+9}=\sqrt{14}\) ; \(AD=\sqrt{0^2+3^2+0^2}=3\)

\(cos\left(SC;AD\right)=\frac{\overrightarrow{SC}\cdot\overrightarrow{AD}}{SC\cdot AD}=\frac{1\cdot0+3\cdot3+\left(-2\right)\cdot0}{\sqrt{14}\cdot3}=\frac{9}{3\sqrt{14}}=\frac{3}{\sqrt{14}}\)

c: S(0;0;2); B(1;0;0); A(0;0;0); C(1;3;0)

\(\overrightarrow{SB}=\left(1-0;0-0;0-2\right)=\left(1;0;-2\right)\) ; \(\overrightarrow{AC}=\left(1-0;3-0;0-0\right)=\left(1;3;0\right)\)

\(SB=\sqrt{1^2+0^2+\left(-2\right)^2}=\sqrt5;AC=\sqrt{1^2+3^2+0^2}=\sqrt{1+9}=\sqrt{10}\)

cos(SB;AC)=\(\frac{\overrightarrow{SB}\cdot\overrightarrow{AC}}{\left|\overrightarrow{SB}\right|\cdot\left|\overrightarrow{AC}\right|}\)

\(\frac{1\cdot1+0\cdot3+\left(-2\right)\cdot0}{\sqrt5\cdot\sqrt{10}}=\frac{1}{\sqrt{50}}=\frac{1}{5\sqrt2}=\frac{\sqrt2}{10}\)

Ta có: \(\sin^2B+cos^2B=1\)

=>\(cos^2B=1-0,8^2=1-0,64=0,36=0,6^2\)

=>cosB=0,6

Ta có: ΔABC vuông tại A

=>cosC =sin B

=>cosC=0,8

C A B D H

a, \(\Delta ABC\) có \(\widehat{C}=90^o\).

Áp dụng pytago có: \(AB=\sqrt{AC^2+BC^2}=\sqrt{\left(12a\right)^2+\left(5a\right)^2}=13a\)

\(\Delta ABC\) có \(\widehat{C}=90^o\)\(\Rightarrow\)\(\left\{{}\begin{matrix}\sin B=\dfrac{AC}{AB}=\dfrac{12a}{13a}=\dfrac{12}{13}\\cosB=\dfrac{BC}{AB}=\dfrac{5a}{13a}=\dfrac{5}{13}\end{matrix}\right.\)

Ta có: \(\dfrac{sinB+cosB}{sinB-cosB}=\dfrac{\dfrac{12}{13}+\dfrac{5}{13}}{\dfrac{12}{13}-\dfrac{5}{13}}=\dfrac{\dfrac{17}{13}}{\dfrac{7}{13}}=\dfrac{17}{7}\)

b, Có S_ABCD= \(\dfrac{CH.AB}{2}=\dfrac{CB.AC}{2}\Rightarrow CH.AB=BC.AC\Rightarrow CH=\dfrac{AC.BC}{AB}=\dfrac{12a.5a}{13a}=\dfrac{60a}{13}\approx4,615a\)

bạn ơi mk gửi nhầm, mk xin lỗi nhé!

Khẳng định đúng: a

b,ta có :\(\frac{sin^2a-cos^2a\left(1-cos^2a\right)}{cos^2a-sin^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)

=>\(\frac{sin^2a-sin^2a.cos^2a}{cos^2a-sin^2a.cos^2a}=\frac{sin^4a}{cos^4a}\)

=>\(\frac{sin^2a\left(1-cos^2a\right)}{cos^2a\left(1-sin^2a\right)}=\frac{sin^4a}{cos^4a}\)

=>\(\frac{sin^4a}{cos^4a}=\frac{sin^4a}{cos^4a}\)luon dung => dpcm