Kẻ \(BH\perp DC\) tại H

=> ABHD là hình chữ nhật (vì tứ giác có 3 góc vuông)

\(\Rightarrow AB=DH=15cm;AD=BH=12cm\)

Có \(CH=DC-DH=20-15=5\left(cm\right)\)

Áp dụng định lí py-ta-go vào tam giác vuông có:

\(BC=\sqrt{BH^2+HC^2}=\sqrt{12^2+5^2}=13\left(cm\right)\)

Vậy BC=13cm

b)Pt \(\Leftrightarrow2sinx.cosx-3\left(sinx+cosx\right)+3=0\)

Đặt \(t=sinx+cosx;t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow t^2-1=2sinx.cosx\)

Pttt: \(t^2-1-3t+3=0\)

\(\Leftrightarrow t^2-3t+2=0\)\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(tm\right)\\t=2\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow t=1\Rightarrow sinx+cosx=1\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\),\(k\in Z\)

Vậy...

e)Pt\(\Leftrightarrow3\left(sinx+cosx\right)=2.2sinx.cosx\)

Đặt \(t=sinx+cosx;t\in\left[-\sqrt{2};\sqrt{2}\right]\)

\(\Rightarrow t^2-1=2sinx.cosx\)

Pttt: \(3t=2\left(t^2-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(ktm\right)\\t=-\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\)

\(t=-\dfrac{1}{2}\Leftrightarrow sinx+cosx=-\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arc.sin\left(-\dfrac{\sqrt{2}}{4}\right)+k2\pi\\x=\dfrac{3\pi}{4}-arc.sin\left(-\dfrac{\sqrt{2}}{4}\right)+k2\pi\end{matrix}\right.\)

Vậy...