\(1,\\ a,\sin13^0=\cos77^0< \sin32^0< \sin42^0=\cos48^0< \sin65^0< \sin75^0=\cos15^0\\ b,A=3\left(\sin^243^0+\sin^247^0\right)-\left(\tan38^0-\cot52^0\right)-\dfrac{\tan28^0}{\tan28^0}\\ A=3\left(\sin^243^0+\cos^243^0\right)-\left(\tan38^0-\tan38^0\right)-1\\ A=3-0-1=2\)
2.
\(\cos\widehat{B}=\cos60^0=\dfrac{AB}{BC}=\dfrac{1}{2}\Leftrightarrow BC=20\left(cm\right)\\ AC=\sqrt{BC^2-AB^2}=10\sqrt{3}\left(cm\right)\left(pytago\right)\)
3.
a, \(AC=\sqrt{BC^2-AB^2}=20\left(cm\right)\left(pytago\right)\)
Áp dụng HTL: \(\left\{{}\begin{matrix}HB=\dfrac{AB^2}{BC}=9\left(cm\right)\\HC=\dfrac{AC^2}{BC}=16\left(cm\right)\\HA=\sqrt{HB\cdot HC}=12\left(cm\right)\end{matrix}\right.\)
b, Vì AIHK là hình chữ nhật nên \(AH=IK\)
Mà \(HA^2=HB\cdot HC\Leftrightarrow IK^2=HB\cdot HC\) (hệ thức lượng)
