Đáp án:

$11)\\ A=2\\ B=2\\ C=1\\ D=1\\ 13)\\ 1)AC=9(cm);BC=15(cm)\\ 2)BC=39(cm);AC=36(cm)\\ 3)AC=2(cm); BC=4(cm)\\ 4)BC=2(cm);AC=\sqrt{3}(cm)$

Giải thích các bước giải:

$11)\\ A=\sin^215^\circ+\sin^235^\circ+\sin^255^\circ+\sin^275^\circ\\ =\sin^215^\circ+\sin^235^\circ+\cos^2(90^\circ-55^\circ)+\cos^2(90^\circ-75^\circ)\\ =\sin^215^\circ+\sin^235^\circ+\cos^235^\circ+\cos^215^\circ\\ =\sin^215^\circ+\cos^215^\circ+\sin^235^\circ+\cos^235^\circ\\ =1+1\\ =2\\ B=\cos^215^\circ+\cos^235^\circ+\cos^255^\circ+\cos^275^\circ\\ =\cos^215^\circ+\cos^235^\circ+\sin^2(90^\circ-55^\circ)+\sin^2(90^\circ-75^\circ)\\ =\cos^215^\circ+\cos^235^\circ+\sin^235^\circ+\sin^215^\circ\\ =\cos^215^\circ+\sin^215^\circ+\cos^235^\circ+\sin^235^\circ\\ =1+1\\ =2\\ C=\tan15^\circ.\tan35^\circ.\tan55^\circ.\tan75^\circ\\ =\tan15^\circ.\tan35^\circ.\cot(90^\circ-55^\circ).\cot(90^\circ-75^\circ)\\ =\tan15^\circ.\tan35^\circ.\cot35^\circ.\cot15^\circ\\ =\tan15^\circ.\cot15^\circ.\tan35^\circ.\cot35^\circ\\ =1.1\\ =1\\ D=\cot15^\circ.\cot35^\circ.\cot55^\circ.\cot75^\circ\\ =\cot15^\circ.\cot35^\circ.\tan(90^\circ-55^\circ).\tan(90^\circ-75^\circ)\\ =\cot15^\circ.\cot35^\circ.\tan35^\circ.\tan15^\circ\\ =\cot15^\circ.\tan15^\circ.\cot35^\circ.\tan35^\circ\\ =1.1\\ =1\\ 13)\\ 1)\tan B=\dfrac{3}{4}\\ \Leftrightarrow \dfrac{AC}{AB}=\dfrac{3}{4}\\ \Rightarrow AC=\dfrac{3}{4}AB=9(cm)$

$\Delta ABC$ vuông tại $A$

$\Rightarrow BC=\sqrt{AB^2+AC^2}=15(cm)\\ 2)\cos B=\dfrac{5}{13}\\ \Leftrightarrow \dfrac{AB}{BC}=\dfrac{5}{13}\\ \Rightarrow BC=\dfrac{13}{5}AB=39(cm)$

$\Delta ABC$ vuông tại $A$

$\Rightarrow AC=\sqrt{BC^2-AB^2}=36(cm)\\ 3)\cot B=\sqrt{3}\\ \Leftrightarrow \dfrac{AB}{AC}=\sqrt{3}\\ \Rightarrow AC=\dfrac{AB}{\sqrt{3}}=2(cm)$

$\Delta ABC$ vuông tại $A$

$\Rightarrow BC=\sqrt{AB^2+AC^2}=4(cm)\\ 4)\sin B=\dfrac{\sqrt{3}}{2}\\ \Leftrightarrow \dfrac{AC}{BC}=\dfrac{\sqrt{3}}{2}\\ \Rightarrow AC=\dfrac{\sqrt{3}}{2}BC$

$\Delta ABC$ vuông tại $A$

$\Rightarrow BC^2=AB^2+AC^2\\ \Leftrightarrow BC^2=1^2+\left(\dfrac{\sqrt{3}}{2}BC\right)^2\\ \Leftrightarrow BC^2=1+\dfrac{3}{4}BC^2\\ \Leftrightarrow \dfrac{1}{4}BC^2=1\\ \Leftrightarrow BC^2=4\\ \Leftrightarrow BC=2(cm)\\ \Rightarrow AC=\sqrt{3}(cm)$