共50条信息
计算:\(4\sin 30^{\circ}-{\left(π-3\right)}^{0}+\left| \sqrt{3}-2\right|+{\left( \dfrac{1}{2}\right)}^{-2} \)
如图,等腰\(\triangle ABC\)是\(⊙O\)的内接三角形,\(AB=AC\),过点\(A\)作\(BC\)的平行线\(AD\)交\(BO\)的延长线于点\(D\).
\((2)\)若\(⊙O\)的半径为\(15\),\(\sin ∠D=\dfrac{3}{5}\),求\(AB\)的长.
如图,已知\(∠MON=25^{\circ}\),矩形\(ABCD\)的边\(BC\)在\(OM\)上,对角线\(AC⊥ON\).
\((1)\)求\(∠ACD\)度数;
\((2)\)当\(AC=5\)时,求\(AD\)的长.
\((\)参考数据:\(\sin 25^{\circ}=0.42\):\(\cos 25^{\circ}=0.91\);\(\tan 25^{\circ}=0.47)\)
已知如图,\(AB\)为\(⊙O\)的直径,\(⊙O\)过\(AC\)的中点\(D\),\(DE⊥BC\)于点\(E\).
\((1)\)求证:\(DE\)为\(⊙O\)的切线;
\((2)DE=4\),\(\tan C=\dfrac{1}{2}\),求\(⊙O\)的直径.
计算:\({{\left( \sqrt{3}-2 \right)}^{0}}+{{\left( \dfrac{1}{3} \right)}^{-1}}+4\cos 30\circ -\left| 4-\sqrt{12} \right|\).
如图,\(C\)地在\(A\)地的正东方向,因有大山阻隔,由\(A\)地到\(C\)地需要绕行附近的\(B\)地,已知\(B\)地位于\(A\)地的北偏东\(67^{\circ}\)方向,距离\(A\)地\(520km\),\(C\)地位于\(B\)地南偏西\(30^{\circ}\)方向,若要打通穿山隧道建高铁,求线段\(AC\)的长\((\)结果保留整数\()\).
如图,在\(⊙O\)中,半径\(OC\)与弦\(AB\)垂直,垂足为\(E\),以\(OC\)为直径的圆与弦\(AB\)的一个交点为\(F\),\(D\)是\(CF\)延长线与\(⊙O\)的交点\(.\)若\(OE=3\),\(OF=6\),求\(⊙O\)的半径和弧\(CD\)的长.
计算:\(\left| -\sqrt{3} \right|+\sqrt{2}\sin 45{}^\circ +\tan 60{}^\circ -{{(-\dfrac{1}{3})}^{-1}}-\sqrt{12}\)
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