1. 在Rt△ABC中,∠C = 90°,AC = 3,AB = 4,则cos A的值为______.
答案:
$\frac{3}{4}$
2. 如图,在平面直角坐标系中,点A的坐标为(4,0),点B的坐标为(0,3),以点A为圆心,以AB的长为半径作弧,交x轴的负半轴于点C,连接BC,则∠ACB的正弦值为________.
答案:
$\frac{3\sqrt{10}}{10}$
3. 如图,直线y = $\frac{1}{2}x+\frac{3}{2}$与x轴交于点A,与直线y = 2x交于点B.
(1)求点B的坐标;
(2)求sin∠BAO的值.
(1)求点B的坐标;
(2)求sin∠BAO的值.
答案:
解:
(1)解方程组$\begin{cases}y = \frac{1}{2}x + \frac{3}{2}\\y = 2x\end{cases}$,得$\begin{cases}x = 1\\y = 2\end{cases}$,
∴B点坐标为(1,2).
(2)作BC⊥x轴于点C,
当y = 0时,$\frac{1}{2}x + \frac{3}{2} = 0$,解得x = - 3,则A(-3,0),
∴OA = 3,而OC = 1,BC = 2,
∴AB = $\sqrt{AC^{2}+BC^{2}} = 2\sqrt{5}$
∴sin∠BAC = $\frac{BC}{AB} = \frac{2}{2\sqrt{5}} = \frac{\sqrt{5}}{5}$,
即sin∠BAO的值为$\frac{\sqrt{5}}{5}$
(1)解方程组$\begin{cases}y = \frac{1}{2}x + \frac{3}{2}\\y = 2x\end{cases}$,得$\begin{cases}x = 1\\y = 2\end{cases}$,
∴B点坐标为(1,2).
(2)作BC⊥x轴于点C,
当y = 0时,$\frac{1}{2}x + \frac{3}{2} = 0$,解得x = - 3,则A(-3,0),
∴OA = 3,而OC = 1,BC = 2,
∴AB = $\sqrt{AC^{2}+BC^{2}} = 2\sqrt{5}$
∴sin∠BAC = $\frac{BC}{AB} = \frac{2}{2\sqrt{5}} = \frac{\sqrt{5}}{5}$,
即sin∠BAO的值为$\frac{\sqrt{5}}{5}$
4. 如图,A,B,C三点在正方形网格线的交点处,若将△ACB绕着点A逆时针旋转得到△AC'B',则tan B'的值为______.
答案:
$\frac{1}{3}$
5. 如图,在△ABC中,∠ACB = 90°,CD⊥AB于点D.若AC = 2$\sqrt{3}$,AB = 3$\sqrt{2}$,则tan∠BCD的值为______.
答案:
$\frac{\sqrt{2}}{2}$
6. 如图,在矩形ABCD中,AB = 5,BC = 4,将矩形折叠,使点B落在边CD上的点F处,折痕为AE,则cos∠EFC = ______.
答案:
$\frac{4}{5}$
7. 如图,边长为1的小正方形构成的网格中,半径为1的⊙O的圆心O在格点上,则cos∠BED = ______.
答案:
$\frac{2\sqrt{5}}{5}$
查看更多完整答案,请扫码查看