gaokao 2010 Q16

gaokao · China · shanghai-science Vectors 3D & Lines Parametric Representation of a Line
16. The parametric equation of line $l$ is $\left\{ \begin{array} { l } x = 1 + 2 t \\ y = 2 - t \end{array} ( t \in \mathbb{R} ) \right.$. Then a direction vector $\overrightarrow { Therefore $\frac { ( 1 - \cos \theta ) ^ { 2 } } { 4 } + \frac { ( 1 + \sin \theta ) ^ { 2 } } { 4 } < 1$ , simplifying gives $\sin \theta - \cos \theta < \frac { 1 } { 2 } , \sin \left( \theta - \frac { \pi } { 4 } \right) < \frac { \sqrt { 2 } } { 4 }$ , Also $0 < \theta < \pi$ , that is $- \frac { \pi } { 4 } < \theta - \frac { \pi } { 4 } < \frac { 3 \pi } { 4 }$ , so $- \frac { \pi } { 4 } < \theta - \frac { \pi } { 4 } < \arcsin \frac { \sqrt { 2 } } { 4 }$ , Thus the range of values for $\theta$ is $\left( 0 , \frac { \pi } { 4 } + \arcsin \frac { \sqrt { 2 } } { 4 } \right)$ .
(5 points) In $\triangle ABC$, $D$ is a point on side $BC$, $BC = 3 BD$, $AD = \sqrt { 2 }$, $\angle ADB = 135 ^ { \circ }$. If $AC = \sqrt { 2 } AB$, then $BD = \_\_\_\_$. 
16. The parametric equation of line $l$ is $\left\{ \begin{array} { l } x = 1 + 2 t \\ y = 2 - t \end{array} ( t \in \mathbb{R} ) \right.$. Then a direction vector $\overrightarrow {
Therefore $\frac { ( 1 - \cos \theta ) ^ { 2 } } { 4 } + \frac { ( 1 + \sin \theta ) ^ { 2 } } { 4 } < 1$ , simplifying gives $\sin \theta - \cos \theta < \frac { 1 } { 2 } , \sin \left( \theta - \frac { \pi } { 4 } \right) < \frac { \sqrt { 2 } } { 4 }$ , \\
Also $0 < \theta < \pi$ , that is $- \frac { \pi } { 4 } < \theta - \frac { \pi } { 4 } < \frac { 3 \pi } { 4 }$ , so $- \frac { \pi } { 4 } < \theta - \frac { \pi } { 4 } < \arcsin \frac { \sqrt { 2 } } { 4 }$ , \\
Thus the range of values for $\theta$ is $\left( 0 , \frac { \pi } { 4 } + \arcsin \frac { \sqrt { 2 } } { 4 } \right)$ .
Paper Questions
Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q16