tan A=-√3 tan(π-A)=√3？ π-A=60 =π/3 A=2π/3
Option d
2. The range of inclination angle is [0, 180].
When 0 ≤ m 2 ≤ 1 and m 2 = 0, the inclination of a straight line is tan a = (0-1)/(1-2) =1,so a = 45.
When m 2 gradually increases to 1, the inclination angle a of the straight line gradually decreases to 0.
When m2 > 1, the inclination angle decreases from 180 (because the inclination angle cannot reach 180, a cannot be equal to 180).
Simplify to M 2 close to infinity, M 2 close to infinity, that is, the straight line is infinitely close to vertical, but it can't reach 90.
So when m 2 >; 1 has (π/2, π), so choose D.
3. Two straight lines are perpendicular to each other, and the product of slopes is-1.
Because the slope of the straight line ax+2y+ 1=0 is -a/2, and the slope of the straight line x+y-2=0 is-1.
So there is
? -a/2×- 1=- 1？ -a/2= 1？ a=-2
So choose d
4. the equation of the circle is x? +y？ -4x-4y- 10=0, and the deformation is (x-2)? +(y-2)？ = 18.
So the center of the circle is a (2,2), the radius is 3 √ 2, and the inclination angle of OA is 45.
Because that distance from at least three different point on the circle to the straight line is 2√2,
Therefore, the distance from the center of the circle A (2,2) to the straight line L is not more than 3√2-2√2=√2.
Take the form that the distance is just √2, and let AH⊥l be on h and l be on o (because the equation of straight line l is y=(-a/b)x, passing through the origin o),?
? AH=√2，OA=2√2
So ∠ AOH = 30 (the angle of the right triangle in which the opposite side is half of the hypotenuse is 30).
? So the inclination angle of the straight line L = 45 ∠ AOH =15 or 75.
That is, the dip angle range is [π/ 12, 5π/ 12], and b is selected.
5. Let the equation of a straight line be y=kx+b, the slope be 1, that is, k= 1, and the intercept be a, so b=a, so the equation of a straight line is
Y=x+a？ x-y+a=0
The equation of a circle is x 2+y 2 = 2, the radius is √2, and the center of the circle is (0,0). Using the distance formula from point to straight line,
? |0-0+a|/[(√( 1^2+ 1^2)]=|a|/√2=√2
? So | a | = 2 and a = 2? Choose C.
6. The moving straight line L passes through the fixed point A and is perpendicular to AB, so the moving straight line can only rotate around the fixed point A, which meets the conditions.
So the area where the straight line L rotates is a plane, which intersects with the plane α. As we know, these two planes intersect in a straight line.
So the answer is a.
7. The topic means to make a circle with a radius of 1 and a center at the origin. Draw two tangents of the circle from point A (-2,0).
Just make sure that point B is outside the angle formed by two tangents, that is, outside the "blind zone" of point A.
Let one of the tangents be y=kx+b, because the tangents pass through point A (-2,0), so -2k+b = 0 and b = 2k.
So the tangent equation is y=kx+2k, that is, kx-y+2k=0. According to the distance formula from point to straight line, there are
|2k|/√[k^2+(- 1)^2]= 1
? 4k^2=k^2+ 1
k^2= 1/3
k= √3/3
? Take a k=√3/3.
y=(√3/3)x+2√3/3？ When x=2, y=4√3/3
? According to the symmetry, when the other tangent is x=2, y=-4√3/3.
? Point B is outside the blind area, so the value range of A is (-∞, -4√3/3)∩(4√3/3,+∞), and C is selected.
8. When is the shortest arc? It is the longest distance to the center of the circle, that is, om is perpendicular to the straight line L.
? Let the equation of the straight line L be y-2=k(x- 1), where
? kx-y-k+2=0
The center o (2,0), M( 1, 2), so the slope of the straight line OM is kOM=2/- 1=-2.
? So the slope k of the straight line L =-1/-2 =1/2.
So the equation of a straight line is? X/2-y+3/2=0, that is, x-2y+3=0.
9. Let x=0, and the coordinates of point A and point B are (0,4) and (0,0) respectively, so AB=4, let point A be (0,4) and point B be (0,0).
Because there are two P points intersecting the X axis, let the left point be P 1 and the right point be P2, so
The radius of a circle is 4. Diameter 8, so sin? ∠AP2B=4/8= 1/2？ ∠AP2B=30
The other P 1 is on the lower arc AB, because the sum of the circumferential angle of one arc on the lower arc and the circumferential angle on the upper arc is 180.
So ∠ AP1b =180-∠ AP2B =180-30 =150.
So the answer is 30 or 150.
It's annoying to do it ......
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