answer:To solve for the eccentricity of the hyperbola given the conditions, we follow these steps:1. Identify the equation of the asymptotes: The asymptotes of the hyperbola frac{x^2}{a^2}-frac{y^2}{b^2}=1 are given by bx pm ay = 0.2. Determine |FQ|: The distance from the focus F to the point Q on the asymptote can be found using the formula |FQ|=frac{|bc|}{sqrt{a^2+b^2}}. Given that |FQ|=b, we have frac{|bc|}{sqrt{a^2+b^2}} = b.3. Use the right triangle QOF: In the right triangle formed by Q, O (the center of the hyperbola), and F, we can express cos angle QFO as cos angle QFO = frac{b}{c}.4. Define |QN| and |QM|: Let |QN|=t, then according to the problem, |QM|=3t, and thus |FN|=b-t.5. Apply the definition of the hyperbola: For the distances to the foci, we have |NF'|=b-t+2a and |MF'|=b+3t-2a.6. Calculate cos angle NFF' and cos angle MFF': Using the cosine rule in triangles FNF' and FMF', we get: - cos angle NFF' = frac{4c^2+(b-t)^2-(b-t+2a)^2}{2 times 2c(b-t)} - cos angle MFF' = frac{4c^2+(b+3t)^2-(b+3t-2a)^2}{2 times 2c(b+3t)}7. Solve for t: Equating the expressions for cos angle QFO with cos angle NFF' and cos angle MFF', we find: - From cos angle QFO and cos angle NFF', solving gives t=frac{ab}{3b-3a}. - From cos angle QFO and cos angle MFF', solving gives t=frac{ab}{a+b}.8. Find the relationship between a and b: Equating the two expressions for t, we get a+b=3b-3a, which simplifies to b=2a.9. Calculate the eccentricity e: The eccentricity of the hyperbola is given by e=frac{c}{a}. Using the relationship between a and b, we find: [ e = sqrt{1+frac{b^2}{a^2}} = sqrt{1+4} = sqrt{5} ]Therefore, the eccentricity of the hyperbola is boxed{sqrt{5}}.

answer:Solution. a) The case alpha=pi is trivial (then x=pi). In the case where 0<alpha<pi, let E be the foot of the perpendicular from point S to AB (diagram on the right). From the right triangles SED and SEB we get operatorname{tg} frac{x}{2}=frac{ED}{SE} and operatorname{tg} frac{alpha}{2}=frac{EB}{SE}, from which it follows that operatorname{tg} frac{x}{2}=frac{1}{3} operatorname{tg} frac{alpha}{2}. Therefore,cos x=frac{1-operatorname{tg}^{2} frac{x}{2}}{1+operatorname{tg}^{2} frac{alpha}{2}}=frac{1-frac{1}{9} operatorname{tg}^{2} frac{alpha}{2}}{1+frac{1}{9} operatorname{tg}^{2} frac{alpha}{2}}=frac{9-frac{1-cos alpha}{1+cos alpha}}{9+frac{1-cos alpha}{1+cos alpha}}=frac{4+5 cos alpha}{5+4 sin alpha}since by assumption 1+cos alpha neq 0.b) Let cos frac{alpha}{3}=t. Thencos x-cos frac{alpha}{3}=frac{4+5 cos alpha}{5+4 sin alpha}-cos frac{alpha}{3}=frac{4+5left(4 t^{3}-3 tright)}{5+4left(4 t^{3}-3 tright)}-t=frac{-4(t-1)^{2}(t+1)(4 t-1)}{16 t^{3}-12 t+5}c) Since according to the conditions of the problem 0<alpha leq pi, or frac{1}{2} leq t<1, the only possibility for cos x-cos frac{alpha}{3}=0 to hold is 4 t-1=0, i.e., t=frac{1}{4}, from which we get alpha=3 arccos frac{1}{4}.

answer:Reflect each of the triangles over its respective side. Then since the areas of the triangles total to the area of the equilateral triangle, it can be seen that the triangles fill up the equilateral one and the vertices of these triangles concur at the circumcenter of the equilateral triangle. Hence the desired answer is just its circumradius, or boxed{dfrac{sqrt3}3textbf{ (B)}}.(Solution by djmathman)

answer:7. sqrt{frac{3-sqrt{5}}{2}}.Since the equation has a repeated root, Delta=0, i.e., left(tan ^{2} theta-sin ^{2} thetaright)^{2}+cos 2 thetaleft(tan ^{2} theta+sec ^{2} thetaright)=0. Let d=cos ^{2} theta. Then left(frac{1-d}{d}+d-1right)^{2}+(2 d-1)left(frac{1-d}{d}+frac{1}{d}right)=0. Therefore, left(d^{2}-3 d+1right)^{2}=0.Solving for d, we get d=frac{3 pm sqrt{5}}{2}.Since theta inleft(frac{pi}{4}, frac{pi}{2}right), we have d<frac{1}{2}.Thus, d=frac{3-sqrt{5}}{2}.Therefore, cos theta=sqrt{frac{3-sqrt{5}}{2}}.

answer:Consider the dual form B=sin ^{2} 10^{circ}+ sin ^{2} 50^{circ}-cos 40^{circ} cos 80^{circ} of the original expression A=, then A+B=2- cos 40^{circ}, A-B=frac{1}{2}+cos 40^{circ}.Thus, A=frac{3}{4}.

answer:15. (1) When x=0, the equation becomesy^{3}-y^{2} z=0 Rightarrow y^{2}(y-z)=0Rightarrow y=0 or y=z.Thus, (0,0, m),(0, m, m)left(m in mathbf{Z}_{+}right) are solutions that satisfy the problem.(2) When y=0, similarly,(0,0, m),(m, 0, m)left(m in mathbf{Z}_{+}right)are solutions that satisfy the problem.(3) When x, y in mathbf{Z}_{+}, let a=x+y, b=x y.Then the original equation becomesb^{2}+3 a b-a^{2}(a-z)=0 text {. }Considering that a, b, z are all integers, thus, the discriminantDelta=9 a^{2}+4 a^{2}(a-z)=a^{2}(4 a+9-4 z)is a perfect square.Since 4 a+9-4 z is odd, we can assume4 a+9-4 z=(2 t+1)^{2}(t geqslant 0) text {. }Then a=t^{2}+t+z-2, b=frac{-3 a+sqrt{Delta}}{2}=a(t-1).Since a, b in mathbf{Z}_{+}, we know t geqslant 2.Next, consider the square numberbegin{array}{l}(x-y)^{2}=(x+y)^{2}-4 x y=a^{2}-4 b =a^{2}-4 a(t-1) . text { Then }[a-2(t-1)-2]^{2} leqslant a^{2}-4 a(t-1) <[a-2(t-1)]^{2},end{array}andbegin{array}{l}a^{2}-4 a(t-1) neq[a-2(t-1)-1]^{2} . text { Hence } a^{2}-4 a(t-1)=[a-2(t-1)-2]^{2} Rightarrow a=t^{2} Rightarrow t+z=2 .end{array}Therefore, t=2, z=0.Thus, a=b=4, x=y=2.At this point, the solution to the equation is (2,2,0).In summary, all non-negative integer solutions to the equation are(2,2,0),(0,0, m),(0, m, m),(m, 0, m) text {, }where m is any non-negative integer.(Provided by Guo Min)