F(x)'= (1/sinx+cosx)' = <span>= (1/sinx)'+(cosx)' = </span><span>= (1' *sinx- 1* (sinx)')/(sinx)^2 -sinx= = -cosx/(sinx)^2 -sinx</span>
Дано:
AB = 0,5 км
∠CAD = 30°
∠CBD = 45°
__________
h - ?
Решение:
По теореме синусов:
AB / Sin(∠ACB) = AC / Sin(∠ABC) = CB / Sin(∠CAB)
∠ABC = 180° - ∠CBD = 180° - 45° = 135°
∠CAD = ∠CAB
∠ACB = 180° - (∠ABC + ∠BAC) = 180° - (135° + 30°) = 15°
0,5 / Sin(15°) = CB / Sin(30°)
CB = 0,5 * Sin(30°) / Sin(15°) = 0,5 * Sin(30⁰) / (sin(45⁰) cos(30⁰) - sin(30⁰) cos(45⁰)) = 1/4 / ((√3/2 - 1/2)√2/2) = 2 / (4*(√3/2 - 1/2)√2) =1/ ((√6 - √2)/2) = 2 / (√6 - √2)
CB / Sin(∠CDB) = CD / Sin(∠CBD)
∠CDB = 90°
∠CBD = 45°
CD = CB * Sin(∠CBD) / Sin(∠CDB) = 2 / (√6 - √2) * Sin(45°) / Sin(90°) = 2 / (√6 - √2) * √2/2 / 1 = 2 / ((√3 - 1)√2) * √2/2 = (2√2) / (2*(√3 - 1)√2) = 1 / (√3 - 1)
Ответ: 1 / (√3 - 1)
5Х2 - 5 =5(х2 -1) =5(Х-1)(5+1)
2а2 -8 =2(a^2 -4) =2(a -2)(a+2)
3ah^2 - 27a = 3a(h^2 -9)= 3a (h -3)(h +3)
2xy^2 -50x =2x(y^2 -25)=2x(y-5)(y+5)