(i) notice that the angle AKD-AKC=θ
=> tan(θ)=tan(AKD-AKC)=(tan(AKD)-tan(AKC))/(1+tan(AKD)tan(AKC))
(where AKD and AKC are angles) and AKD and AKC are right triangles
=> tan(AKD)=(4+a)/x, tan(AKC)=a/x
=> tan(θ)=((4+a)/x-a/x)/(1+((4+a)/x)a/x)=(4/x)/((x^2+4a+a^2)/x^2)=4x/(x^2+4a+a^2)
#Please help
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where we used the identity tan(a-b)=(tan(a)-tan(b))/(1+tan(a)tan(b))
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(ii) simply set the derivative of the function y=tan(θ)=4x/(x^2+4a+a^2) to zero (like you would normally find the maximum point):
y'=4(-x^2+4a+a^2)/(x^2+4a+a^2)^2=0
=> (x^2+4a+a^2)^2≠0 => x≠+-√(-a(a+4))
=> -x^2+4a+a^2=0
=> x=+-√(a^2+4a), a^2+4a≠0 but since we only consider x,a>=0 and we should give the optimal point in terms of a, x=√(a^2+4a) is the optimal point. The corresponding value:
tan(θ)=4x/(x^2+4a+a^2)=4(√(a^2+4a))/((√(a^2+4a))^2+4a+a^2)=4(√(a^2+4a))/(a^2+4a+4a+a^2)=2√(a^2+4a)/(a^2+4a)=2/√(a^2+4a)
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(iv) tan(KDA)=x/(4+a)=√(a^2+4a)/(4+a)=√((a(a+4))/(4+a)^2)=√(a/(a+4)) and the approximate value of KDA will be 0.785 (rad) (just plug a big number for a into the calculator)
(v) the maximum value of θ is at A=C. There tan(θ)=4/x => θ<=atan(4/x).
The minimum value of θ is at A=X. There tan(θ)=4x/(x^2+4a+a^2) and since a=(50-4)/2=23 => tan(θ)=4x/(x^2+4(23)+23^2)=4x/(x^2+621)
=> θ>=atan(4x/(x^2+621))
=> x∈[atan(4x/(x^2+621)); atan(4/x)]
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@cedar abyss
cedar abyss
Tysm
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You're welcome