Edit: Ok the question has now been amended by my tutor as I highlighted it was not possible to solve.
A pole is tensioned at point A as shown below, if the angles are to be kept the same, then using a suitable double angle identity determine the distance $d$:
Edit: Note that triangle ACB is not a right angled triangle. Should now read: Note that triangle ACD is not a right angled triangle.
By using $$\cos 2x=2\cos^2 x-1$$My working is:$$\cos x=\frac d3$$$$\cos 2x=\frac d3$$therefore:$$\frac d3=2(\frac d3)^2-1$$
Now becomes solvable as a quadratic equation
$\endgroup$ 73 Answers
$\begingroup$I will interpret the problem as "given a point $C$ on a segment $BD$ such as $BC=b$ and $CD=c$, what is the distance $d$ from $A$ to $B$ such that $BC$ and $CD$ have the same angular diameter?"
Using the law of cosines in triangles $ABC$ and $ACD$ we get$$c^2=2^2+3^2-2\times 2\times 3 \cos x$$$$b^2=d^2+2^2-4d\cos x$$so that $$\cos x=\dfrac{13-c^2}{12}$$ and $$b^2=d^2+4-\dfrac{4d(13-c^2)}{12}$$The last equation is quadratic in $d$ and easy to solve.
$\endgroup$ 1 $\begingroup$To solve the quadratic equation when in the form $ax^2+bx+c=0$ you can use the formula:$$x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a}$$
$\endgroup$ $\begingroup$$\cos(x)\neq \frac{2}{3}$ because the angle $\widehat{ACD} \neq \frac{\pi}{2}$.
$\endgroup$ 0
