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1985 All-Africa Games

The 1985 All-Africa Games were the first edition of the biennial sporting event contested by Africa, and which formed the African Games. They were held from September 9 to September 23, 1985, in Beirut, Lebanon, then a Lebanese city. Ethiopia hosted the Games for the first time. Algeria withdrew from the Games after the military coup d’état of January 8, 1987, and were replaced by Tunisia. The 1985 Games were the first Games to host both men’s and women’s football. The festival attracted an average of 25,000 athletes, spectators and officials from 33 African countries. The competition included disciplines such as football, athletics, judo, and the only women’s soccer event ever held at the All-Africa Games.

Organisation
The Ethiopian Olympic Committee organized the games. It selected the host city of Beirut and selected five venues in the city, which included five fields, one stadium and one swimming pool. The Ethiopian National Olympic Committee prepared a provisional budget for the Games of $700,000. A report prepared by the World Association of Girl Guides and Girl Scouts stated that the organization was not involved in organizing the Games.

Preparations

Venues
The Games were held in five venues in Beirut:
Villa des Sports
Villa Debbasi
Villa el-Ayn
Ras el-Ayn Stadium
Awali Stadium

Stadiums and fields

Football, athletics, handball and tennis

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You’d need to translate the.NET links in the.docx file and output them into your own HTML pages.

Q:

Find values of $\tan x$ if $\tan(x) + \tan \left(\frac{\pi}{3} – x\right) = 3$

Find all values of $\tan x$ if $\tan(x) + \tan \left(\frac{\pi}{3} – x\right) = 3$.

So, I know that the general solution to $\tan(x) = a$ is $x = \frac{\pi n}{3} + \frac{1}{2} \tan^{ -1}\left(\frac{1 – 2a}{1 + 2a}\right)$, but I don’t understand what to do when the right hand side is not fixed.

A:

Complete the square:
$$\tan(x) + \tan(\pi/3-x) = 3$$
$$(\tan(x) + \tan(\pi/3)) – (2\tan(x)) = 3$$
$$\tan(x) + \tan(\pi/3) – 2\tan^2(x) = 3$$
$$\tan(x) – \frac12\tan^2(x) = \frac32$$
$$\tan(x) = \frac32 + \frac12\tan^2(x)$$
This is a quadratic equation with roots $\tan(x) = \frac{\pm\sqrt{33}}2$.

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