A Geometry Problem

Geometry Level 2

In triangle $DEF$ , $DE = \sqrt{3}$ and $DF = 2$ . What is the value of $\tan x$ ?

\sqrt{3}

\frac{2}{\sqrt{3}}

\frac{1}{2}

2

\frac{\sqrt{3}}{2}

\frac{1}{\sqrt{3}}

1 solution

Terry Yu
May 11, 2017

SOH CAH TOA says that tan(x)= opposite over adjacent. The adjacent can be found using pythagorean theorem which states that if the adjacent side is y, $\sqrt{3}^2+y^2=2^2$ . Then $y^2=4-3=1$ which shows that the adjacent side has a length of 1. When the adjacent is one, tan(x) = $\frac{\sqrt{3}}{1}$ which equals $\sqrt{3}$ .

A Geometry Problem

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