The value of \(x\) is 2.3.
Given:
The figure of a right-angle triangle.
To find:
The value of \(x\).
Explanation:
Let us mark the vertices as shown in the attached figure.
In triangle ABD,
\(\sin \theta=\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}\)
\(\sin A=\dfrac{BD}{AB}\)
\(\sin 30^\circ=\dfrac{2}{AB}\)
\(\dfrac{1}{2}=\dfrac{2}{AB}\)
On cross multiplication, we get
\(AB=4\)
In triangle ABC,
\(\tan \theta=\dfrac{\text{Perpendicular}}{\text{Base}}\)
\(\tan A=\dfrac{BC}{AB}\)
\(\tan 30^\circ=\dfrac{x}{4}\)
\(\dfrac{1}{\sqrt{3}}=\dfrac{x}{4}\)
\(\dfrac{4}{\sqrt{3}}=x\)
\(x\approx 2.3\)
Therefore, the value of \(x\) is 2.3.
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