Find 'x', if :
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tan 30°=PerpendicularBase⇒13=20x⇒x=20×3⇒x=34.64\text{tan 30°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \dfrac{1}{\sqrt3} = \dfrac{20}{x}\\[1em] ⇒ x = 20 \times \sqrt3\\[1em] ⇒ x = 34.64tan 30°=BasePerpendicular⇒31=x20⇒x=20×3⇒x=34.64
Hence, x = 34.64.
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In the given triangle, the length of AB is :
160 cm
120 cm
60 cm
40 cm
Find angle 'A' if :