Find angle 'x' if :
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In Δ ABC,
tan 60°=PerpendicularBase⇒3=CBAC⇒3=30AC⇒AC=303⇒AC=30×33⇒AC=103\text{tan 60°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \dfrac{CB}{AC}\\[1em] ⇒ \sqrt3 = \dfrac{30}{AC}\\[1em] ⇒ AC = \dfrac{30}{\sqrt3}\\[1em] ⇒ AC = \dfrac{30 \times \sqrt3}{3}\\[1em] ⇒ AC = 10\sqrt3tan 60°=BasePerpendicular⇒3=ACCB⇒3=AC30⇒AC=330⇒AC=330×3⇒AC=103
In Δ ADC,
sin x=PerpendicularHypotenuse⇒sin x=ACAD⇒sin x=10320⇒sin x=32⇒sin x=sin 60°\text{sin x} = \dfrac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \text{sin x} = \dfrac{AC}{AD}\\[1em] ⇒ \text{sin x} = \dfrac{10\sqrt3}{20}\\[1em] ⇒ \text{sin x} = \dfrac{\sqrt3}{2}\\[1em] ⇒ \text{sin x} = \text {sin 60°}sin x=HypotenusePerpendicular⇒sin x=ADAC⇒sin x=20103⇒sin x=23⇒sin x=sin 60°
Hence, x = 60°.
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Find angle 'A' if :
Find AD, if :