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# In the figure alongside , AB = AC , $\;\angle{A}=48^{0}\;$ and $\;\angle{ACD} = 18^{0}\;$ , BC equal to

$(a)\;AC\qquad(b)\;CD\qquad(c)\;BD\qquad(d)\;AB$

AB = AC , $\;\angle{A}=48^{0}\;$ and $\;\angle{ACD} = 18^{0}\;$ , BC =CD