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Given a ^@ \Delta ABC^@ in which ^@\angle B = 90^\circ^@ and ^@AB = \sqrt{3} BC^@. Prove that ^@\angle C = 60^\circ^@.
Answer:
- Let ^@D^@ be the midpoint of the hypotenuse ^@AC^@.
Join ^@BD^@. - Now, we have @^ \begin{aligned} &AC^2 = AB^2 + BC^2 &&[\text{ By pythagoras' theorem }] \\ \implies &AC^2 = (\sqrt{ 3 } BC)^2 + BC^2 &&[\because \text{ AB = } \sqrt{3 } \text{ BC (given) } ] \\ \implies &AC^2 = 4BC^2 \\ \implies& AC = 2BC \\ \implies & 2CD = 2BC &&[\because \text{ D is the midpoint of AC}] \\ \implies & CD = BC &&\ldots\text{ (i) } \end{aligned} @^
- We know that the midpoint of the hypotenuse of a right triangle is equidistant from the vertices. @^ \begin{aligned} \therefore \space BD = CD &&\ldots\text{ (ii) } \\ \end{aligned} @^ From ^@\text{(i)}^@ and ^@\text{(ii)}^@, we get @^ BC = BD = CD @^ Therefore, ^@ \Delta BCD^@ is equilateral and hence ^@\angle C = 60^\circ^@.
