A) \[\frac{2BC}{3}\]
B) \[\frac{3BC}{4}\]
C) \[\frac{BC}{\sqrt{3}}\]
D) \[\frac{\sqrt{3}BC}{2}\]
Correct Answer: D
Solution :
[d] Given that, \[\angle A=90{}^\circ \]and \[\angle B=30{}^\circ \] In \[\Delta ABC,\] \[\Rightarrow \] \[\angle C=180{}^\circ -90{}^\circ -30{}^\circ \] \[\Rightarrow \] \[\angle C=60{}^\circ \]and \[BC=2AB\] (i) From Pythagoras theorem, \[B{{C}^{2}}=A{{C}^{2}}+A{{B}^{2}}\] \[\Rightarrow \] \[{{(2\,AB)}^{2}}=A{{C}^{2}}+A{{B}^{2}}\] [from Eq. (i)] \[\Rightarrow \] \[AC=\frac{\sqrt{3}}{2}.BC\] |
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