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Given a Delta ABC in which angle B=90^(@...

Given `a Delta ABC` in which `angle B=90^(@) and AB= sqrt(3) BC`. Prove that `angle C=60^(@)`

Text Solution

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Let D be the midpoint of the hypotenuse AC.
Join BD.
We have
`AC^(2)=AB^(2)+BC^(2)` [ by Pythagoras theorem]
`rArr AC^(2)=(sqrt(3)BC)^(2)+BC^(2) [ :. AB=sqrt(3)BC "given" ]`
`rArr AC^(2)=4BC^(2) rArr AC=2BC`
`rArr 2CD=2BC [ :. D` is the midpoint of AC]
`rArr CD= BC" ...(i)`

Also, we know tha the midpoint of the hypotenus of a right triangle is equidistant from the vertices
` :. BD=CD" "....(ii)`
From (i) and (ii), we get
`BC=BD=CD`
`:. Delta BCD` is equilateral and hence `angle C=60^(@)`
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