Consider a right angle triangle ABC right angle at B such that AC=sqrt8+4sqrt3 and AB=1
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Step-by-step explanation:
AC = 1 :
2
⇒
AC
AB
=
2
1
∴ AB = x and AC =
2
x, for some x.
By Pythagoras theorem, we have
AC
2
=AB
2
+BC
2
⇒(
2
x)
2
=x
2
+BC
2
⇒BC
2
=2x
2
−x
2
=x
2
⇒BC=x
∴tanA=
AB
BC
=
x
x
=1
we have,
1+tan
2
A
2tanA
=
1+1
2×1
=
2
2
, = 1
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