In triangle ABC, if AČ =2i -2j and BC = 4i then show that angleBAC = 2
angleABC.
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Step-by-step explanation:
In right angled triangle ABC
z
1
−z
3
z
2
−z
3
=
AC
BC
e
iπ/2
⇒
(z
1
−z
3
)
(z
2
−z
3
)
=i (∵BC=AC)
⇒(z
2
−z
3
)=i(z
1
−z
3
) squaring
(z
2
−z
3
)
2
=−(z
1
−z
3
)
2
⇒z
2
2
+z
3
2
−2z
2
z
3
=−z
1
2
−z
3
2
+2z
1
z
3
⇒z
1
2
+z
2
2
−2z
1
z
2
=2z
1
z
3
+2z
2
z
3
+2z
2
z
3
−2z
1
z
2
−2z
3
2
⇒(z
1
−z
2
)
2
=2(z
1
−z
3
)(z
3
−z
2
)
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