In ABC, if ABP-ACCB", state with reason whether ABC as a night
angled triangle or not
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We know that
a
2
+b
2
=p
2
AB
.
AC
+
BC
.
BA
+
AC
.
CB
=0
CA
.
CB
=0
AB
⊥
AC
∣
∣
∣
∣
AB
∣
∣
∣
∣
∣
∣
∣
∣
AC
∣
∣
∣
∣
cosθ+
∣
∣
∣
∣
BC
∣
∣
∣
∣
∣
∣
∣
∣
BA
∣
∣
∣
∣
cos(90−θ)+0
pbcosθ+apcos(90−θ)
pbcosθ+apsinθ
p(bcosθ+asinθ)
p(pcosθcosθ+psinθsinθ)
=p
2
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