Let z_1,z_2,z_3 represent vertices A,B,C of a right angled triangle,right angled at C,if AC = 1 unit and (z_1-z_2)^2 = 2 (z_1-z_3)(z_2-z_3)(-1 + i ) ,find angle A
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2 Answers
1\left| z_{1}-z_{2}\right|^{2}=2\left| z_{1}-z_{3}\right|\left| z_{2}-z_{3}\right|\left|-1+i \right|
AB^{2}=2(AC)(BC)(\sqrt{2})
AB2=2√2BC
AB2=BC2+AC2=BC2+1
from these two :
tanA = BC / AC = BC
66Refer to the diagram ("anti-clockwise" is the key word):
We note that
\angle A = \arg \left(\dfrac{z_3-z_1}{z_2-z_1}\right) and \angle B = \arg \left(\dfrac{z_1-z_2}{z_3-z_2}\right)
The given expression can be re-written as
(z_1-z_2)^2=2(z_3-z_1)(z_3-z_2)(-1+i)
Hence, we get
\dfrac{z_1-z_2}{z_3-z_2}=2 \ \dfrac{z_3-z_1}{z_1-z_2}(-1+i)
i.e.
\dfrac{z_1-z_2}{z_3-z_2}=2 \ \dfrac{z_3-z_1}{z_2-z_1}(1-i)
Taking arguments both the side we get
\arg\left(\dfrac{z_1-z_2}{z_3-z_2}\right)=\arg 2 +\arg \left(\dfrac{z_3-z_1}{z_2-z_1}\right)+\arg(1-i)
But \arg 2 =0 and \arg (1-i)=-\dfrac{\pi}{4}, as such we get
\angle B = \angle A -\dfrac{\pi}{4}
But, \angle B =\dfrac{\pi}{2}-\angle A
So we get
2\angle A = \dfrac{3\pi}{4}\quad \Rightarrow \ \angle A=\dfrac{3\pi}{8}