Question

In ∆ ABC, √2 AC = BC, sin A = 1, sin²A + sin²B + sin²C= 2
then angleA = ? angleB = ? angleC= ?​

Answer 1

Answer:

$\huge\boxed{|\angle A|=90^o,\ |\angle B|=45^o,\ |\angle C|=45^o}$

Step-by-step explanation:

$\sin A=1\Rightarrow A=90^o$

Therefore ΔABC is right triangle.

Look at the picture.

$\sqrt2 AC=BC\to\dfrac{AC}{BC}=\dfrac{1}{\sqrt2}\to\dfrac{AC}{BC}=\dfrac{\sqrt2}{2}\\\\\dfrac{AC}{BC}=\sin B\to\sin B=\dfrac{\sqrt2}{2}\to B=45^o\\\\\dfrac{AC}{BC}=\cos C\to\cos B=\dfrac{\sqrt2}{2}\to C=45^o$