Question

1) In the figure, ZABC = 90°, ZBAC = ZBCA 90°, ZBAC = LBCA = 45°. I AC = 2v2, then find AB.​

Answer 1

Solution:-

Given,

AC = 2√2 and angles of the ∆ABC

From the above information,

∆ABC is right angled triangle

We know that,

$\sin \theta = \frac{opposite \: side}{hypotenuse}$

So,

$\sin \: C = \frac{AB}{AC}$

$\sin \: 45 \degree = \frac{AB}{2 \sqrt{2} }$

$\frac{1}{ \sqrt{2} } = \frac{AB}{2 \sqrt{2} }$

$AB = \frac{2 \sqrt{2} }{ \sqrt{2} }$

$AB = 2$

The length of side AB is 2

Answer 2

Answer:

this is you answer

Step-by-step explanation:

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