wewake06298

In 45-45-90 Right triangle, the sides are in ratio \(x : x : \sqrt{2}x\)

i.e. opposite to 45 degree angle, side would be \(x\) whereas opposite to angle 90, side would be \(\sqrt{2}x\)

In the question,
\(BC = 3 = \sqrt{2}x\)

\(x = \frac{3}{\sqrt{2}}\)

Therefore, AB = AC = \(x = \frac{3}{\sqrt{2}}\)

Hi wewake06298

So in Triangle ABC, Angle ABC = angle ACB = 45°, and angle BAC = 90°

Now this is 45-45-90 triangle. so sides are in ratio of 1:1:\(\sqrt{2}\)
WHY?

in triangle ABC , sin C = \(\frac{AB}{BC}\) & cos C = \(\frac{AC}{BC}\)
Now we know that angle C = 45° so sin C = sin 45° = \(\frac{1}{\sqrt{2}}\)

\(\frac{1}{\sqrt{2}}\) = \(\frac{AB}{BC}\) = \(\frac{AC}{BC}\)

AB = AC = \(\sqrt{2}\)BC
so they are in proportion of 1:1:\(\sqrt{2}\)

And if one side given is BC = 3 so other will be \(\frac{3}{\sqrt{2}}\)

Ask if the doubt still remains.
Also please check this : GRE MATH book Special Triangle theory prepared by Carcass

Regards,
R

Notice triangle ABC is 45-45-90 triangle. The sides are always in fixed ratio \(1 : 1 : \sqrt{2}\).
I assume you are not able to understand how to use this fixed ratio in order to find the length of sides for the triangle ABC.

However, this is not very difficult to understand. You may write the fixed ratio as \(k : k : \sqrt{2}*k\).
Sides opposite to 45° are equal, and the length is k here. The side opposite to 90° is \(\sqrt{2}*k\).

Now BC is opposite to 90°, so \(BC = \sqrt{2}*k\)
We already know BC = 3.
Equating both equations for BC,
\(\sqrt{2}*k = 3\)
\(k = \frac{3}{\sqrt{2}}\)

Therefore, the sides opposite to 45°, AC and AB is \(\frac{3}{\sqrt{2}}\).