In triangle ABC, angle C = 90 °. sinB = 1 / √2. Find tgA.

In triangle ABC, angle C = 90 °.

Find tg A if sin B = 1 / √2 is known.

In a right-angled triangle sin b = cos a, so cos a = 1 / √2;

Knowing cos a, we find sin a using the trigonometric formula sin ^ 2 a + cos ^ 2 a = 1.

Substitute the known value of the cosine a, we find the value of the sine of the angle a.

sin ^ 2 a = 1 – cos ^ 2 a;

sin a = √ (1 – (1 / √2) ^ 2) = √ (1 – 1 / √4) = √ (1 – 1/2) = √ (2/2 – 1/2) = √ (1 / 2) = 1 / √2;

Now we find tg a by the formula tg a = sin a / cos a;

tg a = (1 / √2) / (1 / √2) = 1 / √2 * √2 / 1 = √2 / √2 = 1/1 = 1;

Answer: tg a = 1.