In a triangle, 1 of the sides = 12, the other = 10, and the tg of the angle between them = √2 / 4. Find S

Let’s build the height of the ВН in the ABC triangle.

Then in a right-angled triangle ABH tgBAH = BH / AB = √2 / 4.

Let the length of the segment BH = √2 * X cm, then AH = 4 * X cm.

Then, by the Pythagorean theorem, AB ^ 2 = BH ^ 2 + AH ^ 2 = 2 * X ^ 2 + 16 * X ^ 2.

100 = 18 * X ^ 2.

X ^ 2 = 100/18.

X = 10/3 * √2.

Then ВН = √2 * (10/3 * √2) = 10/3.

Then Savs = AC * ВН / 2 = 12 * 10/3 * 2 = 20 cm2.

In triangle a = 8, b = 4, c = 5; find angle alpha

Second way.

We express the sine of this angle through the tangent of the angle BAC.

Cos2BAC = 1 / (1 + tg2BAC) = 1 / (1 + 1/8) = 8/9.

Sin2BAC = 1 – Cos2BAC = 1 – 8/9 = 1/9.

SinBAC = 1/3.

Then Saws = AB * AC * SinBAC / 2 = 10 * 12 * 1/6 = 20 cm2.

Answer: The area of ​​the triangle is 20 cm2.