In triangle ABC, the angle is 90 degrees AB = 10 tga = 0.75 find AC.

Find ac in the triangle abc if it is known:

Angle = 90 °;
AB = 10;
tg a = 0.75.

Decision:

1) 1 / cos ^ 2 a = 1 + tg ^ 2 a;

cos ^ 2 a = 1 / (1 + tg ^ 2 a);

cos a = √ (1 / (1 + tg ^ 2 a)) = √ (1 / (1 + 0.75 ^ 2)) = 1 / √ (1 + (3/4) ^ 2) = 1 / √ (1 + 9/16) = 1 / √ (16/16 + 9/16) = 1 / √ (25/16) = 1 / (5/4) = 1/1 * 4/5 = 4/5;

Hence, cos a = 4/5;

2) cos a = AC / AB;

Let us express the AC leg from here.

AC = AB * cos a;

Substitute the known values of the cosine of the angle a and the hypotenuse and calculate the AC leg.

We get:

AC = 10 * 4/5 = 10/5 * 4 = 2 * 4 = 8;

Answer: AC = 8.