In RT △ ABC, ∠ C = 90 ° AC · BC = quarter AB ^ 2, what is the degree of ∠ a

In RT △ ABC, ∠ C = 90 ° AC · BC = quarter AB ^ 2, what is the degree of ∠ a

In RT △ ABC, ∠ C = 90 ° and AC · BC = one fourth AB ^ 2, then ∠ a degree is 75 degree

In RT △ ABC, ∠ C = 90 °, BC = a, AC = B, and 3a = 4b, then the degree of ∠ a?

∵b=3a/4
tan∠A=a/b
=a/(3a/4)
=4/3.
∠A=arctan(4/3).
∠ A ≈53.13°.

As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC + BC = 17, ab-bc = 8, find the values of cosa and tanb
How to bring it to Pythagorean theorem?

AC = 25-ab is obtained by adding AB ^ 2 = AC ^ 2 + BC ^ 2, AC + BC = 17, ab-bc = 8 to AC = 25-ab. substituting AC = 25-ab, BC = AB-8 into AB ^ 2 = AC ^ 2 + BC ^ 2, AB ^ 2-66ab + 689 = 0 (ab-33) ^ 2 = 400ab = 13 or 53 (rounding off, because AC = 25-ab > 0), so AC = 12, BC = 5, so cosa = AC / AB = 12 / 1

In RT △ ABC, C = 90 ° B = 45 ° then AB: AC: BC

This is an isosceles right triangle. According to Pythagorean theorem, AB: AC: BC = root 2:1:1