In the isosceles 3-angle ABC, ab = AC, if AB = 2BC, find tanb

In the isosceles 3-angle ABC, ab = AC, if AB = 2BC, find tanb

Let AB be 4, then BD be 1. According to Pythagorean theorem, ad is radical 15, so tanb = ad / BD = radical 15

In the triangle ABC, if ∠ C = 90 °, BC = 4, ab = 5, then tanb=

tanB= AC/BC= 3/4

In the triangle ABC, Tana = 1 / 4, tanb = 3 / 5, ab = root 13, find the length of BC
In the triangle ABC, Tana = 1 / 4, tanb = 3 / 5, ab = root 13, find the length of BC
That BC = what, I don't know

Make the perpendicular of AB from C, and the perpendicular foot is h
tanA=CH/AH=1/4,CH=AH/4
tanB=CH/BH=3/5,CH=3*BH/5
So ah / 4 = 3 * BH / 5
AH=12BH/5
AH+BH=AB=13
12BH/5+BH=13
BH=65/17
CH=3*BH/5=39/17
BC^2=BH^2+CH^2=(65^2+39^2)/17^2
BC=13*sqrt(34)/17

As shown in the figure, in △ ABC, ∠ C = 90 °, AC + BC = 9, tanb = 2, find AB and ∠ a, ∠ B

Tanb = AC / BC because AC + BC = 9, so AC = 6, BC = 3, ab = 3 radical, 5 angle, a = arctan0.5 angle, B = arctan2

Given that a, B and C are the three sides of the triangle and satisfy a ^ 2 + B ^ 2 + C ^ 2-ab-ac-bc = 0, try to judge the shape of the triangle ABC

Let the equation x2 be 2A ^ 2 + 2B ^ 2 + 2C ^ 2-2ab-2ac-2bc = 0
Formula: (a-b) ^ 2 + (A-C) ^ 2 + (B-C) ^ 2 = 0
We get a = b = C
So it's an equilateral triangle