If △ ABC ∽ def, ab = 2, BC = 2 times radical 3, de = radical 2, then EF is emergency

If △ ABC ∽ def, ab = 2, BC = 2 times radical 3, de = radical 2, then EF is emergency

Analysis: because △ ABC ∽ def
So AB: BC = de: ef
That is, 2:2 times root 3 = root 2: EF
So EF = root 6

Triangle ABC and triangle def meet the following conditions to judge whether triangle ABC and triangle def are similar, BC = 2, AC = 4: EF = radical 2, DF = radical 3, de = radical 4
It's AB = 4, AC = 3

Because AB = 4, AC = 3, BC = 2; de = radical 4, DF = radical 3, EF = radical 2
So AB / de = 4 / radical 4 = 2,
AC / DF = 3 / radical 3 is not equal to 2,
So triangle ABC is not similar to triangle def

In △ ABC, if acosa = bcosb, judge the shape of △ ABC

∵ cosa = B2 + c2-a22bc, CoSb = A2 + c2-b22ac, ∵ B2 + c2-a22bc · a = A2 + c2-b22ac · B, which is reduced to: a2c2-a4 = b2c2-b4, that is, (A2-B2) C2 = (A2-B2) (A2 + B2), if A2-B2 = 0, a = B, then △ ABC is isosceles triangle; if A2-B2 ≠ 0, A2 + B2 = C2, this

In △ ABC, if | Sina - radical 3 / 2 | + (radical 3-3tanb) ^ 2 = 0, ∠ a, ∠ B are acute angles, find the degree of ∠ C

|Sina - radical 3 / 2 | + (radical 3-3tanb) ^ 2 = 0
Sina radical 3 / 2 = 0, radical 3-3tanb = 0
Sina = (radical 3) / 2, tanb = (radical 3) / 3,
And ∠ a, B are acute angles
A=60°,B=30°
So C = 90 degrees

If the acute angle A. B satisfies (1 + root 3tana) (1 + root 3tab) = 4, then a + B=

(1 + root 3tana) (1 + root 3tanb) = 4
1 + radical 3 (Tana + tanb) + 3tana * tanb = 4
Root 3 (Tana + tanb) + 3tanatanb = 3
Tana + tanb + root 3tanatanb = root 3
Tana + tanb = radical 3 (1-tanatanb) = 0
Tan (a + b) = (Tana + tanb) / (1-tanatanb) = radical 3
B, then 0

In triangle ABC, a = 80 degree, the degree of B is 4 times of C, the degree of B is?, the degree of C is?

A+B+C=180°
B=4C
A=80
B=80
C=20

In the triangle ABC, a = 80 °, the degree of B is three times that of C, B = () ° and C = ()

Because ∠ a = 80 degrees, so ∠ B + C = 180-80 = 100 degrees
Because ∠ B = 3 ∠ C, 4 ∠ C = 100, i.e. ∠ C = 25 degree ∠ B = 75 degree

Triangle ABC C = 2 radical 2 a > b c = π / 4 tanatan B = 6 find a, s Triangle
Such as the title
Sina = 3 / radical 10, be more specific

In the triangle ABC, a + B + C = 180 ',
A＋B＝180’-C,
tan(A+B)=-tanC=-1,
tan(A+B)=tanA+tanB/1-tanAtanB,tanA+tanB=5,
a> B, and tanatanb = 6, Tana = 3, tanb = 2, Sina = 3 / radical 10, SINB = 2 / radical 5, sinc = 1 / radical 2, C = 2 / radical 2,
According to the cosine theorem, a = 6 radical 10 / 5, B = 8 radical 5 / 5, s = 24 / 5

Finding the area of triangle ABC by three points a (4,6) B (0,2) C (6,0) in the plane

Cut complement method, you make a projection of point a on the y-axis, record it as a ', then a'aco (o is the origin), and then calculate the area of trapezoid a'aco, triangle OBC and triangle a'ab. the area of trapezoid a'aco - Triangle OBC - a'ab is the area of ABC

Given the points a (4.5,5), B (6,0) C (- 2,0), find the area of triangle ABC

Draw a coordinate diagram, BC is the length of the triangle, a ordinate is the height of the triangle, according to the area formula, it can be calculated