Triangle ABC cosa = 1 / 2 tanb= ∠c=90°

Triangle ABC cosa = 1 / 2 tanb= ∠c=90°

Because ∠ C = 90 °, the triangle ABC is a right triangle
In right triangle ABC
Cosa = 1 / 2, a is 60 degree
So B is 30 degrees
tanB=tan30°=√3/3

In △ ABC, if cosa = (2 √ 5) / 5, tanb = 1 / 3, find the value of ∠ C

Cos2a + sin2a ≠ 1, because cosa = (2 √ 5) / 5, tanb = 1 / 3, so a and B are acute angles. From (Sina) ^ 2 + (COSA) ^ 2 = 1sina = 1 / √ 5, CoSb = 3 / √ 10, SINB = 1 / √ 10, COSC = cos (π - (a + b)) = - cos (a + b) = - cosacosb + sinasinb = - 5 √ 50 = - 1 / √ 2, so C = 135

In the triangle ABC, cosa = 2 √ 5 / 5, tanb = 1 / 3, find the angle C

cosA=2/√5
sinA=1/√5
tanA=sinA/cosA=1/2
tan(A+B)=(tanA+tanB)/(1-tanAtanB)
=1
A+B=45°
C=180°-(A+B)
=135°

In the triangle ABC, cosa = 2 √ 2 / 5, tanb = 1 / 3, find the size of angle C

If cosa = 2 √ 2 / 5, cos & # 178; if a = 1 / 5, Sina = 2 √ 5 / 5, Tana = 2
Then Tanc = Tan (180 ° - a-b) = - Tan (a + b) = - (Tana + tanb) / (1-tanatanb)
=-5/-5=1
Required ∠ C = 45 °

In the triangle ABC, given cosa = 4 ^ 5, tanb = 5 ^ 12, find the value of COSC

It is known that sina = 3 / 5, SINB = 5 / 13, CoSb = 12 / 13
cosC=-cos(A+B)=sinAsinB-cosAcosB=(3/5)*(5/13)-(4/5)*(12/13)
=-33/65

As shown in the figure. In △ ABC, D is the midpoint of AB, and E is the midpoint of CD. Through point C, make the extension line of CF ‖ AB intersecting AE and connect BF at point F. (1) prove that DB = CF; (2) add a condition in △ ABC______ To make the quadrilateral bdcf______ (fill in: rectangle or diamond)

(1) It is proved that: ∵ CF ∥ AB, ∥ ead = ∥ CFE, ∵ e is the midpoint of CD, ∥ CE = De, ∵ in △ AED and △ FEC, ∥ ead = ∥ CFE ∥ CEF = ∥ deace = ed, ∥ AED ≌ FEC (AAS), ≌ ad = CF, ∥ D is the midpoint of AB, ∥ ad = BD, ∥ BD = cf. (2) add a bar in △ ABC

In triangle ABC, point D is the midpoint of BC, point E is on AB, and AE: EB = 1:2, ad and CE intersect at point F, then what is the ratio of the area of triangle ABC to that of triangle FDC?

DG ‖ AB at g through point d
∵ D is the midpoint of BC, DG ∥ be
The DG is the median line of △ CBE
∴DG＝½BE
∵AE：EB=1：2
∴AE＝½BE
∴AE＝DG
∵DG‖AB
∴∠AEF＝∠DGF,∠EAF＝∠GDF
∵∠AEF＝∠DGF,AE＝DG∠EAF＝∠GDF
∴△AEF≌△DGF
∴AF＝FD
(then you prove that the area of triangle FDC is equal to half the area of triangle ADC, and the area of triangle ADC is equal to half the area of triangle ABC, so the area of triangle FDC is equal to one fourth of the area of triangle ABC.)

In the triangle ABC, ab = AC, BC = BD, ad = AE = EB, find the degree of angle A

Let ∠ abd = x, ∠ CBD = y, because de = EB, so ∠ BDE = abd = x, so ∠ AED = BDE ＋ EBD = 2x, because ad = De, so ∠ a = AED = 2x, because AB = AC, so ∠ C = ABC = x + y, because BC = BD, so ∠ BDC = C = x + y, according to the sum of the inner angles of △ BCD and △ ABC equal to 180 degree, we can get that ∠ abd = x, because AB = AC, so ∠ C = ABC = x + y, because BC = BD, so ∠ BDC = C = x + y, according to the sum of the inner angles of △ BCD and △ ABC equal to 180

In triangle ABC, ad is the middle line, f is the upper point of AD, AF / FD = 1 / 5, connect CF and extend AB to e, calculate AE / EB

Draw a good picture, and make CG / / CE to AB in G
Bgd ~ BEC is obtained
AEF~AGD
So BC / CD = be / Ge = 2
DF/AF=GE/AE=5
So AE / be = AE / Ge * Ge / be = 1 / 5 * 1 / 2 = 1 / 10

The side length of equilateral triangle ABC is a. take point D on the extension line of BC to make CD = 6. Take point E on the extension line of Ba to make AE = a + 6. Try to explain that EC = ed
Can't draw. AC is on the left

You do DF parallel AC intersection be and F
So we get another equilateral triangle BDF
Then we prove that EFD and CAE are congruent by SAS
So EC = ed