2tan15° The value of 1 − tan215 ° is equal to______ .

2tan15° The value of 1 − tan215 ° is equal to______ .

2tan15°
1−tan215°=tan30°=
Three
Three
So the answer is
Three
Three

Let's find Tan (α + quartile) (2) find the square of sin 2 α + cos α of 1 + Cos2 α It is known that Tan half α = 2 (1) Seeking Tan (α + quarter school) (2) Find the square of sin2 α + cos α of 1 + Cos2 α How to find the second question

tana/2=2
tana
=2tan(a/2)/(1-tan^2(a/2))
=2*2/(1-4)
=-4/3
(1)
tan(a+π/4)
=(tana+tanπ/4)/(1-tana*tanπ/4)
=(tana+1)/(1-tana)
=(-1/3)/(7/3)
=-1/7
(2)
(sin2α-cos^2α)/(1+cos2α)
= (2sinαcosα-cos^2α)/(1+2cos^2α-1)
= (2sinαcosα-cos^2α)/(2cos^2α)
= tanα-1/2
=-4/3-1/2
=-11/6

Given Tan α = - 3, find cos squared α

Because Tan α = - 3, that is sin α / cos α = - 3
So sin α = - 3cos α
Square both sides of the above formula at the same time
sinα*sinα＝9cosα*cosα　　　　　（1）
But sin α * sin α = 1-cos α * cos α (2)
From (1) and (2), 10 cos α * cos α = 1
So cos α * cos α = 1 / 10

Given the inscribed quadrilateral ABCD, ∠ A: ∠ B: ∠ C = 2:3:4, find the degrees of a, B, C, D

Let, a, B and C be 2x, 3x 4x, respectively
According to the diagonal complementation of the inscribed quadrilateral of a circle, we get that: 2x + 4x = 180 degrees
Results: x = 30 2x = 60 3x = 90 4x = 120
Angle d = 180-90 = 90
∠A=60 ∠B=90 ∠C=120 ∠D=90
I wonder if it will help you?

If the ratio of degrees in the inscribed quadrilateral ABCD is 1:2:3:4, then ∠ A: ∠ B: ∠ C: ∠ D= If the ratio of degrees in the inscribed quadrilateral ABCD is 1 ∶ 2 ∶ 3 ∶ 4, then ∠ a ∶ B ∶∠ C ∶∠ D= [] A.1∶2∶3∶4 B.4∶3∶2∶1C.3∶5∶7∶5D.5∶7∶5∶3

According to the candidate answers, I wonder if the original question is like this: if the ratio of the degrees of the arcs to the sides of the quadrilateral in the circle is 1:2:3:4, then ∠ A: ∠ B: ∠ C: ∠ D=
Answer: d.5:7:5:3
Because ∠ A: = 1 / 2 (- BC + - CD) = 1 / 2 (2 + 3) = 5 / 2
∠B:=1/2(⌒CD+⌒AD) = 1/2(3+4) = 7/2
∠C:=1/2(⌒AD+⌒AB) = 1/2(4+1) = 5/2
∠D:=1/2(⌒AB+⌒BC) = 1/2(1+2) = 3/2

If the ratio of the degrees of a, B and C in the inscribed quadrilateral ABCD is 1:2:3, then the reading of the largest angle of the quadrilateral is

Circle inscribed quadrilateral diagonal complementarity
Then ∠ a + ∠ C = 180, that is, 4 ∠ a = 180
∠A=45°
∠B=90°
∠C=135°
∠D=90°
So the maximum angle is 135 degrees

The four vertices of the quadrilateral ABCD are all on the circle O, and ∠ A: ∠ B: ∠ C = 2:3:4, calculate the degrees of each angle of the quadrilateral ABCD

It seems that the conditions are insufficient It doesn't matter how much ∠ A: ∠ B: ∠ C = 2:3:4 ∠ D is satisfied
D = 30 ° C

As shown in the figure! In the quadrilateral ABCD, ab = AC = ad = BD, calculate the degree of angle BCD! (picture below)

Because the triangle ADC is an isosceles triangle (AC = AD), so: 2 times angle 1 = 180 degrees - angle 3 because triangle ABC is isosceles triangle (AC = AB), so: 2 (angle 1 + angle 4) = 360 degrees - (angle 3

As shown in the figure, in the quadrilateral ABCD, ab = AC = ad = BD, calculate the degree of ∠ BCD

? AB = ad = BD  abd is an equilateral triangle ? bad = 60 ° and ? AB = AC  ABC = ∠ ACB = (180 ° - ∠ BAC) / 2 ? ad = AC ? ADC = ∠ ACD = (180 ° - ∠ DAC) / 2 ? BCD = = ﹤ ACB + ∠ ACD = = (180 - ∠ BAC) / 2 + (180 - ∠ DAC) / 2 = (360 ° - ∠ bad) / 2 = 150 °

As shown in the figure, in the quadrilateral ABCD, the bisectors of ∠ a + ∠ d = 160 °, the bisectors of ∠ ABC and ∠ BCD intersect with the point O. find the degree of ∠ BOC

80°
The inner angle of quadrilateral and 180 ° * (4-2) = 360 °
And ∠ a + ∠ d = 160 °
So ∠ B + ∠ C = 200 °
In △ BOC,
∠BOC=180°-(∠BCO+∠OBC)
=180°-(∠B+∠C)/2
=80°