In the RT triangle ABC, ∠ BCA = 90 °, CD is the high line, BD = 4, CD = 2, find the trigonometric function value of ∠ a

In the RT triangle ABC, ∠ BCA = 90 °, CD is the high line, BD = 4, CD = 2, find the trigonometric function value of ∠ a

This is a right triangle, CD is the height of the hypotenuse, so CD ^ 2 = BD * Da, (this formula is very useful) 2 ^ 2 = 4 * Da, Da = 1, in RT triangle ADC, AC = radical 5, Sina = 2 radical 5 / 5, cosa = radical 5 / 5, Tana = 2

As shown in the figure, in RT △ ABC, ∠ BCA = 90 ° and ∠ BAC = 30 ° and ab = 6. △ ABC rotates counterclockwise with point B as the center to make point C rotate to C 'on the extension line of edge AB, then the area of the figure (the shadow part in the figure) that edge AC rotates is______ ．

According to the properties of rotation transformation, △ ABC ≌ △ a ′ BC ′, ∵ ∠ BCA = 90 °, ∠ BAC = 30 °, ab = 6, ≌ BC = 12ab = 3, ≌ shadow area = 120 π × 36360-120 π × 9360 = 9 π

The length of three sides of a triangle is ABC. And a > b, the square of a + the square of C < the square of B + 2Ac proves that ABC can form a triangle

a^2+c^2

It is known that a, B and C are three sides of a triangle and satisfy (a + B + C) 2 = 3 (a 2 + B 2 + C 2)

∵ (a + B + C) 2 = 3 (A2 + B2 + C2), ∵ A2 + B2 + C2 + 2Ab + 2BC + 2Ac, = 3a2 + 3B2 + 3c2, A2 + b2-2ab + B2 + c2-2bc + A2 + c2-2ac = 0, that is, (a-b) 2 + (B-C) 2 + (C-A) 2 = 0, ∵ A-B = 0, B-C = 0, C-A = 0, ∵ a = b = C, so △ ABC is an equilateral triangle

Square of sine = 4 / 5 to find cosine
I got a positive 4 / 5, but the answer was negative. Who can tell me what's going on

The answer is obviously wrong. Within 90 degrees, the cosine must be positive,

What is the cosine of the sum of two angles if the sine of one angle multiplied by the cosine of the other angle equals one

0 (sin and COS must only be equal to 1)

Given that the direction vector of the line L is s = (- 1,1,1), and the normal vector of the plane π is n = (1,2, - 3), the cosine value of the angle between the line and the plane is obtained
Such as the title

s. The cosine of the angle n is cos = s * n / (|s | * |n |) = - 1 + 2-3 / (√ 3 * √ 14) = - 2 / √ 42,
So cosine of angle between line and plane = sin = √ [1 - (COS) ^ 2] = √ 38 / √ 42 = √ 399 / 21

[calculation] (2m-5) (2m + 5) - (2m + 1) (2m-3) - 2 (m-1) ^ 2

(2m-5)(2m+5)-(2m+1)(2m-3)-2(m-1)^2
=4m²-25-(4m²-6m+2m-3)-2(m²-2m+1)
=4m²-25-4m²+4m+3-2m²+4m-2
=-2m²+8m-24

Calculation 1, (a-b) ^ 3 * (a-b) ^ 4 * (B-A) ^ 3 2, y ^ (M + 2) * y * y ^ (m-1) + y ^ (2m + 2) 3, (n-m) ^ 3 * (m-n) ^ 2 - (m-n) ^ 5

(a-b)^7*(-(a-b))^3=-(a-b)^10y^(4m+4)-2(m-n)^5

Calculation (m-2) (m ^ 2 + 2m + 4) =?

m^3-8