In a right triangle, the right side to which the 30 degree angle is opposite is equal to half of the hypotenuse. Can we use the trigonometric function of acute angle to prove it

In a right triangle, the right side to which the 30 degree angle is opposite is equal to half of the hypotenuse. Can we use the trigonometric function of acute angle to prove it

In the acute triangle ABC, the angle a = 30, the corresponding side is a, the corresponding side of angle B is B, and the corresponding side of angle c is C. through C, make CD vertical AB, and intersect AB with D. in acute triangle ABC, according to the sine theorem, a / Sina = B / SINB A / 0.5 = B / SINB, in triangle CDB, SINB = CD / A, so: A / 0.5 = CD, a = 2CD

Proof: in a right triangle, if a right angle side is equal to half of the hypotenuse, then the acute angle of the right angle is equal to 30 °

It is known that in △ ACB, ∠ ACB = 90 ° and AC = 1
2AB，
It is proved that: ∠ B = 30 °,
Proof: take AB midpoint D, connect CD,
∵ △ ACB is a right triangle, ∵ ACB = 90 °,
∴CD=1
2AB=AD=BD，
∵AC=1
2AB，
∴AC=AD=CD，
The △ ACD is an equilateral triangle,
∴∠A=60°，
∴∠B=180°-90°-60°=30°．

It is proved that the center line of the hypotenuse of a right triangle is equal to half of the hypotenuse In RT △ ABC, O is the midpoint on the edge of AC. how to prove 2BO = AC? (it seems to extend Bo to point d so that point Bo = do)

Let ABC be the midpoint of a right triangle
Do ed parallel AB cross BC and E
The angle Dec is a right angle
D is the midpoint, ED is parallel to ab
So e is the midpoint of BC
In the triangle DBC. It is easy to see
Triangle DBE is equal to triangle Dec
Then BD = DC
And D is the midpoint of the hypotenuse AC
So BD = AC = DC
OVER

Trigonometric relation of right triangle with 15 degrees Reasoning process (trigonometric function is not allowed)

If the right triangle with 15 degrees is BCD, let AB = x, then AC = 2x is obtained according to the side opposite to 30 degree angle equal to half of the hypotenuse. According to the Pythagorean theorem, BC = √ 3x CE = BC = √ 3 x = ac-ce = (2 - √ 3) x ad = 2ae = (4-2 √ 3) XBD = ab-ad = x - (4-2 √ 3) x = (2 √ 3) x can be obtained according to the Pythagorean theorem

In a right triangle, one angle is 15 degrees and the other 75 degree angle is 48. Find the length of the other two sides Very urgent

48 (2-radical 3) 64 (radical 6-radical 3)

A right triangle, its three angles are 90 degrees, 15 degrees, 75 degrees, the oblique side is 4 meters long, ask how high is the right triangle?

Let the height of the hypotenuse C be H
The lengths of the two right angles are a = 4sin 15 ° and B = 4cos 15 ° respectively
From the equal area method, (1 / 2) AB = (1 / 2) ch
So h = (AB) / C
=(4sin15°)(4cos15°)/4
=4sin15°cos15°
=2sin30°
=1

Given that a right triangle has a right angle side length of 2.3m, an oblique angle side length of 2.4m, and an angle of 15 and 75 degrees respectively, calculate the bottom side length? Right angle side 2.3m is a, oblique side 2.4m is B. The angle between a and B is 15 degrees, and that between B and C is 75 degrees

Are you sure it's a right angle? A right angle, a right angle, a hypotenuse square, a right angle side square, and then a root sign, 2.4 * 2.4-2.3 * 2.3

The angle of the triangle is 90, 15 and 75 degrees, and the length of one side is 24.5?

It's easy to remember the sine of these two special angles
(sin2) / sin2
Sin75 = (Radix 6 + Radix 2) / 4

A right triangle ABC, the inner angle B is 90 degrees, the inner angle a is 15 degrees, and the inner angle c is 75 degrees. AB is 27 cm long. Find the length of the other right angle side BC

54-27√3

If α and β are known to be acute angles, and COS (α + β) = sin (α - β), then Tan α=______ .

∵cos（α+β）=sin（α-β），
∴cosαcosβ-sinαsinβ=sinαcosβ-cosαsinβ，
That is cos β (sin α - cos α) + sin β (sin α - cos α) = 0,
∴（sinα-cosα）（cosβ+sinβ）=0，
∵ α and β are acute angles,
∴cosβ+sinβ＞0，
∴sinα-cosα=0，
∴tanα=1．
So the answer is: 1