In RT △ ABC, ∠ C = 90 ° s = 18 root sign 3, a

In RT △ ABC, ∠ C = 90 ° s = 18 root sign 3, a

From the question, s = 1 / 2C × H = 1 / 2 × 3 √ 3C = 18 √ 3 ν C = 12
Then, AC = 2S = 36 √ 3 a? + B? = C? = 144
Then (a + b) 2 = 144 + 72 √ 3 (B-A) 2 = 144-72 √ 3 ∵ a

In the RT triangle ABC, the angle c is equal to 90 degrees, and the right triangle is solved according to the following conditions;

(2) Ad is the center line on the side of BC, AC = 3 times the root sign 3, the angle ADC = 60 degrees. The first question about how to find the triangle ABC is not understood

As shown in the figure, in the RT triangle ABC, ∠ ACB = 90, CD ⊥ AB, if BC = radical 10, Tan ∠ BCD = 1 / 3. Find the values of BD and AC

Because tanbcd = 1 / 3 = BD / CD, CD = 3bd, and BD + CD = BC, then 10bd = 10, BD = 1
If the middle angle of a right triangle BCD = angle BAC, similarly, tanbac = tanbcd = 1 / 3 = BC / AC, then AC = 3bC = 3, root number 10

As shown in the figure, ∠ ACB = 90 ° and ab = 8，BC＝ 2. Find the height CD on the hypotenuse ab

AC=
AB2−BC2=
8−2=
6，
∵S△ABC=1
2AC•BC=1
2CD•AB，
∴CD=AC•BC
AB=
6 x
Two
8=
Six
2．

As shown in the figure, ∠ ACB = 90 ° and ab = 8，BC＝ 2. Find the height CD on the hypotenuse ab

AC=
AB2−BC2=
8−2=
6，
∵S△ABC=1
2AC•BC=1
2CD•AB，
∴CD=AC•BC
AB=
6 x
Two
8=
Six
2．

In the triangle ABC, angle c = 90 degrees, angle a = 60 degrees, BC = 3 + 2, root sign 3, BD bisect angle ABC intersects AC with D, and finds the length of AD

Pass D as de ⊥ AB, cross AB to E
∵ BD bisection ∵ ABC, ∵ C = 90 °, de ⊥ ab
∴CD=DE
In RT △ BCD and RT △ bed, CD = De, BD = dB
∴Rt△BCD≌Rt△BED
∴BE=BC
∵ in RT △ ABC, ∵ a = 60 ° BC = 3 + 2 √ 3
∴AB=BC/(√3/2)=4+2√3
∴AE=AB-BE=1
∴AD=2AE=2

In the triangle ABC, A.B.C is the opposite side of the angle A.B.C respectively, and the vector M = (2b is C, COSC, which is three times the root sign), Vector n = (a, cosa, 3 times the root sign), and vector m is parallel to vector n. find the size of angle A

From the vector m parallel vector n, we get: (2B - √ 3C) / √ 3A = COSC / cosa. According to the sine theorem, (2sinb - √ 3sinc) / (√ 3sina) = COSC / cosa, 2sinbcosa = √ 3sinacosc + √ 3sinccosa, 2sinbcosa = √ 3sin (a + C), 2sinbcosa = √ 3sinb = √ 3sinb

The maximum area of △ ABC satisfying the condition AB = 2, AC = radical 2BC

Maximum 2 pieces
Make CE vertical AB crossing AB (or AB extension line) at E
Let CE = h, be = x (x is negative in AB extension)
S=AB*CE/2 =2*h/2=h
To maximize s, that is, to make h, that is, to maximize CE
Because CE is perpendicular to AB, according to Pythagorean theorem, there are
Be ^ + CE ^ = BC ^ = = > x ^ + H ^ = a ^ (1) (^ denotes Square)
AE ^ + CE ^ = AC ^ = = > (2-x) ^ + H ^ = (root 2 * a) ^ (2)
(1) (2) x = (4-A ^) / 4 is substituted by (1)
h^=a^-[(4-a^)/4]^=-(a^-12)^/16+8
When a ^ = 12, H ^ has a maximum value of 8 and h has a maximum value of 2
Therefore, the maximum area of △ ABC is 2
If you have learned the coordinate method, see the following tips
Let a (0,0), B (2,0), C (x, y)
Root (x ^ + y ^) = root 2 * root ((X-2) ^ + y ^)
x^+y^=2((x-2)^+y^)
y^=x^-2(x-2)^=-x^+8x-8=-(x-4)^+8
When x = 4, y has a maximum of 2
The area has a maximum of 2

The maximum area of △ ABC satisfying the condition AB = 2, AC = radical 2BC How do I want to know 128

The maximum area of a B = 2, AC = √ 2 * BC, △ ABC
According to: the sum of the two sides of a triangle is greater than the third
AB＝2,
Suppose BC > 2
Then 2 + BC > AC
2+BC>√2*BC
BC(√2-1)

In the triangle ABC, a is equal to the root 3, and the angle a is equal to 60 °. When B times C is the maximum, what is the form of the triangle

Because (B-C) is greater than or equal to 2B * C, when B = C, b * C is maximum
Then the triangle ABC is an equilateral triangle