As shown in the figure, ad = de = EC, f is the midpoint of BC, and G is the midpoint of FC

As shown in the figure, ad = de = EC, f is the midpoint of BC, and G is the midpoint of FC

The problem is incomplete
And there's no picture

In the triangle ABC, ab = AC BC = X. if the circumference of triangle ABC is 24, then the value range of X is A 1 ≤ x ≤ 12 B 0 less than x less than or equal to 12 C 0 less than x less than 12 D 6 less than x less than 12

AB=AC =(24-X)/2
AB+AC>BC
0 less than x less than 12
C

In △ ABC, if a = 6, B = 30 ° and C = 120 °, then the area of △ ABC is___ .

∵ in ᙽ ABC, a = 6, B = 30 °, C = 120 °, i.e., a = 30 °,
From the sine theorem a
sinA=b
SINB: B = asinb
sinA6×1
Two
One
2=6，
Then s △ ABC = 1
2absinC=9
3．
So the answer is: 9
3．

In the triangle ABC, we know that a and B are equal to 10, C is equal to 6, and B is equal to 30 degrees

Cos30? = (a? + 36-b?) / 12a = √ 3 / 2, and a + B = 10
So a = (320 + 96 √ 3) / 73
So area = 1 / 2acsinb = 1 / 2 * [(320 + 96 √ 3) / 73] * 6 = (960 + 288 √ 3) / 73

In a right triangle, angle c is equal to 90 degrees, angle a, angle B, and the opposite side of angle c is a, B, C. If a ratio C equals 15 to 7 and B equals 24, calculate the area of triangle ABC

Let a = 7K, C = 15K, k0a ^ 2 + B ^ 2 = C ^ 2, (7K) ^ 2 + 24 ^ 2 = (15K) ^ 2, K ^ 2 = 24 ^ 2 / (15 ^ 2-7 ^ 2) = 24 ^ 2 / (22x8) = 24 ^ 2 / (11x4 ^ 2) k = 24 / (4 root number 11) s = AB / 2 = 7kx24 / 2 = 7x24 / (4 root number 11) X12 = 24x21 / root number 11

In the triangle ABC, the vector AB * AC = 9, SINB = cosasinc, the area is 6, P is the point on the segment AB, and the vector CP = x * vector Ca / / Ca / module + y * vector CB / / CB / module, then the minimum value of 1 / X + 1 / y is obtained Answer; (7 / 12 + 3 / 3 under radical) Newspapers 11 I will give you extra points for the college entrance examination this year!

sinB=sin（A+C）=cosAsinC
So sinacosc = 0, COSC = 0, C = 90 degrees
And the vector AB * AC = | AC ^ 2 = 9, | AC | = 3
Area = 1 / 2 * | Ca | * | CB | = 6, so | CB | = 4
If C is the origin, CA is the x-axis, CB is the y-axis, then the coordinates of point P are (x, y)
The point P is on the line AB, so x / 3 + Y / 4 = 1
So (1 / x + 1 / y) = (1 / x + 1 / y) * (x / 3 + Y / 4)
=1/3+1/4+1/3(x/y)+1/4（y/x）
>=7 / 12 + 3 / 3 under radical

In the triangle ABC, the vector AB * AC = 9, SINB = cosasinc, the area is 6, and P is the line segment In the triangle ABC, the vector AB * AC = 9, SINB = cosasinc, the area is 6, P is the point on the segment AB, and the vector CP = x * vector Ca / / Ca / module + y * vector CB / / CB / module, then the minimum value of 1 / X + 1 / y is obtained Answer; (7 / 12 + 3 / 3 under radical)

In the triangle ABC, the sides of a, B, C pair are a, B, C
sinB=cosAsinC==>sin(A+C)=cosAsinC
==>sinAcosC+cosAsinC=cosAsinC
==>sinAcosC=0==>C=π/2
AB*AC=9==>cbcosA=9 (1)
SΔ=6==>1/2 cbsinA=1/2ab=6 (2)
SΔ=1/2ab=6==>ab=12
(2)/(1):tanA=4/3,sinA=4/5,cosA=3/5
bc=15,c^2=a^2+b^2,
==>c=5,b=3,a=4
Vector CP = x * vector Ca / | Ca | + y * vector CB / | CB|
=X / b * vector Ca + Y / A * vector CB
∵ a, P and B are collinear
∴x/b+y/a=1
1/x+1/y=(1/x+1/y)(x/b+y/a)
=1/3+1/4+x/(3y)+y/(4x)
≥1/3+1/4+2√(1/12)=7/12+√3/3
When x / (3Y) = Y / (4x), take the equal sign
The minimum value of 1 / x + 1 / y is 7 / 12 + √ 3 / 3

In △ ABC, a, B and C are opposite sides of angles a, B and C respectively AB• AC = 9, SINB = cosasinc, (I) find the length of edge AC; (II) if BC = 4, find the size of angle B

（I）
AB•
AC = 9 {cbcosa = 9, SINB = cosasinc} Cosa · C = B, B = 3,
（II）cbcosA＝9⇒cosA＝9
bc＝b2+c2−a2
2 BC, substituting BC = 4 = a, B = 3 to get AB = 5 {C2 = B2 + A2} SINB = B
c＝3
5⇒B＝arcsin3
Five

Given that the area s of triangle ABC satisfies 3 ≤ s ≤ 3 * root 3 and vector AB * vector BC = 6, the angle between vector AB and vector BC is a. find the value range of A Find the minimum value of F (a) = sin ^ 2A + 2sinacosa + 3cos ^ 2A

Record | ab | = C; | BC | = a;
3 ≤ s = a * c * SINB / 2 ≤ 3 * radical 3; (1)
Vector AB * vector BC = 6 = a * c * cos (180 degrees - b),
So a * c * CoSb = - 6; (2)
(1) (2) by simplification, the following results are obtained
-Radical 3 ≤ tanb ≤ - 1;
So the value range of B is: 120 degrees ≤ B ≤ 135 degrees
The angle is the complement of B, so 45 degree ≤ a ≤ 60 degree!
After simplification, f (a) = radical 2 * sin (2 * a + 45 degree) + 2 (45 degree ≤ a ≤ 60 degree);
Therefore, when a = 60 degrees, the minimum value is (3 + radical 3) / 2;

In triangle ABC, ab = 1 AC = 2 (vector AB + vector AC) * vector AB = 2 area of triangle ABC

Triangle area = 1 / 2 * | ab | * | AC | sin ∠ BAC = (1 / 2) * 1 * 2 * sin ∠ bac
(vector AB + vector AC) * vector AB = AB] + | ab | * | AC | cos ∠ BAC = 2, so cos ∠ BAC = 1 / 2
Sin ∠ BAC = root three / 2, so we can get the area = root three / 2