In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and meet 2C − B a＝cosB cosA． (I) find the size of angle a; (II) if a = 2 5. Find the maximum value of △ ABC area

In △ ABC, the opposite sides of angles a, B and C are a, B and C respectively, and meet 2C − B a＝cosB cosA． (I) find the size of angle a; (II) if a = 2 5. Find the maximum value of △ ABC area

（Ⅰ）∵2c−b
a＝cosB
cosA，
So (2C-B) • cosa = a • CoSb
From the sine theorem, (2sinc SINB) • cosa = Sina • CoSb
2sinc • Cosa SINB • cosa = Sina • CoSb
∴2sinC•cosA=sin（A+B）=sinC．
In △ ABC, sinc ≠ 0
∴cosA＝1
2，∠A＝π
3．
(II) cosine theorem cosa = B2 + C2 − A2
2bc＝1
2，a＝2
5．
∴b2+c2-20=bc≥2bc-20
‡ BC ≤ 20, if and only if B = C, "="
‡ area of triangle s = 1
2bcsinA≤5
3．
The maximum value of triangle area is 5
3．

In triangle ABC, it is known that the square of B minus BC minus 2 times the square of C equals 0, and a = root 6, cosa = 7 / 8, then the area of the triangle =?

B ^ 2-bc-2c ^ 2 = 0, then B = 2C or B = - C (rounded off) according to the cosine theorem, cosa = (b ^ 2 + C ^ 2-A ^ 2) / 2BC is substituted into a = √ 6, cosa = 7 / 8, and the equation is as follows: (5C ^ 2-6) / 4C ^ 2 = 7 / 8, then C = 2 or - 2 (rounded off) because cosa = 7 / 8, Sina = √ 15 / 8, Sina / a = √ 10 / 16, according to the positive

B is known in triangular ABC ²;- bc-2c ²;= 0 and a = root 6, cosa = 7 / 8, then the area of triangle ABC is?

Cosine theorem: A ^ 2 = B ^ 2 + C ^ 2-2bccosaa6 = B ^ 2 + C ^ 2-7bc / 4 known B ²- bc-2c ²= 0 eliminate B ^ 2: C ^ 2 = 2 + BC / 4 and substitute in: B ²- bc-2c ²= 0b ^ 2 = 4 + 3bC / 2 (BC) ^ 2 = (2 + BC / 4) * (4 + 3bC / 2) 5 (BC) ^ 2-32bc-64 = 0bc = 8 area of triangle ABC s = Sina * BC / 2 = (1

The lengths of the two sides of the triangle are 4 and 6 respectively. If the length of the third side is a real number following the square of the univariate quadratic equation x - 16x + 60 = 0, then the area of the triangle is—————————————————— The two roots of the equation are 6 and 10

Solution x ²- 16x+60=0
(x-6)(x-10)=0
x1=6 x2=10
① When x = 6, the three sides of the triangle are 4,6,6 (can form a triangle)
The triangle is an isosceles triangle
(I'll write more)
(the base of the triangle is 4 and the waist is 6
Make a high h on the bottom
At this time, half of the bottom edge, waist and height form a right triangle,
h ²= six ²- two ² (Pythagorean theorem)
h=√32=4√2）
S triangle = 1 / 2 · 4 · 4 √ 2 = 8 √ 2
② When x = 10, the three sides of the triangle are 4, 6 and 10 respectively
Cannot form a triangle
To sum up, the area of the triangle is 8 √ 2

The lengths of both sides of the triangle are 8 and 6 respectively. The length of the third side is a real root of the univariate quadratic equation x ^ 2-14x + 48 = 0. Find the area of the triangle quickly

x^2-14x+48=0
(x-6)(x-8)=0
X = 6 or 8
When the third side is 6
The triangle is an isosceles triangle
The height on the bottom edge of isosceles triangle is √ (6 ^ 2 - (8 / 2) ^ 2) = 2 √ 5
So the triangle area is 1 / 2 * 8 * 2 √ 5 = 8 √ 5
When the third side is 8
The height on the bottom edge of isosceles triangle is √ (8 ^ 2 - (6 / 2) ^ 2) = √ 55
Therefore, the area of the triangle is 1 / 2 * 6 * √ 55 = 3 √ 55

The lengths of both sides of the triangle are 8 and 6 respectively. If the length of the third side is a real root of the univariate quadratic equation x2-16x + 60 = 0, the area of the triangle is （　　） A. 24 B. 24 or 8 five C. 48 D. 8 five

X2-16x + 60 = 0 ⇒ (X-6) (X-10) = 0, ‡ x = 6 or x = 10. When x = 6, the triangle is an isosceles triangle with 6 as the waist and 8 as the bottom. Height h = 62 − 42 = 25 and s △ = 12 × eight × 25=85； When x = 10, the triangle is a right triangle with 6 and 8 as right angles and 10 as hypotenuse × six ×...