In △ ABC, Tanc = 3, radical 7 (1) find COSC (2) if vector CB · vector CA = 5 / 2 and a + B = 9 find C Process

In △ ABC, Tanc = 3, radical 7 (1) find COSC (2) if vector CB · vector CA = 5 / 2 and a + B = 9 find C Process

1. Tanc = 3 √ 7 > 0, C is an acute angle, and COSC is positive,
tanC=sinC/cosC,
Let COSC = x, sinc = √ [1 - (COSC) ^ 2],
√（1-x^2)/x=3√7,
1-x^2=63x^2,
64x^2=1,
x=1/8,
cosC=1/8,
2. Vector CB · vector CA = 5 / 2,
|CB|*|CA|*cosC=5/2,
a*b*(1/8)=5/2,
ab=20,(1)
a+b=9,
b=9-a,(2)
(1) And (2),
a(9-a)=9,
a^2-9a+20=0,
(a-5)(a-4)=0,
a=5,b=4,
Or, a = 4, B = 5,
According to the cosine theorem,
c^2=a^2+b^2-2abcosC
=25+16-2*5*4*(1/8)
c=6.

In △ ABC, a = 1, C = radical 2, COSC = 3 / 4, find the value of vector CB * vector ca

First, the cosine theorem is used to calculate the value of B
c^2=a^2+b^2-2abcosC
That is: 2 = 1 + B ^ 2-1.5b, B = - 0.5 (omitted), B = 2
So: vector CB * vector CA = | CB | * | Ca | COSC = 1 * 2 * 3 / 4 = 3 / 2
That's it,

Then the root sign 7 (1) of Tan C = 3 times in the triangle ABC is used to find COSC (2) if the vector CB is multiplied by the vector CA = 5 / 2 and a + B = 9, find C

1, COSC = 1 / 8; take C as the acute angle vertex of a right triangle, the opposite side is 3 times the root sign 7, and the adjacent side is 1, then the hypotenuse is 8; (the original ABC is not necessarily a right triangle, but once the tangent is determined, the angle C is determined, the cosine is determined, and it can be assumed to be a simple right angle triangle to deal with) 2

In the triangle ABC, the sides of angles a, B and C are a, B, C respectively, a = π / 6, (1 + radical 3) * C = 2B, if the product of CB vector and Ca vector = 1 + √ 3, find a, B,

C = π / 4, B = 7 / 12 * π B = 7 / 12 * π CB * CA = AB * COSC = 1 + √ 3 (1 + √ 3) * C = 2bac * COSC = 1 + √ 3 (1 + √ 3) * C = 2bab * COSC * C = 2bac * COSC = 2bac * COSC = 2Ac = 2Ac = 2 √ 2Ab * cos (7 / 12 * π) = 1 + √ 3 (1 + √ 3) * C = 2B according to the above three equations, we can get: a = root number, 2C = 2C = 2C = 2C = 2C = 2baccording to the above three equations: a = root number, 2C = 2C = 2C = 2C = 2C = 2C = 2baccording to the above three equations: a = a = root number 2C = 2C = 2C = 2C = 2C = b = 1 + radical 3

It is known that the three sides of the triangle ABC are A.B.C, and A.B.C satisfies the square of the root of A-3 + │ B-4 │ + c-10c + 25 = 0,

√(a-3)≥0,
│b-4│≥0,
c²-10＋25=(c-5)²≥0,
∵√ A-3 + │ B-4 │ + C? - 10C + 25 = 0, so its independent three values are all 0
∴a=3,b=4,c=5

In the triangle ABC, the opposite sides are a, B and C. if the area of triangle AB C is s, it is equal to the root of 4 [a square plus b square minus C square] One Find the size of angle C

S= (root sign 3) /4 × (a? +b? -c?) =1/2 × ab sinC
The cosine theorem C 2 = a 2 + B 2 a B COSC leads to a conclusion
(radical 3) / 4 × 2Ab COSC = 1 / 2 × AB sinc
Therefore, Tanc = radical 3, that is, C = 60 degrees

In △ ABC, we know AB = 8, radical 2, BC = 14, AC = 10, and find the high utility Pythagorean theorem on the edge of BC

Let CD = x, then BD = 14-x, in ACD of right triangle, we can get ad ^ 2 = AC ^ 2-CD ^ 2 = 10 ^ 2-x ^ 2, in abd of right triangle, we get ad ^ 2 = AB ^ 2-bd ^ 2 = (8 √ 2) ^ 2 - (14-x) ^ 2, so 10 ^ 2-x ^ 2 = (8 √ 2) ^ 2 - (14-x) ^ 2

In the triangle ABC, ab = 8 times root sign 6, angle B = 45 degrees, angle c = 60, how long is AC and BC (to solve the problem process) Need to solve the whole process of the problem, AC and BC are required to come out

Make a high line ad on BC and cross it to point D,
Because ∠ B = 45 °,
So ad = BD = AB divided by root 2 = 8 times root 3
Because ∠ C = 60 °,
So CD = ad divided by root 3 = 8,
AC = 2AD = 16,
BC = BD + CD = 8 times root 3 + 8,
AC = 16.

As shown in the figure, it is known that in △ ABC, ∠ a = 60 °, B = 45 ° and ab = 8. Find the area of △ ABC. (the root sign can be retained in the result)

In RT △ ADC, ∵⊥ CDA = 90 °, dacd = cot ∠ DAC = cot 60 ° = 33, i.e. ad = CD × 33. In RT △ BDC, ? B = 45 °, BCD = 45 °, CD = BD. ∵ AB = DB + Da = CD + CD × 33 = 8,  CD = 12-43. ? ABC = 12ab × CD = 12 × 8 × (12-4

In △ ABC, if ∠ a = 60 °, B = 45 ° and BC = 3 2, then AC = () A. 4 Three B. 2 Three C. Three D. Three Two

According to the sine theorem, BC
sinA＝AC
sinB，
Then AC = BC · SINB
sinA＝3
2 x
Two
Two
Three
2＝2
Three
Therefore, B is selected