It is known that: A, B, C are the three sides of the triangle ABC, and the square of a + the square of B + the square of C - 10a-24b-26c = - 338 Proof: this triangle is a right triangle

It is known that: A, B, C are the three sides of the triangle ABC, and the square of a + the square of B + the square of C - 10a-24b-26c = - 338 Proof: this triangle is a right triangle

A ^ 2-10a + B ^ 2 + C ^ 2 + C ^ 2 + 338 = 10A + 24B + 26c,  a ^ 2-10a + B ^ 2-24b + C ^ 2-26c + 338 = 0,, (a ^ 2-10a + 25) + (b ^ 2-24b + 144) + (C ^ 2-26c + 169) = 0,, (a-5) ^ 2 + (B-12) ^ 2 + (C-13) ^ 2 = 0, and (a-5) ^ 2 ≥ 0, (B-12) ^ 2 ≥ 0, (C-13) ^ 2 ≥ 0, (a-a-5) ^ 2 ≥ 0, (C-13) ^ 2 ≥ 0, (a-a-a-a-5) ^ 2 ≥ 0,, (A-13) ^ 2 ≥ 5) ^ 2 = 0 and (B-12) ^ 2 = 0

If the area of the triangle ABC s = 1 / 4 root sign 3 (b square + C square - a square), then the value of angle a is equal to

First of all, from the cosine theorem
b²+c²-a²=2bccosA ①
Because s = 1 / 2bcina
How to get the formula of (1) and (2) into the question
1/2bcsinA=√3/4×2bccosA
Tana = √ 3 is obtained
So a = π / 3

In the triangle ABC, if its area s = (the square of a + the square of B - the square of C) / 4, then the angle C=

∵ in the triangle ABC, s = (a + B-C) / 4 √ 3 and S = absinc ᙽ 4 √ 3 = absinc, then a + B-C = 2 √ 3absinc ∵ and ∵ from cosine theorem, COSC = (a + B-C) / 2Ab

If the area of the triangle ABC is (a + B + C) / (4), what is ∠ C equal to?

Cosine theorem: if C ^ 2 = a ^ 2 + B ^ 2-2abcosc, then 2abcosc = a ^ 2 + B ^ 2-C ^ 2 is divided by 4 at the same time, then 1 / 2abcosc = 1 / 4 (a ^ 2 + B ^ 2-C ^ 2) is obtained. According to the meaning of the title, the triangle area s = 1 / 4 (a ^ 2 + B ^ 2-C ^ 2) so s = 1 / 2abcosc, and the formula of triangle area is s = 1 / 2absc, so the COSC = sinc solution of 1 / 2abcosc = s = 1 / 2absc gives C = 45 degrees

In △ ABC, if a, B and C are opposite sides of a, B and C respectively, and area s = a2 - (B-C) 2, then Sina = () A. 15 Seventeen B. 13 Fifteen C. 8 Seventeen D. 13 Seventeen

By substituting s = 12bcsina, A2 = B2 + c2-2bccosa into the known equation, 12bcsina = a2-b2-c2 + 2BC = - 2bccosa + 2BC can be obtained: 12sina = - 2cosa + 2, i.e. Sina = 4 (1-cosa)

In the triangle ABC, Sina = CoSb, a = 6sina, the square of C = 2Ab, find the area of the triangle

sinA = cosB,A+B=90
C=90
c^2 = a^2 + b^2 = 2ab
(a-b)^2 = 0
a = b
A=B=45
a/sinA = 6 = b/sinB
a = 6sinA
b = 6sinB
S = ab/2 = 18 sinA sinB = 9 sin2B = 9

In the triangle ABC, the square of sina + the square of SINB = 1, and the longest side C = 12

Square of sina + square of SINB = 1
SINB squared = cosa squared
The longest side C = 12, a and B are acute angles
sinB=cosA
B = 90 degrees - A
C = 90 degrees, right triangle
a^2+b^2=c^2=144
(a-b)^2>=0
a^2+b^2>=2ab
AB

As shown in Figure RT, the area of triangle ABC is 20cm Λ 2. On the same side of AB, make three semicircles with the diameter of AB, BC and AC respectively

S (shadow) = 1 / 2 * π * (1 / 2Ac) ^ 2 + 1 / 2 * π (1 / 2BC) ^ 2 + s (triangle ABC) - 1 / 2 * π * (1 / 2Ab) ^ 2
=1 / 8 * π * (AC ^ 2 + BC ^ 2-AB ^ 2) + s (triangle ABC)
AC ^ 2 + BC ^ 2 = AB ^ 2,
So s (shadow) = s (triangle ABC) = 20

The area of RT △ ABC is 20 cm 2. On the same side of AB, three semicircles with diameters of AB, BC and AC are made to calculate the area of shadow part

It can be seen from the figure that the area of the shadow part is 1
2π（1
2b）2+1
2π（1
2a）2+S△ABC-1
2π（1
2c）2，
= Pi
8（a2+b2-c2）+S△ABC，
In RT △ ABC, A2 + B2 = C2,
The area of shadow part = s △ ABC = 20cm2

As shown in the figure, in RT △ ABC, ∠ ACB = 90 ° AB = 5, AC = 3, take AC bcab as the diameter to make a semicircle to calculate the area of shadow part in the circle In RT △ ABC, ∠ ACB = 90 ° AB = 5, AC = 3, take AC bcab as the diameter to make a semicircle to find the area of shadow part in the circle fast

BC=6
The square π of (BC / 2)
The square π of (AC / 2)
The square π of (AB / 2)