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

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

Set s = 1
2bcsina, A2 = B2 + c2-2bccosa, substitute into the known equation to obtain: 1
2bcsinA=a2-b2-c2+2bc=-2bccosA+2bc，
Finishing: 1
2sina = - 2cosa + 2, that is, Sina = 4 (1-cosa),
Simplified: 2sina
2cosA
2=4 × 2sin2A
2，
∴tanA
2=1
4，∴sinA=8 × tan2A
two
tan2A
2+1=8
seventeen
Therefore: C

In triangle ABC, ab = 8, BC = 17, AC = 15, find the area of triangle ABC

It can be found that: 15 * 15 + 8 * 8 = 289
17*17=289
So it's a right triangle
The right angle side is ab, AC
Area: 8 * 15 divided by 2 = 60

In the right triangle ABC, it is known that angle c is equal to 90 degrees, and the opposite sides of angle a, angle B and angle c are a.b.c. let the area of triangle ABC be s, the perimeter be l and the three sides be a

Right triangle area s = AB / 2, so AB = 2S; Perimeter L = a + B + C, a + B-C = M. proof: A ^ 2 + B ^ 2 = C ^ 2A ^ 2 + 2Ab + B ^ 2 = C ^ 2 + 2Ab (a + b) ^ 2 = C ^ 2 + 2Ab (a + b) ^ 2 － C ^ 2 = 2Ab (a + B + C) × (a + B-C) = 2Ab, because a + B + C = L, a + B-C = m, ab = 2S, substitute into the above formula to obtain L × M＝2 × 2...

Given that △ ABC perimeter is 11, ∠ C = 60 °, C = 3, the maximum triangular area is

The three sides are represented by ABC
a+b=8
According to the basic inequality
A + B is greater than or equal to ab under 2 root signs
So when you take the equal sign, you meet the meaning of the question
So AB = 16
According to the Zhengxuan theorem, s △ ABC = 1 / 2 into AB sinc = 4 three roots

It is known that triangle ABC and triangle a'b'c are homomorphic figures. The area of triangle a'b'c is 4cm2 and the perimeter is twice that of triangle ABC For——

Answer:
It is known that triangle ABC and triangle a'b'c are homomorphic figures. The area of triangle a'b'c is 4cm2 and the perimeter is twice that of triangle ABC
Is - 1cm2

In RT triangle ABC, ∠ C = 90 °, a = 8, C = 17, then the perimeter of triangle ABC is -----, and the area is——————

8+15+17=40 8 × fifteen ×½= sixty