In the triangle ABC, a, B and C are the opposite sides of angles a, B and C respectively, and satisfy b square + C square - a square = BC 1. Find the value of angle a 2. Find the value of sin (a + 15 degrees)
The cosine theorem A2 = B2 + c2-2cosabc satisfies b square + C square - a square = BC so cosa = 1 / 2, a = 60 sin75 = sin (45 + 30) = sin45cos30 + cos45sin30 = (gen6 + Gen2) / 4
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- 1. In △ ABC, we know that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10 What's the math problem In △ ABC, it is known that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10 Find the value of a + B?
- 2. In △ ABC, we know that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10, C = √ In △ ABC, it is known that a, B and C are angles, a, B and C are opposite sides, a and B are acute angles, and cos2a = 3 / 5, SINB = 10 / √ 10 C = √ 5 for ABC area
- 3. It is known that a, B and C are the three sides of △ ABC. It is proved that | a + B-C | + | a-b-c | = 2B Such as the title
- 4. The length of the three sides of a triangle is ABC. If a and B are equal to 10, AB is equal to 18, and C is equal to 8, judge the shape of the triangle,
- 5. If three sides a, B and C of triangle ABC satisfy (a-b): (C-B): (a + b) = 7:1:18, the shape of the triangle is judged There is a reasoning process
- 6. If the three sides a, B and C of a triangle are suitable for A2 (B-C) + B2 (C-A) + C2 (a-b) = 0, then the shape of △ ABC is () A. Right triangle B. isosceles triangle C. isosceles right triangle D. equilateral triangle
- 7. If A-B = 5, the square of a + the square of B = 17, then AB = a + B=
- 8. (1 + 2 + 3 + 4 +... + 98 + 99 + 100) / 5 is equal to?
- 9. Mathematical problem of senior one: it is known that y = f (x) is an increasing function in the domain (- 1,1) and f (1 + a) is less than f (A2-1) (A2 represents the square of a, then the range of a is
- 10. 1. Given that the domain of F (x + 1) is [- 1,1], then the domain of F | x + 1 | is the process and method of? - (3,1) 2. Given the function f (x) = 1 / 2x ^ 2-x + 3 / 2, whether there is a real number m, so that the domain of definition and range of value of the function are [1, M] (M > 1)? If there is, request m m = - 3, 1 / 2x ^ 2 is the square of half X
- 11. In the triangle ABC, A.B.C is the three sides opposite to the angle A.B.C, a square - (B-C) square = BC, 1. Find the angle A 2. If BC = 2 times root sign 3, angle B is equal to X and perimeter is y, find the value range of function y = f (x)
- 12. Given that A.B.C is a positive integer, what is the answer to prove that (a + b) (B + C) (c + a) is greater than or equal to 8abc
- 13. (a + b) * (B + C) * (a + C) greater than or equal to 8abc
- 14. It is proved that (a + b) (B + C) (c + a) is greater than or equal to 8abc, and (a, B, c) are positive real numbers
- 15. Proof: if ABC is positive, then (a + b) (B + C) (c + a) > = 8abc Online solution, get rid of~~
- 16. How to prove that (A's square + 1) times (B's square + 1) times (C's square + 1) is greater than or equal to 8abc
- 17. Positive numbers a, B, C satisfy a + B + C-2 = 0, and prove that (2-A) (2-B) (2-C) is greater than or equal to 8abc
- 18. Prove the inequality, if a, B, C are all positive numbers, then (a + b) (B + C) (c + a) ≥ 8abc
- 19. If a, B and C are all positive numbers, then (a + b) (B + C) (c + a) > = 8abc
- 20. In the triangle ABC, if three sides are known to be continuous positive integers, The cosine value of the maximum angle is - 1 / 4. Find the maximum area of the parallelogram with the maximum angle as the inner angle and the sum of the two sides of the angle as 4