In 4abc c2 125 b2 325 cot c 3 2 find tan a
WebJan 4, 2024 · To find - 4 abc. (a+b+c)² = a² + b² + c² + 2ab + 2bc +2ca ------------> 1. Substitute the values of (a+b+c) and a²+b²+c² in the equation 1. (5)² = 27 + 2 (ab + bc+ ca) 25 = 27 + … WebJul 2, 2024 · If a2 + 4b2 + 49c2 + 18 = 2 (2b + 28c – a), then the value of (3a + 2b + 7c) is: Q5. If a + b + c = 5, a2 + b2 + c2 = 27 and a3 + b3 + c3 = 125, then the value of 4abc is: Q6. …
In 4abc c2 125 b2 325 cot c 3 2 find tan a
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WebIf tan A 2,tan B 2,tan C 2 are in H.P., then the minimum value of cot B 2 is. Q. If ABC is triangle and tan A 2,tan B 2,tan C 2 are in H.P the find the minimum value of cot B 2. Q. If … WebWhat is the sum of cubes formula? The sum of cubes formula is a³ + b³ = (a+b) (a² - ab + b²) What is the difference of squares formula? The difference of squares formula is a² - b² = (a+b) (a-b) What is the difference of cubes formula? The difference of cubes formula is a³ - b³ = (a-b) (a² + ab + b²)
WebJun 20, 2024 · 2 Answers Sorted by: 2 Solving this problem by Law of Cosines isn't short but not that unmanageable. The key is the observation of pattern within the expression: ( b 2 − c 2) cot A = ( b 2 − c 2) b 2 + c 2 − a 2 2 b c / a 2 R = R a b c [ b 4 − c 4 − a 2 b 2 + b 2 c 2] This has the form ϕ ( a, b, c) − ϕ ( b, c, a) where WebFree math problem solver answers your trigonometry homework questions with step-by-step explanations.
WebJul 3, 2024 · 4abc = -20. Given To find value of 4abc : = ? Formula : ( 1 ) ( 2 ) Using formula ( 1 ), Substitute the values, -----> ( A ) Using formula ( 2 ), 25 - 3abc = ( 5 ) ( 27 - (ab+bc+ca) ) Now apply ( A ) value in above equation, 125 - 3abc = ( 5 ) ( 27 - (ab+bc+ca)) 125 - 3abc = ( 5 ) ( 27 - ( - 1 ) ) 125 - 3abc = ( 5 ) ( 27 + 1 ) WebSymbolab is the best step by step calculator for a wide range of math problems, from basic arithmetic to advanced calculus and linear algebra. It shows you the solution, graph, detailed steps and explanations for each problem.
WebUsing the Law of Cosines we also have: a 2 = b 2 +c 2 -2bccosA b 2 = a 2 +c 2 -2accosB c 2 = a 2 +b 2 -2abcosC The trick is to add these three equations together, simplify, divide by 4, and then replace 1/2 ac with K/sinB (for example). a 2 +b 2 +c 2 =2a 2 +2b 2 +2c 2 -2bccosA-2accosB-2abcosC This can be rewritten as:
the practice of accountancy includesWebJul 3, 2024 · 4abc = -20. Given To find value of 4abc : = ? Formula : ( 1 ) ( 2 ) Using formula ( 1 ), Substitute the values, -----> ( A ) Using formula ( 2 ), 25 - 3abc = ( 5 ) ( 27 - (ab+bc+ca) ) … the practice of adaptive leadership amazonWebThe identity 1 + cot2θ = csc2θ is found by rewriting the left side of the equation in terms of sine and cosine. Prove: 1 + cot2θ = csc2θ 1 + cot2θ = (1 + cos2θ sin2θ) Rewrite the left side. = (sin2θ sin2θ) + (cos2θ sin2θ) Write both terms with the common denominator. = sin2θ + cos2θ sin2θ = 1 sin2θ = csc2θ sift bake shop nianticWeb3cot B 2 = cot A 2cot B 2cot C 2 ⇒ cot A 2cot C 2 = 3 Now, A.M.≥G.M. ⇒ cot A 2 +cot C 2 ≥ 2√cot A 2cot C 2 ⇒ 2cot B 2 ≥2√3 ∴ cot B 2 ≥√3 Suggest Corrections 1 Similar questions Q. If ABC is a triangle and tan A 2,tan B 2,tan C 2 are in H.P., then find the minimum value of cotB/2. Q. Let A,B and C represents the angles of ABC. the practice of adaptive leadership audiobookWebIntroduction to Systems of Equations and Inequalities; 9.1 Systems of Linear Equations: Two Variables; 9.2 Systems of Linear Equations: Three Variables; 9.3 Systems of Nonlinear Equations and Inequalities: Two Variables; 9.4 Partial Fractions; 9.5 Matrices and Matrix Operations; 9.6 Solving Systems with Gaussian Elimination; 9.7 Solving Systems with … sift bakery plainfield ilWebClick here👆to get an answer to your question ️ In ABC, a^2 + c^2 = 2002b^2 , then cotA + cotCcotB equals to ... Solve Study Textbooks Guides. Join / Login. Question . In A B C, a 2 … sift bakery san franciscoWebJan 4, 2024 · 4abc = -20 Step-by-step explanation: Given data a+b+c = 5 a²+ b² + c² = 27 a³ + b³ + C³ = 125 To find - 4 abc (a+b+c)² = a² + b² + c² + 2ab + 2bc +2ca ------------> 1 Substitute the values of (a+b+c) and a²+b²+c² in the equation 1 (5)² = 27 + 2 (ab + bc+ ca) 25 = 27 + 2 (ab + bc+ ca) -2 = 2 (ab + bc+ ca) Eliminate 2 on both sides of the equation the practice of bhakti yoga by sivananda