{"id":9104,"date":"2021-12-28T21:32:47","date_gmt":"2021-12-28T16:02:47","guid":{"rendered":"https:\/\/mathemerize.com\/?p=9104"},"modified":"2021-12-28T21:48:24","modified_gmt":"2021-12-28T16:18:24","slug":"what-is-the-value-of-cos-60-degrees","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/what-is-the-value-of-cos-60-degrees\/","title":{"rendered":"What is the Value of Cos 60 Degrees ?"},"content":{"rendered":"
The value of Cos 60 degrees<\/strong> is \\(1\\over 2\\)<\/strong>.<\/p>\n Proof :<\/strong><\/p>\n Consider an equilateral triangle ABC with each side of length of 2a. Each angle of \\(\\Delta\\) ABC is of 60 degrees. Let AD be the perpendicular from A on BC.<\/p>\n \\(\\therefore\\) AD is the bisector of \\(\\angle\\) A and D is the mid-point of BC.<\/p>\n \\(\\therefore\\) BD = DC = a and \\(\\angle\\) BAD = 30 degrees.<\/p>\n In \\(\\Delta\\) ADB, \\(\\angle\\) D is a right angle, AB = 2a and BD = a<\/p>\n By Pythagoras theorem,<\/p>\n \\(AB^2\\) = \\(AD^2\\) + \\(BD^2\\) \\(\\implies\\) \\(2a^2\\) = \\(AD^2\\) + \\(a^2\\)<\/p>\n \\(\\implies\\) \\(AD^2\\) = \\(4a^2\\) – \\(a^2\\) = \\(3a^2\\) \\(\\implies\\) AD = \\(\\sqrt{3}a\\)<\/p>\n Now, In triangle ADB, \\(\\angle\\) B = 60 degrees<\/p>\n