Solution : The value of cot 60 degrees is \(1\over \sqrt{3}\). Proof : 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. \(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point …
Solution : The value of tan 60 degrees is \(\sqrt{3}\). Proof : 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. \(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point …
Solution : The value of Cos 60 degrees is \(1\over 2\). Proof : 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. \(\therefore\) AD is the bisector of \(\angle\) A and D is the …
Solution : The value of sin 60 degrees is \(\sqrt{3}\over 2\). Proof : 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. \(\therefore\) AD is the bisector of \(\angle\) A and D is the …
Solution : The value of cosec 30 degrees is 2. Proof : 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. \(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point …
Solution : The value of sec 30 degrees is \(2\over \sqrt{3}\). Proof : 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. \(\therefore\) AD is the bisector of \(\angle\) A and D is the …
Solution : The value of cot 30 degrees is \(\sqrt{3}\). Proof : 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. \(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point of …
Solution : The value of tan 30 degrees is \(1\over \sqrt{3}\). Proof : 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. \(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point …
Solution : The value of cos 30 degrees is \(\sqrt{3}\over 2\). Proof : 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. \(\therefore\) AD is the bisector of \(\angle\) A and D is the mid-point …
Solution : The value of sin 30 degrees is \(1\over 2\). Proof : 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. \(\therefore\) AD is the bisector of \(\angle\) A and D is the …
