A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 min from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then, the time taken (in minutes) by him, from B to reach the pillar, is
In \(\triangle ABC\), if incircle touches the sides \(AB\), \(BC\) and \(CA\) at \(P\), \(Q\) and \(R\) respectively and \(s - a = 3\), \(s - b = 5\) and \(s - c = 7\), then area of the quadrilateral \(QCRI\) is, where \(I\) is incentre of \(\triangle ABC\):[Note: Symbols used have usual meaning in \(\triangle ABC\).]
Find the value of \(\cos 3A + \cos 3B + \cos 3C\) given that \(A + B + C = 180°\) (angles of a triangle), and determine under what conditions the expression equals \(1 + \cos(3A + 3B)\). Specifically, evaluate: \(\cos 3A + \cos 3B = 1 - \cos(3C)\), i.e., \(2\cos\dfrac{3}{2}(A+B)\cos\dfrac{3}{2}(A-B) = 2\cos^2\dfrac{3}{2}(A+B)\). If \(\cos\dfrac{3}{2}(A+B) = 0\), then \(\dfrac{3}{2}(A+B) = 90°\), \(A + B = 60°\), so \(C = 120°\). What is the answer?