Angles of elevation of the top of a 20 m high pole from two points in

1.	Angles of elevation of the top of a 20 m high pole from two points in a line are 30 and 60 find the distance between the points
Answer» Answer: 23.094 m Step-by-step EXPLANATION: Here, a png file is attached. length of the POLE AB = 20 m ELEVATION of point A from point C = 60° ∠ACB = 60° elevation of point A from point D = 30° ∠ADB = 30° Now, $tan\angle ACB= \frac{AB}{BC}$ ► $BC= \frac{AB}{tan\angle ACB} = \frac{20}{tan60} = \frac{20}{\sqrt3}$ m Again, $tan\angle ADB= \frac{AB}{BD}$ ► $BD= \frac{AB}{tan\angle ADB} = \frac{20}{tan30} = 20\sqrt3$ m Finally, DISTANCE between the points C & D = CD = BD-BC =( $20\sqrt3-\frac{20}{\sqrt{3}}$ ) m = 23.094 m