A square tower stands upon a horizontal plane. From a point in this plane, from which three of its upper corners are visible, their angular elevations are respectively 45 dergrees, 60 degrees and 45 degrees. Show that the height of the tower is to the breadth of one of its sides as √6(√5+1) to 4.
please answer fast...all help appreciated :-D
Asked by reetatyagi8 | 31st May, 2015, 12:53: PM
One face of the tower with two upper corners 'D' and 'E' is shown in the figure.
'D' makes 60o with a point 'A' in the plane. Hence,
The corner 'E' makes an angle of 45o with point 'A' in the plane. Hence,
In triangle ABC, the angle ACB=135o (using symmetry, since the angle is same 45o with two upper corners)
Triangle ABC is redrawn below:
Answered by satyajit samal | 1st Jun, 2015, 11:59: AM
Kindly Sign up for a personalised experience
- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions
Verify mobile number
Enter the OTP sent to your number