Q. 134.2( 6 Votes )

From the top of a tower 100 m high, a man observes two cars on the opposite sides of the tower with angles of depression 30° and 45° respectively. Find the distance between the cars. [Take √3 = 1.73.]

Answer :

Untitled19.jpg


In the figure, let AC be the tower of height 100m. The angles of depression are BAE and DAF. Given, AC = 100m. Since the line EF is parallel to the line BD, we have BAE = BAC = 30°, DAF = DAC = 45°, where B, D are the positions of the cars. Join B, C, and C, D. We get two right-angled triangles ∆ABC and ∆ADC, both right angled at C. We use trigonometric ratio tan for both the triangles, where BC is the base and Ac is the height (for ∆ABC) and CD is the base and AC is the height in ∆ACD. We are to find the distance between the two cars, i.e. BD.


In ∆ABC,



or,



In ∆ADC,



or,


DC = 100


Therefore, the distance between the two cars is


BD = BC + CD = 100√3 + 100 = 100(√3 + 1) = 273 m.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Heights and Distances33 mins
Quiz | Imp. Qs. on Heights and Distances37 mins
NCERT | Basics of Heights and Distances37 mins
Idioms and Phrases43 mins
Goprep Genius Quiz | Analogy and Classification48 mins
Acid - Types and Nomenclature52 mins
Agriculture and its Importance35 mins
Foundation | Permutation and Combination45 mins
Goprep Genius Quiz | Direction and Distance58 mins
Light and it's Reflection58 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses