From the top of tower. 100 …

CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Related Questions
Posted by Sahil Sahil 1 year, 3 months ago
- 2 answers
Posted by Parinith Gowda Ms 2 months, 2 weeks ago
- 1 answers
Posted by Hari Anand 5 months, 1 week ago
- 0 answers
Posted by Vanshika Bhatnagar 1 year, 3 months ago
- 2 answers
Posted by Kanika . 3 days, 9 hours ago
- 1 answers
Posted by Parinith Gowda Ms 2 months, 2 weeks ago
- 0 answers

myCBSEguide
Trusted by 1 Crore+ Students

Test Generator
Create papers online. It's FREE.

CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
myCBSEguide
Gaurav Seth 4 years, 11 months ago
Let the Tower be AC = 100m
Distance between cars is BD
BD=BC+CD
Let the BC be 'x' and CD be 'y'
In Traingle ACD
Cot45°= Base/Perpendicular
Cot45°=y/100
1=y/100
y=100m
In Traingle ACB
Cot30°=x/100
1.732=x/100
1.732×100=x
x=173.2m
Distance between cars = BC+CD
=x+y
=173.2+100
=273.2m
Or
0Thank You