Q.There are two temples, one on each bank of a river, just opposite to each other. One temple is 50m high. From the top of this temple, the angles of depression of the top &the foot of the other temple are 30 0 & 60 0 respectively. Find the width of the river and the hight of the other temple.
Answers
Answered by
3
Let AB and CD be the two temples and BC be the width of the river
Then , ADE=30° and ACB= 60°
Given: AB=50m
In ∆ ABC
tan 60°=AB/BC
= √3=50/BC
= BC =50/√3=50√3/3=28.867
In ∆ AED
tan 30°=AE/ED =AB-BE/BC=AB-DC/BC
= 1/√3=50-DC/50/√3
=50/3=50-DC
=DC=50-50/3=150-50/3=100/3
=33.33m
Hence, the width of the river is 28.867m and height of the other temple is 33.33m
Then , ADE=30° and ACB= 60°
Given: AB=50m
In ∆ ABC
tan 60°=AB/BC
= √3=50/BC
= BC =50/√3=50√3/3=28.867
In ∆ AED
tan 30°=AE/ED =AB-BE/BC=AB-DC/BC
= 1/√3=50-DC/50/√3
=50/3=50-DC
=DC=50-50/3=150-50/3=100/3
=33.33m
Hence, the width of the river is 28.867m and height of the other temple is 33.33m
Attachments:
Similar questions