The height of the cliff is
I. The angle of elevation of the cliff from a fixed point F is 45∘
II. After going up towards a distance of 1000 m at an inclination of 30∘, the angle of elevation is 60∘
Both statements I and II are required
In ΔFPL,
sin30∘=PLPF⇒PL=PF sin30∘=(1×12)km=12 km∴OM=PL=12 km⇒MS=OS−OM=(h−12) km
Also, cos30∘=FLPF⇒FL=PFcos30∘=(1×√32)=√32 km
Now, h=OS=OL+LF⇒h=OL+√32⇒OL=(h−√32) km⇒PM=(h−√32) km
In ΔSPM,
tan60∘=SMPM⇒SM=PM tan60∘⇒(h−12)=(h−√32)√3⇒(h−12)=(√3n−32)⇒(√3h−h)=32−12
⇒h(√3−1)=1⇒h=1√3−1
Both the statements are required