The angles of depression of the top and the bottom of a 15 m tall building from the top of a multistoried building are 30 and 60°, respectively. Find the height of the multistoried building and the distance between the two buildings. (Take V3 = 1.732).
Answers
Answer:
AB and CD be the multi-storied building and the building respectively.
Let the height of the multi-storied building be hm and the distance between the two building be xm
AE=CD=8m [ Given ]
BE=AB−AE=(h−8)m and
AC=DE=xm [ Given ]
Now, in △ACB,
⇒ tan45o=ACAB
⇒ 1=xh
∴ x=h ---- ( 1 )
In △BDE,
⇒ tan30o=EDBE
⇒ 31=xh−8
∴ x=3(h−8) ------ ( 2 )
⇒ h=3h−83
⇒ 3h−h=83
⇒
Step-by-step explanation:
AB and CD be the multi-storied building and the building respectively.
Let the height of the multi-storied building be hm and the distance between the two building be xm
AE=CD=8m [ Given ]
BE=AB−AE=(h−8)m and
AC=DE=xm [ Given ]
Now, in △ACB,
⇒ tan45o=ACAB
⇒ 1=xh
∴ x=h ---- ( 1 )
In △BDE,
⇒ tan30o=EDBE
⇒ 31=xh−8
∴ x=3(h−8) ------ ( 2 )
⇒ h=3h−83
⇒ 3h−h=83
⇒
Answer:
the correct answer is 1 by 2