P 24
From a point P on the ground level, the angle of elevation of the roof of the
building is 45" The angle of elevation of the centre of logo is 30° from same point
The point P is at a distance of 24 m from the base of the building
On the basis of the above information, answer any four of the following questions:
(i) What is the height of the building logo from ground?
(h) 8V3 m (b) 8/2 m (c) 473 m (d) 4/2 m
(ii) What is the height of the building from ground?
(a) 243-V3)m (178(3-V3)m (c) 24m (d) 32m
(iii) What is the nerial distance of the point P from the top of the building (Flint PC )?
(WY
(24√3 m (6) 24V2 m (0)32/3 m (d) 3272 m
(iv) If the point of observation Pis moved 9 m towards the base of the building, then the
angle of elevation 0 of the logo on building is given by
(a) tand) = v3 (b) tan0 = 7 (c) ton0 (d) tan0 =
(v) In above case the angle of elevation of the top of building is given by 21
(o) tan o 1.6 (b) tan o 1.5 (c) tan o 0.75 V6 tano 0.8
3
Answers
1:- A) 8√3
2:- A) 24m
3:- C) 24√3
4:- D) tan=8√3/15
5:- A) tan = 1.6
Answer:
i. The height of the building logo from ground is 8V3 m
ii. The height of the building from ground is 24 m
iii.
The aerial distance of the point P from the top of the building is 24V2 m
iv. If the point of observation moved 9 m towards the base of the building, then the angle of elevation 0 of the logo on building is given by 42.73
Step-by-step explanation:
Given:
Point P is at a distance = d = 24 m
Angle of elevation to roof = 45 degree
Angle of elevation to logo = 30 degree
i.
Height of the building logo = BC
tan 30 = BC/d
So,
BC = d*tan30 = 24*tan 30 = 13.856 m = 8V3 m
ii.
Height of the building to Roof = AC
tan 45 = AC/d)
So,
AC = d*tan30 = 24*tan 45 = 24 m
iii.
The Aerial distance of the point P from the top of the building= S
S = [(AC^2) +(d^2)]^0.5 = 24V2 m
iv.
If the point of observation moved 9 m towards the base of the building, then the angle of elevation of the logo on building is given by theta
tan theta = BC/(24-9) = 8V3/15
So, theta = 42.73 degree