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Let’s assume the length of the shadow of the tree to be x m.
So, the height of the tree = √3 x m
If θ is the angle of elevation of the sun, then we have
tan θ = √3 x/ x = √3 = tan 60
o
Therefore, θ = 60
o
Let’s take the height of the tower to be h m.
Given that, the angle of elevation is 60
o
So, tan 60
o
= h/160
√3 = h/160
h = 160√3 = 277.12 m [For √3 = 1.732]
Thus, the height of the tower is 277.12 m.
Let’s take the height upto which the ladder reaches as ‘h’ m.
Given that, the angle of elevation is 68
o
So, tan 68
o
= h/ 2.4
2.475 = h/2.4
h = 2.475 x 2.4 = 5.94 m
Thus, the ladder reaches upto a height of 5.94 m.
Let one of the persons, A be at a distance of ‘x’ m and the second person B be at a distance of ‘y’ m from the foot of the tower.
Given that angle of elevation of A is 30
o
tan 30
o
= 50/x
1/√3 = 50/x
x = 50√3 = 86.60 m
And the angle of elevation of B is 38
o
So, tan 38
o
= 50/y
0.7813 = 50/ y
y ≈ 64 m
Thus, the distance between A and B is x + y = 150.6 m
Let’s assume the length of the rope to be x m.
Now, we have
sin 30
o
= 60/x
½ = 60/x
x = 120 m
Thus, the length of the rope is 120 m.
Solution:
In ∆ABC,
AB/BC = tan 45
o
= 1
So, BC = AB = X
In ∆ABD,
AB/BC = tan 45
o
= 1
So, BC = AB = X
In ∆ABD,
AB/BD = tan 30
o
X/(30 + X) = 1/ √3
30 + X = √3X
X = 30/ (√3 – 1) = 30/ (1.732 – 1) = 30/ 0.732
Thus, X = 40.98 m
Let’s assume AB to be the height of the tree, h m.
Let the two points be C and D be such that CD = 20 m, ∠ADB = 30
o
and ∠ACB = 60
o
In ∆ABC,
AB/BC = tan 60
o
= √3
BC = AB/ √3 = h/ √3 ….. (i)
In ∆ABD,
AB/BD = tan 30
o
h/ (20 + BC) = 1/ √3
√3 h = 20 + BC
√3 h = 20 + h/ √3 …… [From (i)]
h (√3 – 1/√3) = 20
h = 20/ (√3 – 1/√3) = 20/ 1.154 = 17.32 m
Therefore, the height of the tree is 17.32 m.
Let’s assume AB to be the building of height h m.
Let the two points be C and D be such that CD = 40 m, ∠ADB = 30
o
and ∠ACB = 45
o
In ∆ABC,
AB/BC = tan 45
o
= 1
BC = AB = h
And, in ∆ABD,
AB/ BD = tan 30
o
h/ (40 + h) = 1/√3
√3h = 40 + h
h = 40/ (√3 – 1) = 40/ 0.732 = 54.64 m
Therefore, the height of the building is 54.64 m.
Let’s consider AB to be the lighthouse.
And, let the two ships be C and D such that ∠ADB = 36
o
and ∠ACB = 48
o
In ∆ABC,
AB/BC = tan 48
o
BC = 100/ 1.1106 = 90.04 m
In ∆ABD,
AB/BD = tan 36
o
BD = 100/ 0.7265 = 137.64 m
Now,
(i) If the ships are on the same side of the light house,
Then, the distance between the two ships = BD – BC = 48 m
(ii) If the ships are on the opposite sides of the light house,
Then, the distance between the two ships = BD + BC = 228 m
Let AB and CD be the two towers of height h m each.
And, let P be a point in the roadway BD such that BD = 150 m, ∠APB = 60
o
and ∠CPD = 30
o
In ∆ABP,
AB/BP = tan 60
o
BP = h/ tan 60
o
BP = h/ √3
In ∆CDP,
CD/DP = tan 30
o
PD = √3 h
Now, 150 = BP + PD
150 = √3h + h/√3
h = 150/(√3 + 1/√3) = 150/ 2.309
h = 64.95 m
Thus, the height of the pillars are 64.95 m each.
Now,
The point is BP/ √3 from the first pillar.
Which is a distance of 64.95/ √3 from the first pillar.
Thus, the position of the point is 37.5 m from the first pillar.
Solution:
In ∆AED,
AE/ DE = tan 22
o
AE = DE tan 22
o
= 15 x 0.404 = 6.06 m
In ∆ABC,
AB/BC = tan 47
o
AB = BC tan 47
o
= 15 x 1.072 = 16.09 m
Thus,
CD = BE = AB – AE = 10.03 m
(i) In ∆ADE,
AE/DE = tan 45
o
= 1
AE = DE
In ∆ABC,
AB/BC = tan 60
o
= √3
AE + 30 = √3 BC
BC + 30 = √3 BC [Since, AE = DE = BC]
BC = 30/(√3 -1) = 30/ 0.732 = 40.98 m
Thus,
AB = 30 + 40.98 = 70.98 m
Thus, the height of the tower is 70.98 m
(ii) The horizontal distance from the points of observation is BC = 40.98 m
Given, CA = CB = 15 cm and ∠ACB = 131
o
Construct a perpendicular CP from centre C to the chord AB.
Then, CP bisects ∠ACB as well as chord AB.
So, ∠ACP = 65.5
o
In ∆ACP,
AP/AC = sin (65.5
o
)
AP = 15 x 0.91 = 13.65 cm
(i) AB = 2 AP = 2 x 13.65 = 27.30 cm
(ii) CP = AP cos (65.5
o
) = 15 x 0.415 = 6.22 cm
Let’s assume AB to be the vertical tower and C and D be the two points such that CD = 192 m.
And let ∠ACB = θ and ∠ADB = α
Given,
tan θ = 5/12
AB/BC = 5/12
AB = 5/12 BC …… (i)
Also, tan α = ¾
AB/BD = ¾
(5/12 x BC)/ BD = ¾
(192 + BD)/ BD = ¾ x 12/5
BD = 240 m
BC = (192 + 240) = 432 m
By (i), AB = 5/12 x 432 = 180 m
Therefore, the height of the tower is 180 m.
Let AB be the tower of height x metre, surmounted by a vertical flagstaff AD. Let C be a point on the plane such that ∠ACB = α, ∠ACB = β and AD = h.
In ∆ABC,
AB/BC = tan α
BC = x/ tan α ……. (i)
In ∆DBC,
BD/BC = tan β
BD = (x/tan α) x tan β …… [From (i)]
(h + x) tan α = x tan β
x tan β – x tan α = h tan α
Therefore, height of the tower is h tan α/ (tan β – tan α)
Solution:
Let take AD to be the height of the man, AD = 2 m.
So, CE = (10 – 2) = 8 m
(i) In ∆CED,
CE/DE = tan x = 2/5
8/DE = 2/5
⇒ DE = 20 m
And, as AB = DE we get,
AB = 20 m
(ii) Let A’D’ be the new position of the man and θ be the angle of elevation of the top of the tower.
So, D’E = 15 m
In ∆CED,
tan θ = CE/ D’E = 8/15 = 0.533
θ = 28
o
Let’s assume AB to be the tower of height h meters.
And, let C and D be two points on the level ground such that BC = b meters, BD = a meters, ∠ACB = α, ∠ADB = β.
Given, α + β = 90
o
In ∆ABC,
AB/BC = tan α
h/b = tan α ….. (i)
In ∆ABD,
AB/BD = tan β
h/a = tan (90
o
– α) = cot α ….. (ii)
Now, multiplying (i) by (ii), we get
(h/a) x (h/b) = 1
h
2
= ab
So, h = √ab meter
Therefore, height of the tower is √ab meter.
