.

Solution:
Let MN be the height of the tower and NO be the distance between the tower and man respectively.
Here,
O is the point of observation.
∠MON =30°, NO = 30m and MN = ?

From the right angled triangle MNO,
or, tan30° = \(\frac{MN}{NO}\)
or, \(\frac{1}{\sqrt{3}}\) = \(\frac{MN}{30}\)
or, MN = \(\frac{30}{\sqrt{3}}\)
∴ The height of the tower is\(\frac{30}{\sqrt{3}}\)m.

Solution:

/



Let EF be a height of a building and FG be the distance between the point on the ground level and foot of the building.
Here,
∠EFG = 60°, EF = 15m and EG = ?

From the right angled triangle EFG
or, tan60° = \(\frac{EF}{FG}\)
or, \(\sqrt{3}\) = \(\frac{15}{FG}\)
or, FG = \(\frac{15}{\sqrt{3}}\)
∴ The required distance is \(\frac{15}{\sqrt{3}}\)m.

Solution:

.

Let XY be the height of a pole and ZY be the length of its shadow. Let ∠XYZ =α be the altitude of the sun.
Here,
XY = 10ft, ZY = 10\(\sqrt{3}\) ft, α = ?

From the right angled triangle XYZ,
or, tanα = \(\frac{XY}{ZY}\)
or, tanα = \(\frac{10}{10\sqrt{3}}\)
or, tanα = \(\frac{1}{\sqrt{3}}\)
or, tanα = tan30°
∴ α = 30°
Hence, the altitude of te sun is 30°.

Note: altitude of the sun = angle of elevation of the sum.

Solution:

.

Let OP be the height of the tree and QP be the distance between the distance the object and the tree.
Here,
O is the point of observation.
∠AOQ = ∠OQP =45°, OP = 20m and QP = ?

From the right angled triangle OQP,
or, tan45° = \(\frac{OP}{QP}\)
or, 1 = \(\frac{20}{QP}\)
or, QP = 20
Hence, the distance between the object and the tree is 20m.

Solution:

.



Let, BC be the distance between tree, AC be height of tower and point is 30m and angle of elevation is 60°
Given,
∠B = 60° and BC = 30m
Now,
or, tan60° = \(\frac{AC}{BC}\)
or, \(\sqrt{3}\) = \(\frac{AC}{30}\)
∴ AC = 30\(\sqrt{3}\)m
The height of tree is 30\(\sqrt{3}\)m.

Solution:

/



Let, YZ be the distance between the point to the foot of a tree and XY is the height of tree which is 15m and the angle of elevation is 45°
Here,
XY = 15m, ∠XZY = 45° and YZ = ?
Now,
In right angle triangle XYZ,
or, tan45° = \(\frac{XY}{YZ}\)
or, 1 = \(\frac{15}{YZ}\)
or, YZ = 15m
∴ The distance of the point from the foot of the tree is 15m.

Solution:

.

Let, RQ be the point between the point of the foot of a tree and PR be the height of the tree which is 15m and angle of elevation is 30°
Here,
PR = 15m, ∠PQR = 30° and QR = ?
Now,
In right angled triangle
or, tan30° = \(\frac{PR}{QR}\)
or, \(\frac{1}{\sqrt{3}}\) = \(\frac{15}{QR}\)
or QR = 15\(\sqrt{3}\)
∴ The height of a kite is15\(\sqrt{3}\)m.

Solution:

.

Let, AB be the height of a tower and BC the distance between the tower and a point is 100m and angle of elevation is 45°
Here,
BC = 100m, ∠ABC = 45° and AB = ?
In right angle triangle
or, tan45° = \(\frac{p}{b}\)
or, 1 = \(\frac{AC}{100}\)
or, AC = 100m
∴ The height of the tower is 100m.

Solution:

.

Let, AB be the height of a pillar and AC is the shadow of pillar of height 30ft and angle of elevation is θ
Here,
AB = 30ft, BC = 30\(\sqrt{3}\) and ∠C = ?
Now,
or, tanθ = \(\frac{AB}{BC}\)
or, tanθ = \(\frac{30}{30\sqrt{3}}\)
or, tanθ = \(\frac{1}{\sqrt{3}}\)
or, tanθ = tan30°
∴ θ =30°
Hence,the altitude of the sun is 30°.

Solution:

.

Let AC be height of pole, BC be distance between pole of point is200\(\sqrt{3}\)m and angle of elevation is 30°
Given,
∠B = 30°, BC = 200\(\sqrt{3}\)m and AC = ?
Now,
or, tan30° = \(\frac{AC}{BC}\)
or, \(\frac{1}{\sqrt{3}}\) = \(\frac{AC}{200\sqrt{3}}\)
or, AC =\(\frac{200\sqrt{3}}{\sqrt{3}}\)
or, AC = 200
∴ The height of the pole is 200m.

Solution:

.

Height of tower =AB - BD = (61.6 − 1.6)m = 60m
Now,
or, tan60° = \(\frac{p}{b}\)
or, \(\sqrt{3}\) = \(\frac{60}{BC}\)
or, \(\sqrt{3}\)BC = 60
or, BC = \(\frac{60}{\sqrt{3}}\)
or,BC = \(\frac{60}{\sqrt{3}}\) × \(\frac{\sqrt{3}}{\sqrt{3}}\) (Multiplying both sides by\(\sqrt{3}\) )
or, BC = \(\frac{60\sqrt{3}}{3}\)
or, BC = 20\(\sqrt{3}\)
∴The distance between man and tower is 20\(\sqrt{3}\)m

Solution:

.

Let, height of a pole be AD, angle of elevation is 30°, height of man is 1.7m
Given,
∠C = 30°, CE = BD = 1.7m and CB = ED = 20m

From right angle triangle ABC
or, tan30° = \(\frac{AB}{BC}\)
or, \(\frac{1}{\sqrt{3}}\) = \(\frac{AB}{20}\)
or, \(\sqrt{3}\) AB = 20
or, AB = \(\frac{20}{\sqrt{3}}\)
or, AB = \(\frac{20}{\sqrt{3}}\)×\(\frac{\sqrt{3}}{\sqrt{3}}\) (Multiplying both side by \(\sqrt{3}\)
or, AB = \(\frac{20\sqrt{3}}{3}\)
∴ AB = 11.54m
Now,
AD = AB + BD
= 11.54 + 1.7 m
= 13.24m
∴ Height of pole is 13.24m.

Solution:

.

Let lenght of the tree be CA
From right angle triangle
or, sinθ = \(\frac{p}{h}\)
or, sin60° = \(\frac{14 − x}{x}\)
or, \(\frac{\sqrt{3}}{2}\) = \(\frac{14 − x}{x}\)
or, \(\sqrt{3}\) x = 28 − 2x
or, 1.73x = 28 − 2x
or, (1.73 + 2) x = 28
or, 3.37x = 28
or, x = \(\frac{28}{3.37}\)
∴ x = 7.5m
∴ The lenght of a tree is 7.5m.

Solution:

.

Let, AB be the angle made by the ladder with the horizontal ground and AC is a ladder of 200ft and BC is the base of wall.
Here,
AC = 200ft, CB = 100\(\sqrt{3}\) ft and AB = ?
By using Pythagoros Theorem
or, AB = \(\sqrt{AC^2− CB^2}\)
or, AB = \(\sqrt{200^2− (100\sqrt{3})^2}\)
or, AB = \(\sqrt{40000− 30000}\)
or, AB = \(\sqrt{10000}\)
or, AB = 100ft
Again,
or, sinC = \(\frac{AB}{AC}\)
or, sinC = \(\frac{100}{200}\)
or, sinC = sin 30°
∴ C = 30°
Hence, the angle made by the ladder with the horizontal ground is 30°.

Solution:

.

Let, height of tree be AB
From right angle triangle
or, tan45° = \(\frac{p}{b}\)
or, 1 = \(\frac{AB}{9}\)
∴ AB = 9m
Again,
or, cos45° = \(\frac{b}{h}\)
or, \(\frac{1}{\sqrt{2}}\) = \(\frac{9}{BC}\)
∴ BC = 9\(\sqrt{2}\)
Then,
AC = AB + CB
= 9 + 9\(\sqrt{2}\)
= 9 + 12.72
= 21.72m
Hence, the height of the tree before it was broken is 21.72m