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