Subject: Optional Mathematics

This note explains about the angle of depression and angle of elevation.

One of the main uses of trigonometry is to find the height of an object or the distance between two points. The instruments called theodolite and sextant are used to measure certain angles and then the method of solution of triangles is used to find the required height or distance.

**Angle of elevation**

In the adjoining figure, O is the position of the observer, P is the position of the object and OP is the line of vision. OA is the horizontally through the observation point O. the angle AOP formed when the observer observes the object at P above horizontal line is called an angle of elevation.

**Angle of depression**

In the adjoining figure, O is the position of the observer, is the position of the object and P is the line of vision.

OP is the horizontal line through the observation point. The angle AOP formed when the observer observes the object at P below the horizontal line is called angle of depression.

- Angle of elevation
- Angle of depression

- It includes every relationship which established among the people.
- There can be more than one community in a society. Community smaller than society.
- It is a network of social relationships which cannot see or touched.
- common interests and common objectives are not necessary for society.

The angle of elevation of the top of a tower from a point was observed to be 45°. On walking 30m, away from the point it was found to be 30°. Find the height of the tower.

Here,

Height of the tower (PQ) = ?

Distance between R and S (RS) = 30m

Angle of elevation (\(\angle\) PRQ) = 45°

Angle of elevation (\(\angle\) PSQ) = 30°

We have,

In right angled triangle \(\triangle\) PQR:

tan 45° = \(\frac {PQ}{PR}\)

1 = \(\frac {PQ}{PR}\)

∴ PR = PQ

In right angled triangle \(\triangle\) PQS:

tan 30° = \(\frac {PQ}{PS}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {PQ}{30 + PR}\) [\(\because\) PQ = PR]

or, 30 + PQ = \(\sqrt 3\) PQ

or, \(\sqrt 3\) PQ - PQ = 30

or, PQ (\(\sqrt 3\) - 1) = 30

or, PQ = \(\frac {30}{\sqrt 3 - 1}\)

or, PQ = \(\frac {30}{1.732 - 1}\)

or, PQ = \(\frac {30}{0.732}\)

∴ PQ = 40.98 m

∴ The height of tower (PQ) = 40.98 m _{Ans}

The angle of elevation of a tower was observed to be 60° from a point. On walking 200 m away from the point it was found to be 30°. Find the height of the tower.

Let: AB be the tower and C be initial point of observation.

∴ \(\angle\) ACB = 60°

D is the point 200 m away from C.∴ \(\angle\) ADB = 30°

Height of the tower (AB) = ?

In right angled triangle ABC,

tan 60° = \(\frac {AB}{BC}\)

or, \(\sqrt 3\) = \(\frac {AB}{BC}\)

∴ BC = \(\frac {AB}{\sqrt 3}\) .............................(i)

In right angled triangle ABD,

tan 30° = \(\frac {AB}{BC}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {AB}{BC + CD}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {AB}{\frac {AB}{\sqrt 3} + 200}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {AB}{\frac {AB + 200\sqrt 3}{\sqrt 3}}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {\sqrt 3 AB}{AB + 200\sqrt 3}\)

or, AB + 200 \(\sqrt 3\) = 3 AB

or, 3 AB - AB = 200\(\sqrt 3\)

or, 2 AB = 200\(\sqrt 3\)

or, AB = \(\frac {200\sqrt 3}{2}\)

or, AB = 100× 1.732

∴ AB = 173.2 m

Hence, the height of tower = 173.2 m _{Ans}

Two men are on the opposite side of a tower of 30 m high. They observed the angle of elevation of the tower and found to be 30° and 60°. Find the distance between them.

Let:Height of tower (PS) = 30 m

Position of two men Q and R.

Angle of elevation \(\angle\)PQS = 60° and \(\angle\)PRS = 30°

Distance between two men (QR) = ?

In right angled triangle PQS,

tan 60° = \(\frac {PS}{QS}\)

or, \(\sqrt 3\) = \(\frac {30}{QS}\)

or, QS = \(\frac {30}{\sqrt 3}\)

or, QS = \(\frac {10.\sqrt 3.\sqrt 3}{\sqrt 3}\)

∴ QS = 10\(\sqrt 3\) m

In right angled triangle PRS,

tan 30° = \(\frac {PS}{RS}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {30}{RS}\)

∴ RS = 30\(\sqrt 3\)

QR

= QS + SR

= 10\(\sqrt 3\) + 30\(\sqrt 3\)

= 40\(\sqrt 3\)

= 40× 1.732

= 69. 28 m _{Ans}

The angle of elevation of the top of a tower as observed for a point the ground is found to be 60°. On walking 60m, away from the point, the angle of the elevation was found to be 45°. Find the height of the tower.

Let: AB be the height of the tower. The angle of elevation observed from a point C to the top of the tower B is \(\angle\)ACB = 60° and from the D is \(\angle\)ADB = 45°.

Distance of CD = 60 m

In the right angled triangle ABC,

tan 60° = \(\frac {AB}{AC}\)

or, \(\sqrt 3\) = \(\frac {AB}{AC}\)

∴ AB = \(\sqrt 3\) AC.........................(i)

In right angled triangle ABD,

tan 45° = \(\frac {AB}{AD}\)

or, 1 = \(\frac {AB}{AD}\)

∴ AB = AD...............................................(ii)

From equation (i) and (ii)

AD = \(\sqrt 3\) AC

or, AC + CD = \(\sqrt 3\) AC

or, \(\sqrt 3\)AC - AC = CD

or, AC (\(\sqrt 3\) - 1) = 60

or, AC (1.732 - 1) = 60

or, AC = \(\frac {60}{0.732}\)

∴ AC = 81.97 m

AD = AC + CD = 81.97 + 60 = 141.97 m

AB = AD = 141.97 m

∴ The height of tower = 141.97 m _{Ans}

A pole is surmounted on its top by a flagstaff. The angles of elevation of the top and the bottom of the flagstaff as observed from a point 30 meter away from the bottom of the pole and found to be 45° and 30° respectively. Find the height of the flagstaff.

Let:

The height of the pole = AB

The height of the flagstaff = BC = ?

The angle of elevation bottom and top of the flagstaff are 30° and 45°.

The distance between the pole and observation point = 30 m

In right angled triangle ACD,

tan 45° = \(\frac {AC}{AD}\)

or, 1 = \(\frac {AC}{30}\)

∴ AC = 30 m

In right angled triangle ABD,

tan 30° = \(\frac {AB}{AD}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {AB}{30}\)

or, AB \(\sqrt 3\) = 30

or, AB = \(\frac {30}{\sqrt 3}\)

or, AB = \(\frac {30}{1.732}\)

∴ AB = 17.32 m

Hence, the height of flagstaff BC = AC - AB = (30 - 17.32) = 12.68 m _{Ans}

Two lamps posts of equal heights are standing opposite to each other on either side of a road which is 80m wide from a point between them on the road, the angles of elevation of the tops are 30° and 60°. Find the position of the point and the height of the posts.

Let: AB and CD be two lamp posts of equal heights. E be the point of observation.

BD = 80 m

\(\angle\)AEB = 60°

\(\angle\)CED = 30°

BE = ?

AB = CD = ?

In right angled triangle ABE,

tan 60° = \(\frac {AB}{BE}\)

or, \(\sqrt 3\) = \(\frac {AB}y\)

or, AB = y\(\sqrt 3\) .........................(i)

In right angled triangle CDE,

tan 30° = \(\frac {CD}{DE}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {CD}{80 - y}\)

or, CD = \(\frac {80 - y}{\sqrt 3}\)..........................(ii)

We have,

AB = CD

y\(\sqrt 3\) = \(\frac {80 - y}{\sqrt 3}\)

or, 3y = 80 - y

or, 3y + y = 80

or, 4y = 80

or, y = \(\frac {80}4\)

∴ y = 20

Putting the value of y in eq^{n} (i)

AB = CD = \(\sqrt 3\)y = 20\(\sqrt 3\) = 20× 1.732 = 34.65 m

The point is 20 m from the second post and height of posts each is 34.65 m. _{Ans}

from the top of a building 30 m high a man observes two persons sitting on the ground both due east on the same line at angles of depression of 45° and 30°. How far apart are the two persons?

Let:

Height of the building (AB) = 30 m

C and D be the positions of two men standing due to east from the building. The angles depression are \(\angle\)MBC = 45° and \(\angle\)MBD = 30°.

The distance between the two persons (CD) = ?

\(\angle\)MBD = \(\angle\)ADB = 30°

\(\angle\)MBC = \(\angle\)ACB = 45°

In the right angled triangle ABC,

tan 45° = \(\frac {AB}{AC}\)

or, 1 = \(\frac {30}{AC}\)

∴ AC = 30 m

In right angled triangle ABD,

tan 30° = \(\frac {AB}{AD}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {30}{AC + CD}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {30}{30 + CD}\)

or, 30 + CD = 30\(\sqrt 3\)

or, CD = 30× 1.732 - 30

or, CD = 51.96 - 30

∴ CD = 21.96 m

Hence, the distance between two men = 21.96 m _{Ans}

The angles of elevation of the top of a tower from two points 'a' meters and 'b' meters from the box in the same straight line with it, are complementary. Prove that the height of the tower is \(\sqrt {ab}\).

Let: AB be the tower and C and D be two points on the ground such that BC = a meters BD = b meters and \(\angle\)ACB = \(\theta\), then:

\(\angle\) ADB = 90 - \(\theta\)

In right angled triangle ABC,

tan \(\theta\) = \(\frac {AB}{BC}\)

or, tan\(\theta\) = \(\frac {AB}a\)............(i)

In right angled triangle ABD,

tan(90 - \(\theta\)) = \(\frac {AB}{BD}\)

or, cot\(\theta\) = \(\frac {AB}{BD}\)

or, \(\frac 1{tan\theta}\) = \(\frac {AB}b\)

or, \(\frac 1{\frac {AB}a}\) = \(\frac {AB}b\) [\(\because\) From equation (i)]

or, \(\frac a{AB}\) = \(\frac {AB}b\)

or, (AB)^{2} = ab

∴ AB = \(\sqrt {ab}\)

Hence, the height of the tower is \(\sqrt {ab}\) m. _{Proved}

The upper part of a tree broken by the wind in two parts, make an angle 30° with the ground. The top of the tree touches the ground at a distance 9 m from the foot of the tree. Find the height of the tree.

Suppose: AB be tree which is broken at the point C and the upper part makes an angle 30° with ground 9 m apart from the foot of the tree.

where,

Ac = CD

BD = 9 m

AB = ?

In right angled triangle CBD,

tan 30° = \(\frac {CB}{BD}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {CB}9\)

or, \(\sqrt 3\) CB = 9

or, BC = \(\frac 9{\sqrt 3}\)

or, BC = \(\frac 9{\sqrt 3}\)× \(\frac {\sqrt 3}{\sqrt 3}\)

or, BC = 3\(\sqrt 3\) m

Again,

In right angled triangle,

sin 30° = \(\frac {BC}{CD}\)

or, \(\frac 12\) = \(\frac {3\sqrt 3}{CD}\)

or, CD = 6\(\sqrt 3\) m

Now,

The height of the tree:

= AB

= AC + BC

= CD + BC

= 6\(\sqrt 3\) + 3\(\sqrt 3\)

= 9\(\sqrt 3\) m _{Ans}

From the top of 21 m high cliff, the angles of depression of the top and the bottom of a tower are observed to be 45° and 60° respectively. Find the height of the tower.

Here,

Height of the cliff (AB) = 21 m

Height of the tower (CD) = ?

Angle of depression (\(\angle\)MBD) = 45° = \(\angle\)BDE

Angle of depression (\(\angle\)MBC) = 60° = \(\angle\)BCA

In right angled triangle BDE,

tan 45° = \(\frac {BE}{DE}\)

or, 1 = \(\frac {BE}{DE}\)

or, BE = DE............................(i)

In right angled triangle ABC

tan 60° = \(\frac {AB}{AC}\)

or, \(\sqrt 3\) = \(\frac {21}{AC}\)

or, AC = \(\frac {21}{\sqrt 3}\)

∴ AC = 12.12 m

BE = AC = DE = 12.12

AE = CD = AB - BE = 21 - 12.12 = 8.88 m

∴ The height of tower = 8.9 m _{Ans}

From the top of pole 25 m high the angle of elevation of the top of a tower is 45° and the angle of depression of the foot of the pole from the top of the towers 60°. Find the height of the tower.

Let:

Height of pole (AB) = 25 m

Height of tower = CE

Angle of elevation \(\angle\)EAD = 45°

Angle of depression \(\angle\)MEB = \(\angle\)CBE = 60°

AB = CD

AD = BC

In right angled \(\triangle\)ADE,

tan 45° = \(\frac {DE}{AD}\)

or, 1 = \(\frac {DE}{AD}\)

or, DE = AD = BC.........................................(i)

In right angled \(\triangle\)BCE,

taan 60° = \(\frac {CE}{BC}\)

or, \(\sqrt 3\) = \(\frac {DE + CD}{BC}\)

or, \(\sqrt 3\)DE = DE + 25

or, \(\sqrt 3\)DE - DE = 25

or, DE (\(\sqrt 3\) - 1) = 25

or, DE (1.732 - 1) = 25

or, DE = \(\frac {25}{0.732}\)

∴ DE = 31.15 m

CE = CD + DE = 25 + 31.15 = 59.15 m

∴ The height of the tower = 59.15 m _{Ans}

A man 2 m high observes that the angle of elevation of the tops of a house and a window on its side are 45° and 30° respectively. If the window tops is at the height of 12 m above the ground. Find the height of the house.

Let:

EB be the height of man = 2 m

\(\angle\)ABC and \(\angle\)DBC are the angle of elevation.

\(\angle\)ABC = 45°

\(\angle\)DBC = 30°

DF is the height of window above the ground.

DF = 12 m

CF = BE = 2 m

CD = DF - CF = 12 - 2 = 10 m

In the right angled triangle BCD,

tan 30° = \(\frac {CD}{BC}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {10}{BC}\)

or, BC = 10\(\sqrt 3\) m

Again,

In right angled triangle ABC,

tan 45° = \(\frac {AC}{BC}\)

or, 1 = \(\frac {AD + CD}{10\sqrt 3}\)

or, 10\(\sqrt 3\) = AD + 10

or, AD = 10\(\sqrt 3\) - 10

or, AD = 10× 1.732 - 10

or, AD = 17.32 - 10

∴ AD = 7.32 m

Hence, the height of the house (AF) = AD + DF = 7.32 + 12 = 19.32 m _{Ans}

A man of 1.68 m height observed the angles of elevation of the top of a house and its window from a place and found to be 45° and 30° respectively. If the height of the window from the ground is 11.68 m, calculate the height of the house.

Let:

Height of man (MN) = 1.68 m

Height of the house (PR) = ?

Height of the window (QR) = 11.68 m

Angle of elevation: \(\angle\)QMS = 30° and \(\angle\)PMS = 45°

From figure,

MN = SR = 1.68 m

QS = 11.68 - 1.68 = 10 m

In the right angled \(\triangle\)QMS

tan 30° = \(\frac {QS}{MS}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {10}{MS}\)

or, MS = 10\(\sqrt 3\) m

Again,

In right angled triangle PMS,

tan 45° = \(\frac {PS}{MS}\)

or, 1 = \(\frac {PS}{10\sqrt 3}\)

or, PS = 10\(\sqrt 3\)

or, PS = 10× 1.732

∴ PS = 17.32 m

Hence, the height of house (PR) = PS + SR = 17.32 + 1.68 = 19 m _{Ans}

Find height of CD, from the adjoining figure.

From the figure,

Height of CD = ?

AF = BD

AB = DF = 10 m

CE = 30 m

In the right angled triangle AFE,

tan 30° = \(\frac {EF}{AF}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {EF}{AF}\)

or, AF = \(\sqrt 3\)EF....................................(i)

In the right angledtriangle ACF,

\(\angle\)CAF = 30° + 15° = 45°

tan 45° = \(\frac {CE}{AF}\)

or, 1 = \(\frac {CE + EF}{AF}\)

or, AF = CE + EF..........................(ii)

From equation (i) and (ii)

\(\sqrt 3\)FE = CE + EF

or, \(\sqrt 3\)EF - EF = 30

or, EF (\(\sqrt 3\) - 1) = 30

or, EF (1.732 - 1) = 30

or, EF = \(\frac {30}{0.732}\)

∴ EF = 40.98 m

CD = CE + EF + DF = 30 + 40.98 + 10 = 80.98 m

Hence, the height of CD is 80.98 m. _{Ans}

From the top of 42 m high cliff, the angle of depression of the top and bottom of a tower are observed to be 30° and 45° respectively. Find the height of the tower.

Let:

Height of the cliff (AB) = 42 m

Height of the tower (CD) = ?

\(\angle\)OAC = \(\angle\)ACE = 30°

\(\angle\)OAD = \(\angle\)ADB = 45°

In the right angled triangle ABD,

tan 45° = \(\frac {AB}{BD}\)

or, 1 = \(\frac {AB}{BD}\)

∴ BD = AB...................................(1)

In the right angled triangle AEC,

tan 30° = \(\frac {AE}{CE}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {AB - BE}{BD}\) = \(\frac {AB - CD}{AB}\)

or, \(\frac {42 - CCD}{42}\) = \(\frac 1{\sqrt 3}\)

or, 42 - CD = \(\frac {42}{\sqrt 3}\)

or, CD = 42 - \(\frac {42}{\sqrt 3}\)

or, CD = 42 - 24.25

∴ CD = 17.75 m

Hence, the height of the tower is 17.75 m. _{Ans}

The angle of depression and elevation of the top of a pole 25 m high from the top and bottom of a tower are 60° and 30° respectively, find the height of the tower.

Let:

Height of pole (AB) = 25 m

Height of tower (CE) = ?

Angle of depression (\(\angle\)MEA) = 60°

\(\angle\)DAE = \(\angle\)MEA = 60°

Angle of elevation (\(\angle\)ACB) = 30°

In right angled \(\triangle\)ABC,

tan 30° = \(\frac {AB}{BC}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {25}{BC}\)

∴ BC = 25\(\sqrt 3\)

Again,

AD = BC = 25\(\sqrt 3\)

In right angled \(\triangle\)ADE,

tan 60° = \(\frac {DE}{AD}\)

or, \(\sqrt 3\) = \(\frac {DE}{25\sqrt 3}\)

or, DE = 25\(\sqrt 3\)× \(\sqrt 3\)

∴ DE = 75 m

∴ The height of the tower = 100 m _{Ans}

From the top of a tower, the angle of depression to the bottom of a pillar on the same surface of a height 20 m is 60°. If the angle of depression from the top of the pillar to the bottom of the tower is 22°. Find the height of the tower.

Let Ab and CD be the height of the pillar and tower respectively.

Let AC be the distance between them.

From B, draw BE//FD//AC.

Ab = 60 m

\(\angle\)DBE = 60°

\(\angle\)BCA = 22°

In the right angled\(\triangle\)ABC,

tan 22° = \(\frac {AB}{BC}\)

or, 0.4 = \(\frac {60}{AC}\)

or, AC = \(\frac {60}{0.4}\)

∴ AC = 150 m

Also,

BE = AC = 150 m

In right angled \(\triangle\)BDE,

tan 60° = \(\frac {DE}{BE}\)

or, \(\sqrt 3\) = \(\frac {DE}{150}\)

∴ DE = 150\(\sqrt 3\) m

∴ Height of tower = CD = CE + DE = AB + DE = 60 + 150\(\sqrt 3\) = 319.81 m _{Ans}

The angles of depression and elevation of the top of a tower, 20 meters high, from the top and bottoms of a second tower are 60° and 30° respectively. Find the height of the second tower.

Let:

AB be the tower 20 m height and CD be another tower.

\(\angle\)ADB = 30°

\(\angle\)XCA = \(\angle\)CAE = 60°

AB = ED = 20 m

CD =?

In right angled triangle ABD,

tan 30° = \(\frac {AB}{BD}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {20}{BD}\)

or, BD = 20\(\sqrt 3\) m

∴ BD = AE = 20\(\sqrt 3\) m

In right angled triangle CEA,

tan 60° = \(\frac {CE}{AE}\)

or, \(\sqrt 3\) = \(\frac {CE}{20\sqrt 3}\)

or, 20\(\sqrt 3\)× \(\sqrt 3\) = CE

∴ CE = 60 m

∴ The height of the second tower = CD = CE + ED = 60 + 20 = 80 m _{Ans}

From the top of the tower 96 m high, the angle of depression, of two vehicles on a road at the same level as the base of the tower and on the same side of it, are x° and y° where tan x° = \(\frac 34\) and tan y° = \(\frac 13\). Calculate the distance between the vehicles.

Let:

AB be the tower and C and D are two points on the road where vehicles are stand. \(\angle\)ACB = x° and \(\angle\)ADB = y°.

Now,

In right angled triangle ABC,

tan x° = \(\frac {AB}{BC}\)

or, \(\frac 34\) = \(\frac {96}{BC}\)

or, 3 BC = 96× 4

or, BC = \(\frac {96 × 4}3\)

∴ BC = 128 m

In right angled triangle ABD,

tan y° = \(\frac {AB}{BD}\)

or, \(\frac 13\) = \(\frac {96}{BD}\)

or, BD = 3× 96

∴ BD = 288 m

∴ The distance between two vehicles = CD = BD - BC = 288 - 128 = 160 m _{Ans}

From a distance of 200 m, the elevation of the top of a pillar is found to be an angle. If it is observed by approaching 125 m

towers the pillar from the points, it is found to be double of the angle. Find the height of the pillar.

Let:

Height of the pillar (AB) = ?

The foot of the pillar and observation point (AC) = 200 m

The length between two observation point (CD) = 125 m

The angle of elevation from point C = \(\theta\) and the angle of elevation from point D = 2\(\theta\), \(\angle\)ACB = \(\theta\) and \(\angle\)ADB = 2\(\theta\), AD = 200 - 125 = 75 m

In right angled \(\triangle\)ABC,

tan\(\theta\) = \(\frac {AB}{AC}\)

or, tan \(\theta\) = \(\frac {AB}{200}\).....................................(i)

In right angled\(\triangle\)ABD,

tan 2\(\theta\) = \(\frac {AB}{AD}\)

or, \(\frac {2 tan\theta}{1 - tan^2\theta}\) = \(\frac {AB}{75}\)..................................(ii)

Putting the value of tan\(\theta\) from equation (i)

or, \(\frac {2 × \frac {AB}{200}}{1 - (\frac {AB}{200})^2}\) = \(\frac {AB}{75}\)

or, \(\frac {AB}{100}\)× 75 = AB (1 - \(\frac {{AB}^2}{40000}\))

or, \(\frac {3AB}4\) = AB (\(\frac {40000 - {AB}^2}{40000}\))

or, \(\frac {3 × 40000}4\) = 40000 - AB^{2}

or, AB^{2} = 40000 - 30000

or, AB^{2} = 10000

∴ AB = 100 m

∴ The height of the pillar = 100 m _{Ans}

A pole is painted in two colours in the ratio of 1 : 9 from the base. If both of the parts of the pole subtend equal angle at the point 20 m from it. Find the height of the pole.

Let:Height of the pole ACbe derived into two parts BC and AB from the base. If D be the point at which they subtend equal angles.

Let:

AB = 9x

BC = x

∴ AC = 10x

\(\angle\)ADB = \(\angle\)BDC = \(\theta\)

CD = 20 m

Height of the pole (AC) = ?

In right angled\(\triangle\)BCD,

tan\(\theta\) = \(\frac {BC}{CD}\)

or, tan\(\theta\) = \(\frac x{20}\).......................(i)

In right angled \(\triangle\)ACD,

tan (\(\theta\) + \(\theta\)) = \(\frac {AC}{CD}\)

or, tan 2\(\theta\) = \(\frac {10x}{20}\)

or, \(\frac {2 tan\theta}{1 - tan^2\theta}\) = \(\frac x2\)..........................(ii)

Putting the value of tan\(\theta\) on eq^{n} (ii),

\(\frac {2 (\frac x{20})}{1 - (\frac x{20})^2}\) = \(\frac x2\)

or, \(\frac {\frac x{10}}{1 - \frac {x^2}{400}}\) = \(\frac x2\)

or, \(\frac {\frac x{10}}{\frac {400 - x^2}{400}}\) = \(\frac x2\)

or, \(\frac x{10}\)× \(\frac {400}{400 - x^2}\) = \(\frac x2\)

or, \(\frac {40x × 2}x\) = 400 - x^{2}

or, 80 = 400 - x^{2}

or, x^{2} = 400 - 80

or, x^{2} = 320

∴ x = 17.89 m

AC = 10x = 10× 17.89 = 178.9 m

∴ The height of the pole = 17.89 m _{Ans}

A radio transmitter antenna of height h stands of the top of a building. At a point on the ground, the angle of elevation of the bottom of the antenna is \(\alpha\) and that of the top of the antenna is \(\theta\). Prove that the height of the building is: \(\frac {h tan\alpha}{tan\theta - tan\alpha}\)

Let:

Height of the building = BC

The height of the antenna = CD = h. The distance between observer and foot of the building = AB. The angle of elevation of the antenna = \(\theta\) and the angle of elevation bottom of the antenna = \(\alpha\).

In right angled \(\triangle\)ABD,

tan\(\theta\) = \(\frac {BD}{AB}\) = \(\frac {BC + CD}{AB}\)

∴ AB = \(\frac {BC + CD}{tan\theta}\).................................(i)

In right angled \(\triangle\)ABC,

tan\(\alpha\) = \(\frac {BC}{AB}\)

or, AB = \(\frac {BC}{tan\alpha}\).........................................(ii)

From equation (i) and (ii)

\(\frac {BC + CD}{tan\theta}\) = \(\frac {BC}{tan\alpha}\)

or, BC tan\(\alpha\) + CD tan\(\alpha\) = BC tan\(\theta\)

or, BC tan\(\theta\) - BC tan\(\alpha\) = CD tan\(\alpha\)

or, BC (tan\(\theta\) - tan\(\alpha\)) = h tan\(\alpha\) [\(\because\) CD = h]

∴ BC =\(\frac {h tan\alpha}{tan\theta - tan\alpha}\)

Hence, Proved.

From a point on the horizontal plane, the angle of elevation of the top of a pillar standing on the same plane was observed and found to be 60° and the angle of elevation of a point 20 m below the top of the pillar was found to be 30°. Find the height of the pillar.

Let:

Height of the pillar=BD = ?

angle of elevation top of the pillar = \(\angle\)BAD = 60°

Angle of elevation 20 m below the top of the pillar = \(\angle\)BAC = 30°

∴ The distance between pillar and observation point = AB

In the right angled \(\triangle\)ABC,

tan 30° = \(\frac {BC}{AB}\)

or, \(\frac 1{\sqrt 3}\) = \(\frac {BC}{AB}\)

∴ AB = \(\sqrt 3\)BC

In right angled \(\triangle\)ABD,

tan 60° = \(\frac {BD}{AB}\)

or, \(\sqrt 3\) = \(\frac {BC + CD}{\sqrt 3 BC}\)

or, 3 BC = BC + 20

or, 3 BC - BC = 20

or, 2 BC = 20

or, BC = \(\frac {20}2\)

∴ Bc = 10 m

BD = BC + CD = 10 + 20 = 30 m

∴ The height of the pillar = 30 m _{Ans}

A ladder 9 m long reaches to a point below the top of building. From the foot of the ladder, the angle of elevation of the building is 60°. Find the height of the building. Here, the ladder makes the angle of 45° with the ground.

Let:

Height of building (AC) = ?

Length of ladder (BD) = 9 m

Angle of elevation on the top of building \(\angle\)ABC = 60°

Angle of elevation makes by ladder with ground \(\angle\)CBD = 45°

In the right angled \(\triangle\)BCD,

cos 45° = \(\frac {BC}{BD}\)

or, \(\frac 1{\sqrt 2}\) = \(\frac {BC}9\)

or, BC = \(\frac 9{\sqrt 2}\)

∴ BC = 6.36 m

In the right angled \(\triangle\)BCD,

tan 45° = \(\frac {CD}{BC}\)

or, 1 = \(\frac {CD}{6.34}\)

∴ CD = 6.34 m

In right angled \(\triangle\)ABC,

tan 60° = \(\frac {AC}{BC}\)

or, \(\sqrt 3\) = \(\frac {AD + CD}{BC}\)

or, \(\sqrt 3\)× 6.36 = AD + 6.36

∴ AD = 4.66 m

∴ Height of building = AD + CD = 6.36 + 4.66 = 11.02 m _{Ans}

Two points are 120 meters apart and the height of one is double of other from the middle points of the joining their feet and angle observe finds the angular elevation of their tops to be complementary. Find the height of the ports.

Let: The height of posts AB and DE distance between the two post is 120 m. Angle of elevation \(\angle\)BCA = \(\alpha\) and \(\angle\)DCE = 90 - \(\alpha\), CD = 60 m = BC

DE = 2 AB

In the right angled \(\triangle\)ABC,

tan\(\alpha\) = \(\frac {AB}{BC}\) = \(\frac {DE}{2× 60}\)

∴ DE = 120 tan\(\alpha\)............................(i)

In the right angled \(\triangle\)CDE,

tan (90 - \(\alpha\)) = \(\frac {DE}{Cd}\) = \(\frac {120 tan\alpha}{60}\) = 2 tan\(\alpha\)

cot\(\alpha\) = 2 tan\(\alpha\)

or, \(\frac 1{tan\alpha}\) = 2 tan\(\alpha\)

or, tan^{2}\(\alpha\) = \(\frac 12\)

∴ tan\(\alpha\) = \(\frac 1{\sqrt 2}\)

Putting the value of tan\(\alpha\) in equation (i),

DE = 120×\(\frac 1{\sqrt 2}\) = 84.85 m

AB = \(\frac {DE}2\) = \(\frac {84.85}2\) = 42.43 m

∴ Height of the posts are 42.43 m and 84.85 m. _{Ans}

Two poles are such that one is double the height of the other and are at a distance of 40 m. If the angles of elevation of the top of the poles from a point midway between them are complementary, find the height of the poles.

Let: AB and DE are the height of the poles where, DE = 2 AB.

Let: angle of elevation \(\angle\)BCA = \(\theta\), \(\angle\)DCE = 90 - \(\theta\)

Distance between the two poles BD = 40 m

BC = CD = 20 m

In the right angled \(\triangle\)ABC,

tan\(\theta\) = \(\frac {AB}{BC}\) = \(\frac {DE}{2× 20}\) = \(\frac {DE}{40}\)

or, DE = 40 tan\(\theta\)....................................(i)

In the right angled \(\triangle\)CDE,

tan (90 - \(\theta\)) = \(\frac {DE}{CD}\)

or, cot\(\theta\) = \(\frac {40 tan\theta}{20}\)

or, \(\frac 1{tan\theta}\) = tan\(\theta\)× 2

or, \(\frac 12\) = tan^{2}\(\theta\)

or, tan\(\theta\) = \(\frac 1{\sqrt 2}\)

POutting the value of tan\(\theta\) in equation (i)

DE = \(\frac 1{\sqrt 2}\)× 40 = 28.28 m

AB = \(\frac {DE}2\) = \(\frac {28.28}2\) = 14.14 m

∴ Height of two poles are 28.28 m and 14.14 m. _{Ans}

The angle of elevation of the top of a tower as observed from the distances of 36 m and 16 m from the foot of the tower are found to be complementary. Find the height of the tower.

Let:

The tower's height (AB) = ?

Distance of AD = 16 m

Distance of Ac = 36 m

\(\angle\)ACB = \(\theta\)

\(\angle\)ADB = 90 - \(\theta\)

CD = 36 - 16 = 20 m

In right angled \(\triangle\)ABD,

tan (90 - \(\theta\)) =\(\frac {AB}{AD}\)

or, cot\(\theta\) = \(\frac {AB}{16}\)

or, AB = 16 cos\(\theta\)...........................(i)

In right angled \(\triangle\)ABC,

tan\(\theta\) = \(\frac {AB}{AC}\) = \(\frac {AB}{36}\)

or, AB = 36 tan\(\theta\)............................(ii)

From equation (i) and (ii),

16 cot\(\theta\) = 36 tan\(\theta\)

or, 16 \(\frac 1{tan\theta}\) = 36 tan\(\theta\)

or, \(\frac {16}{36}\) = tan^{2}\(\theta\)

or, tan^{2}\(\theta\) = \(\frac 49\)

or, tan^{2}\(\theta\) = \(\frac 23\)^{2}

∴ tan\(\theta\) = \(\frac 23\)

Putting the value of tan\(\theta\) in equation (ii)

AB = 36 tan\(\theta\) = 36× \(\frac 23\) = 12× 2 = 24 m

∴ The height of tower = 24 m _{Ans}

A flagstaff of height 7 meters stands on the top of a tower. The angles subtended by the tower and the flagstaff to a point on the ground are 45° and 15° respectively. Find the height of the tower.

Let: Height of the tower (PQ) = ?

The height of flagstaff (QR) = 7 m

Angle of elevation:

\(\angle\)PSQ = 45° and \(\angle\)QSR = 15°

∴ \(\angle\)PSR = 45° + 15° = 60°

In right angled \(\triangle\)PRS,

tan 60° = \(\frac {PR}{PS}\)

or, 1 = \(\frac {PR}{PS}\)

∴ PQ = PS...............................(i)

In right angled \(\triangle\)PRS,

tan 60° = \(\frac {PR}{PS}\)

or, \(\sqrt 3\) = \(\frac {PQ + QR}{PQ}\) [From eq^{n} (i)]

or, PQ \(\sqrt 3\) = PQ + 7

or, PQ \(\sqrt 3\) - PQ = 7

or, PQ (\(\sqrt 3\) - 1) = 7

or, PQ = \(\frac 1{\sqrt 3 - 1}\)

or, PQ = \(\frac 1{1.732 - 1}\)

or, PQ = \(\frac 1{0.732}\)

∴ PQ = 9.56 m

∴ Height of the tower = 9.56 m _{Ans}

A flagstaff stands on the top of a tower. the angle subtended by the tower and the flagstaff at a point 100 meters away from the foot of the tower are 45° and 15° respectively. Find the length of the flagstaff.

Suppose, AD be the flagstaff and DB be the tower. C be the point of observation. \(\angle\)DCB = 45°, \(\angle\)ACD = 15° so that \(\angle\)ABC = 60°, \(\angle\)ACD = 60° and BC = 100 m, AD = ?

In the right angled\(\triangle\)DBC,

tan 45° = \(\frac {DB}{BC}\)

or, 1 = \(\frac {DB}{100}\)

∴ DB = 100 m

In right angled \(\triangle\)ABC,

tan 60° = \(\frac {AB}{BC}\)

or, \(\sqrt 3\) = \(\frac {AD + DB}{100}\)

or, \(\sqrt 3\) = \(\frac {AD + 100}{100}\)

or, 100\(\sqrt 3\) = AD + 100

or, AD = 100\(\sqrt 3\) - 100

or, AD = 173.2 - 100

∴ AD = 73.2 m

∴ The length of the flagstaff is 73.2 m. _{Ans}

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