Heights and Distances

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 method of solution of triangles is used to find the required height or distance.

Angle of elevation

Angle of elevation
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.

The angle of depression

angle of depression
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.

sdvfg

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

hgf

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

ge

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

f

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

ds

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

d

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 eqn (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

zc

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

cd

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

df

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

fv

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

xc

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

x

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

 zcx

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



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

d

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

d

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

c

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

sa

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

f

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

xc

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 - AB2

or, AB2 = 40000 - 30000

or, AB2 = 10000

∴ AB = 100 m

∴ The height of the pillar = 100 m Ans

czx

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 eqn (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 - x2

or, 80 = 400 - x2

or, x2 = 400 - 80

or, x2 = 320

∴ x = 17.89 m

AC = 10x = 10× 17.89 = 178.9 m

∴ The height of the pole = 17.89 m Ans

d

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.

xc

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

sc

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

sc

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, tan2\(\alpha\) = \(\frac 12\)

∴ 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

f

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, \(\frac 12\) = tan2\(\theta\)

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

ghb

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, \(\frac {16}{36}\) = tan2\(\theta\)

or, tan2\(\theta\) = \(\frac 49\)

or, tan2\(\theta\) = \(\frac 23\)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

xc

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 eqn (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

c

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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krishna

it is the best and simple way to teach height and distance everyone have to try it