# Height and Distance

From the roof and foot of a house, the angles of depression and elevation of the top of a tree are 60° and 30° respectively. If the height of the tree is 15 ft. find the height of the house.

From figure,

AE = x, EB = CD = 15 ft, ED = BC = y, ∠FAD = ∠ADE =
60° (alternate angle), ∠DBC = 30°

From right angle triangle AED,

tan(∠ADE) = $\frac {AE}{ED}$

or, tan60° = $\frac {x}{y}$

or, y = $\frac{x}{\sqrt 3}$ ---- (A)

Also, From right angle triangle BCD,
tan(∠DBC) = $\frac {CD}{BC}$
or, tan(∠30) = $\frac {15}{y}$

or, $\frac {1}{\sqrt 3}$ = $\frac {15\sqrt 3}{x}$ (from A)

so, x = 45 ft

And, AB = x + 15 = 60 ft

Hence, the height of the house is 60 ft.

From the roof of a building 16 meter high, the angles of elevation and depression of the top and the foot of an electric pole are observed to be 60° And 30° respectively. Find the height of the pole.

From figure,

AE = x, EB = CD = 15 ft, ED = BC = y, ∠DAE = ∠ADB =
30° (alternate angle), ∠CAE = 60°

From right angle triangle ABD,
tan(∠ADB) = $\frac {AB}{BD}$
or, tan30° = $\frac {16}{BD}$

or, BD = 16$\sqrt{3}$ ---- (A)

Also, From right angle triangle ACE,
tan(∠CAE) = $\frac {CE}{AE}$
or, tan(∠60) = $\frac {CE}{16\sqrt{3}}$ (BD = AE)

or, CE = 48 m

And, CD = CE + ED = 48 + 16 = 64 m (ED = AB = 16m)

Hence, the height of the pole is 64 m.

The shadow of a tower standing on a plain is found to be 60 ft longer when the sun's altitude is 30° than when it is 45°. What will be the height of the tower? Find it.

Let AB = x, BC = y, then according to question,

BD = 60 + y

From right angle triangle ABC,
tan(∠ACB) = $\frac {AB}{BC}$
or, tan45° = $\frac xy$

so, x = y ---- (A)

Also, From right angle triangle ABD,
tan(∠ADB) = $\frac {AB}{BD}$
or, tan30° = $\frac {AB}{BD}$

or, $\frac {1}{\sqrt 3}$ = $\frac {x}{\text{60 + y}}$

or, 60 + x = x√3 (from A)

or, 60 = x(√3 - 1)

so, x = 81.96 feet

Hence, the height of the tower is 81.96 feet.

The angles of elevations of the roof of a house and window are 45° and 30° respectively. If the window is 18 ft below the roof if the house, find the height of the house.

From figure,

BD = x, BC = y, AD = 18 ft, ∠BCD = 30°, ∠ACB =
45°

From right angle triangle BCD,
tan(∠BCD) = $\frac {BD}{BC}$
or, tan30° = $\frac xy$

so, y = x√3 ---- (A)

Also, From right angle triangle ABC,
tan(∠ACB) = $\frac {AB}{BC}$
or, tan45° = $\frac {\text{x + 18}}{y}$

or, x√3 - x = 18 (from A)

so, x = 24.588 feet

Finally,

AB = 24.588 + 18

so, AB = 42.588 feet

Hence, the height of the house is 42.588 feet.

The angles of elevation of the top of a tower observed from 27 m and 75 m away from its foot on the same side are found to be complementary. Find the height of the tower.

From figure,

AB = x, ∠ACB = 90 - θ, ∠ADB = θ

From right angle triangle ABC,
tan(∠ACB) = $\frac {AB}{BC}$
or, tan(90 - θ) = $\frac {x}{75}$

or, cotθ = $\frac {x}{75}$

or, tanθ = $\frac {75}{x}$ ---- (A)

Also, From right angle triangle ABD,
tan(∠ADB) = $\frac {AB}{BD}$
or, tanθ = $\frac {x}{27}$

or, $\frac {75}{x}$ = $\frac {x}{27}$

or, x^{2} = 2025

or, x = ±45

so, x = 45 m (height cannot be negative)

A vertical pole AB is divided by a point C such that AC : CB = 2 : 1. The angle of elevation of C from a point on the ground is 30°. What is the angle of elevation of A from the same pont? Find it where B is the foot of the pole.

From figure,

AC = 2x, BC = x, ∠BDC = 30°, ∠BDA = θ

From right angle triangle BDC,
tan(∠BDC) = $\frac {BC}{BD}$
or, tan30° = $\frac {x}{BD}$

or, $\frac {1}{\sqrt 3}$ = $\frac {x}{BD}$

so, BD = x√3 ---- (A)

Also, From right angle triangle ABD,
tan(∠BDA) = $\frac {AB}{BD}$
or, tanθ = $\frac {3x}{x\sqrt 3}$

or, tanθ = $\frac {\sqrt 3 × \sqrt 3}{\sqrt 3}$

or, tanθ = √3

or, tanθ = tan60°

so, θ = 60°

Hence, the angle of elevation of A from the point 'D' is 60°.

A tower and flagstaff on its top subtend angles of 30° and 15° respectively at a point 100 m away from the foot of the tower, find the height of flagstaff.

From figure,

AB = x, BC = y, CD = 100 m, ∠BDC = 30°, ∠ADC = 30 +
15 = 45°

From right angle triangle BCD,
tan(∠BDC) = $\frac {BC}{CD}$
or, tan30° = $\frac {y}{100}$

or, y = $\frac {100}{\sqrt 3}$

so, y = 57.735 m ---- (A)

Also, From right angle triangle ADC,
tan(∠ADC) = $\frac {AC}{CD}$
or, tan45° = $\frac {\text{x + y}}{100}$

or, 100 = x + 57.735 (from A)

so, x = 42.624 m

From the top of mountain 21 m high, 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.

From figure,

AB = 21 m, ∠EAD = ∠ADF = 45° (alternate angle),
∠EAC = ∠ACB = 60°, FD = BC, CD = FB = x

From right angle triangle ABC,
tan(∠ACB) = $\frac {AB}{BC}$
or, tan60° = $\frac {21}{BC}$

so, BC = $\frac {21}{\sqrt 3}$ ---- (A)

Also, From right angle triangle AFD,
tan(∠ADF) = $\frac {AF}{FD}$
or, tan45° = $\frac {AF}{BC}$

or, 1 = $\frac {\text {21 - FB}}{BC}$

or, 1 = (21 - x) × $\frac {\sqrt 3}{21}$ (from A)

or, 21 = 21√3 - x√3

or, x√3 = 15.373

so, x = 8.875 m

From the roof and basement of a house 20 m high, the angles of elevation of the top of a tower are 45° amd 60° respectively. Find the height of the tower.

From figure,

AB = EC = 20 m, AE = BC, DE = x, ∠DAE = 45°, ∠DBC =
60°

From right angle triangle BCD,
tan(∠DBC) = $\frac {CD}{BC}$
or, tan60° = $\frac {\text {20 + x}}{BC}$

so, BC = $\frac {\text{20 + x}}{\sqrt 3}$ ---- (A)

Also, From right angle triangle AED,
tan(∠DAE) = $\frac {DE}{AE}$
or, tan45° = $\frac {x}{BC}$

or, x = $\frac {\text{20 + x}}{\sqrt 3}$ (from A)

or, x√3 = 20 + x

so, x = 27.32 m

Finally,

DC = x + 20 = 47.32 m

Hence, the height of the tower is 47.32 m.

The angle of elevation of the top of a tower from a point was observed to be 45° and on walking 60 m away from that point it was found to be 30°. Find the height of the tower.

From figure,

AB = x, BC = y, CD = 60 m, ∠ACB = 45°, ∠ADB =
30°

From right angle triangle ABC,
tan(∠ACB) = $\frac {AB}{BC}$
or, tan45° = $\frac {x}{y}$

so, x = y ---- (A)

Also, From right angle triangle ABD,
tan(∠ADB) = $\frac {AB}{BD}$
or, tan30° = $\frac {x}{\text{y + 60}}$

or, $\frac {1}{\sqrt 3}$ = $\frac {x}{\text{x + 60}}$ (from A)

or, x + 60 = x√3 so, x = 81.96 m

Hence, the height of the tower is 81.96 m.

The angles of elevation of the top of a tower observed from two points on the same plane are found to be complementary. If the points are at a distance of 20 m and 45 m from the foot of the tower, find the height of the tower.

From figure,

AB = x, ∠ACB = 90 - θ, ∠ADB = θ

From right angle triangle ABC,
tan(∠ACB) = $\frac {AB}{BC}$
or, tan(90 - θ) = $\frac {x}{45}$

or, cotθ = $\frac {x}{45}$

or, tanθ = $\frac {45}{x}$ ---- (A)

Also, From right angle triangle ABD,
tan(∠ADB) = $\frac {AB}{BD}$
or, tanθ = $\frac {x}{20}$

or, $\frac {45}{x}$ = $\frac {x}{20}$ (from A)

or, x^{2} = 900

or, x = ±30

so, x = 30 m (height cannot be negative)

Hence, the height of the tower is 30 m.

The angles of elevation of the top of a tower as observed from the points at the distances of 144 meter and 121 metre from the foot of the tower are found to be complementary. Find the height of the tower.

From figure,

AB = x, ∠ACB = 90 - θ, ∠ADB = θ

From right angle triangle ABC,
tan(∠ACB) = $\frac {AB}{BC}$
or, tan(90 - θ) = $\frac {x}{144}$

or, cotθ = $\frac {x}{144}$

or, tanθ = $\frac {144}{x}$ ---- (A)

Also, From right angle triangle ABD,
tan(∠ADB) = $\frac {AB}{BD}$
or, tanθ = $\frac {x}{121}$

or, $\frac {144}{x}$ = $\frac {x}{121}$ (from A)

or, x^{2} = 17424

or, x = ±132

so, x = 132 m (height cannot be negative)

Hence, the height of the tower is 132 m.

The angles of depression and elevation of the pinnacle of a temple 20 meter high is found to be 60° and 30° from the top and foot of a tower respectively. Find the height of the tower.

From figure,

AB = x, ∠EAD = ∠ADB = 60° (alternate angle),
∠CBD = 30°

From right angle triangle CBD,
tan(∠CBD) = $\frac {CD}{BD}$
or, tan30° = $\frac {20}{BD}$

or, BD = $\frac {20}{tan30\degree}$

or, BD = 20√3 ---- (A)

Also, From right angle triangle ABD,
tan(∠ADB) = $\frac {AB}{BD}$
or, tan60° = $\frac {x}{20\sqrt 3}$ (from A)

or, √3 = $\frac {x}{20\sqrt 3}$

so, x = 60 m

Hence, the height of the tower is 60 m.

The angle of elevation of the top of a tower from the roof of a building 10 m high is found to be 60°. From the same place, the angle of elevation at the point 18 m below the top of the tower is observed to be 30°. Find the height of the tower and the distance between the tower and the building.

From figure,

CD = x = 10 m, FC = 18 + x, AD = BC = y, ∠FAD = 60°,
∠EAD = 30°

From right angle triangle FAD,
tan(∠FAD) = $\frac {FD}{AD}$
or, tan60° = $\frac {\text{18 + ED}}{AD}$

or, √3 = $\frac {\text{18 + ED}}{AD}$

or, y = $\frac {\text{18 + ED}}{\sqrt 3}$ ---- (A)

Also, From right angle triangle EAD,
tan(∠EAD) = $\frac {ED}{AD}$
or, tan30° = $\frac {ED}{y}$

or, y = ED × √3

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or, $\frac {\text{18 + ED}}{\sqrt 3}$ = ED × √3 (from A)

or, 18 + ED = 3 × ED

so, ED = 9 m ---- (B)

Now, From (A),

y = $\frac {\text{18 + 9}}{\sqrt 3}$ (from B)

so, y = $\frac{27}{\sqrt 3}$ = $\frac {9 \times 3}{\sqrt 3}$ =
$\frac {9 \times \sqrt 3 \times \sqrt 3}{\sqrt 3}$ = 9√3 m

Hence, the height of the tower = FE + ED + DC = 18 + 9 +10 = 37 m

Distance between the tower and building = BC = y = 9√3 m.

The angle of depression of the top of a 5 m high pole observed from the top of a house is 60° and the angle of elevation of the top of the pole from the foot of the house is 30° then find the height of the house.

From figure,

AE = x, EB = CD = 5 m, ED = BC = y, ∠FAD = ∠ADE =
60° (alternate angle), ∠DBC = 30°

From right angle triangle AED,
tan(∠ADE) = $\frac {AE}{ED}$
or, tan60° = $\frac {x}{y}$

or, y = $\frac{x}{\sqrt 3}$ ---- (A)

Also, From right angle triangle BCD,
tan(∠DBC) = $\frac {CD}{BC}$
or, tan(∠30) = $\frac {5}{y}$

or, $\frac {1}{\sqrt 3}$ = $\frac {5\sqrt 3}{x}$ (from A)

so, x = 15 m

And, AB = x + 5 = 20 m