# 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.

Let AB and CD be the height of the house and tree respectively. ∠FAD and ∠DBC be the angles of depression and elevation made from the roof and foot of the house to the top of a tree respectively. 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.

Let AB and CD be the height of the building and electric pole respectively. ∠DAE and ∠CAE be the angles of depression and elevation made from the roof of the building to the top and the foot of an electric pole respectively. 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 be the height of a tower and BC and BD be the length if its shadow when sun's altitude is 45° and 30° respectively. 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.

Let AB be the height of a house and BD be the height from window to the floor of the house. ∠ACD and ∠BCD be the angle of elevation made while observing the roof and window of a house from point C respectively. 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.

Let AB be the height of the tower. ∠ADB and ∠ACB be the angle of elevation observed from 27 m and 75 m away from its foot respectively. 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, x2 = 2025
or, x = ±45
so, x = 45 m (height cannot be negative)

Hence, the height of the tower is 45 m.

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.

Let AB be the height of a pole in which AC and CB are divided in the ratio of 2:1. Angle of elevation of 30° is made while observing point 'C' from a ground 'D'. 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.

Let AB and BC be the height of the flagstaff and tower respectively. ∠BDC and ∠ADB be the angles subtended by tower and flagstaff respectively at point D. 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

Hence, the height of the flagstaff is 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.

Let AB and CD be the height of mountain and tower respectively. ∠EAD and ∠EAC be the angle of depression made while observing from top of mountain to the top and bottom of the tower respectively. 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

Hence, the height of the tower is 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.

Let AB and CD be the height of a house and tower respectively. ∠DAE and ∠DBC be the angle of elevation of the top of tower from top and basement of a house respectively. 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.

Let AB be the height of the tower. ∠ACB and ∠ADB be the angle of elevation from point C and point D respectively to 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.

Let AB be the height of the tower. ∠ADB and ∠ACB be the angle of elevation observed from 20 m and 45 m away from its foot respectively. 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, x2 = 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.

Let AB be the height of the tower. ∠ADB and ∠ACB be the angle of elevation observed from 121 m and 144 m away from its foot respectively. 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, x2 = 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.

Let AB and CD be the height of the tower and temple respectively. ∠EAD and ∠CBD be the angles of depression and elevation of the pinnacle of a temple from the top and foot of a tower respectively. 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.

Let AB and FC be the height of the building and tower respectively. ∠FAD and ∠EAD be the angles of elevation from the top of a building to the top and 18 m below from the top of the tower respectively. 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.

Let AB and CD be the height of the house and pole respectively. ∠FAD and ∠DBC be the angles of depression and elevation made from the top and bottom of the house to the tower respectively. 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

Hence, the height of the house is 20 m.