Height and Distance

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.

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.

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.

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.

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.

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°.

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.

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.

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.

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.

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.

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.

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.

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.

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.