Kuulchat - Senior High School (S.H.S) Mathematics Angle of Elevation & Depression Practice Test Questions

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ANGLE OF ELEVATION & DEPRESSION

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1.

The angle of elevation of the top of a tower from the top of a building, 5 m high is 30°. If on the horizontal ground, the building is 40 m away from the foot of the tower:

(i)

illustrate the information in a diagram.

(ii)

calculate, correct to three significant figures, the height of the tower.

Show Solution

(i)

Let h = the height of the tower.

(ii)

SOHCAHTOA

SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (sin), cosine (cos) and tangent (tan).

SOH

sin(θ) = $\frac{Opposite}{Hypothenuse}$

CAH

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

TOA

tan(θ) = $\frac{Opposite}{Adjacent}$

The longest side (side opposite/facing the right-angle) is the hypotenuse

tan 30° = $\frac{y}{40}$

y = 40 x tan 30° = 23.094 m ≈ 23.1 m

h = 5 m + y = 5 m + 23.1 m = 28.1 m

2.

From the top, X of a building 320 m high, the angles of depression of the top, Y and bottom, Z of another building on the same horizontal ground are 29° and 41° respectively.

(i)

Illustrate the information in a diagram.

(ii)

Calculate, correct to the nearest metre, the height of the other building.

Show Solution

(i)

(ii)

h = the height of the other building

x + h = 320 m

SOHCAHTOA

SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (sin), cosine (cos) and tangent (tan).

SOH

sin(θ) = $\frac{Opposite}{Hypothenuse}$

CAH

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

TOA

tan(θ) = $\frac{Opposite}{Adjacent}$

The longest side (side opposite/facing the right-angle) is the hypotenuse

tan(θ) = $\frac{Opposite}{Adjacent}$

tan 41° = $\frac{x + h}{y}$ = $\frac{320}{y}$

tan 41° = $\frac{320}{y}$

Multiply both sides by y

y x tan 41° = 320

Divide both sides by tan 41°

y = $\frac{320}{tan 41°}$ = 368.11789031 m

tan 29° = $\frac{x}{y}$

Multiply both sides by y

x = y x tan 29°

But y = 368.11789031 m

x = 368.11789031 m x tan 29°

x = 204.0510786 m ≈ 204.0511 m

x + h = 320 m

204.0511 m + h = 320 m

h = 320 m - 204.0511 m

h = 115.94 m ≈ 116 m (nearest metre)

3.

Two points X and Y, 7 metres apart are on the same horizontal ground. The angles of elevation of a point P from X and Y are 50° and 70° respectively.

Q is a point on XY produced such that ∠ YQP = 90°.

(i)

Illustrate the information in a diagram.

(ii)

Calculate, correct to two decimal places, the length:

(α)

XP;

(β)

YQ.

Show Solution

(i)

(ii)

∠ XYP + ∠ PYQ = 180° (Sum of angles on a straight line)

∠ XYP = 180° - ∠ PYQ

∠ XYP = 180° - 70°

∠ XYP = 110°

∠ PXY + ∠ XYP + ∠ XPY = 180° (Sum of angles in a triangle)

50° + 110° + ∠ XPY = 180°

160° + ∠ XPY = 180°

∠ XPY = 180° - 160°

∠ XPY = 20°

(α)

Apply the sine rule

$\frac{7}{sin 20°}$ = $\frac{XP}{sin 110°}$

Cross multiply.

XP x sin 20° = 7 x sin 110°

Divide both sides by sin 20°

XP = $\frac{7 x sin 110°}{sin 20°}$ = 19.2323 m ≈ 19.23 m (two decimal places)

(β)

XY + YQ = XQ

YQ = XQ - XY

XY = 7 m

YQ = XQ - 7 m

∠ PXY + ∠ PQY + ∠ XPQ = 180°(Sum of angles in a triangle)

50° + 90° + ∠ XPQ = 180°

140° + ∠ XPQ = 180°

∠ XPQ = 180° - 140°

∠ XPQ = 40°

Apply the sine rule

$\frac{XQ}{sin 40°}$ = $\frac{XP}{sin 90°}$

But XP = 19.23

$\frac{XQ}{sin 40°}$ = $\frac{19.23}{sin 90°}$

Cross multiply.

XP x sin 90° = 19.23 x sin 40°

Divide both sides by sin 90°

XP = $\frac{19.23 x sin 40}{sin 90°}$ = 12.36 m

YQ = 12.36 m - 7m

YQ = 5.36 m

4.

The angle of depression of a point P on the ground from the top, T, of a building is 23.6°.

If the horizontal distance from P to the base of the building is 50 m, calculate, correct to three significant figures, the height of the building.

Show Solution

Let h = height of the building.

tan(θ) = $\frac{Opposite}{Adjacent}$

tan 23.6° = $\frac{h}{50}$

h = 50 tan 23.6° = 21.8444 m ≈ 21.8 m(three significant figures)

5.

The angle of elevation of the top, X, of a vertical pole from a point, W, on the same horizontal ground as the foot, Z, of the pole is 60°. If W is 15 km from X and 12 km from a point Y on the pole,

(a)

illustrate this information with a diagram.

(b)

calculate, correct to two decimal places, the:

(i)

angle of elevation of Y from W;

(ii)

length, XY.

Show Solution

(a)

(b)

(i)

SOHCAHTOA

SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (sin), cosine (cos) and tangent (tan).

SOH

sin(θ) = $\frac{Opposite}{Hypothenuse}$

CAH

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

TOA

tan(θ) = $\frac{Opposite}{Adjacent}$

The longest side (side opposite/facing the right-angle) is the hypotenuse

cos(θ) = $\frac{Adjacent}{Hypothenuse}$

cos60° = $\frac{|WZ|}{15}$

Get rid of the fraction by multiplying both sides by the denominator, 15

15 x cos60° = $\frac{|WZ|}{15}$ x 15

|WZ| = 15 x cos60°

|WZ| = 7.5 km

cosx = $\frac{|WZ|}{12}$

cosx = $\frac{7.5}{12}$

x = cos^-1( $\frac{7.5}{12}$ ) = 51.3178° ≈ 51.32° (two decimal places)

(ii)

|XY| = |XZ| - |YZ|

sin(θ) = $\frac{Opposite}{Hypothenuse}$

sin60° = $\frac{XZ}{15}$

Get rid of the fraction by multiplying both sides by the denominator, 15

15 x sin60° = $\frac{XZ}{15}$ x 15

|XZ| = 15 x sin60° = 12.99 km

sinx = $\frac{YZ}{WY}$

x = 51.32°

|WY| = 12 km

sin51.32° = $\frac{YZ}{12}$

|YZ| = sin51.32° x 12 km = 9.37 km

|XY| = 12.99 km - 9.37 km = 3.62 km (two decimal places)

6.

From two points P and Q, 15 m apart and on the same horizontal line as the foot of a tower, the angles of elevation of the top of the tower are 35° and 45° respectively.

If P and Q are on the side of the tower,

(a)

represent the information in a diagram.

(b)

find, correct to the nearest metre, the height of the tower.

Show Solution

(a)

(b)

tan 35° = $\frac{h}{15 + y}$

h = (15 + y)tan 35° ---- eqn 1

tan 45° = $\frac{h}{y}$

h = y x tan 45°

h = ytan 45° --- eqn 2

eqn 1 = eqn 2

(15 + y)tan 35° = ytan 45°

(15 + y)0.7002 = y

Note: tan 45 = 1

10.503 + 0.7002y = y

10.503 = y - 0.7002y

10.503 = 0.2998y

Divide both sides by 0.2998

y = $\frac{10.503}{0.2998}$ = 35.033

From eqn 2,

h = ytan 45°

h = 35.033 x tan 45°

h = 35.033 m

The height of the tower is 35 m (nearest metre)

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