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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A building is 12 m high. A football on the ground floor s 30 m away from the foot of the building.

Find, correct to the nearest degree, the angle of depression of the ball from the top of the building.

22°

68°

66°

24°

Angle of depression

The angle of depression is defined as an angle constructed by a horizontal line and the line joining the object and observer's eye.

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

Using the known sides, tan θ = $\frac{12 m}{30 m}$ where 12 m is the opposite side of θ and 30 m is the adjacent side of θ

tan θ = $\frac{12 m}{30 m}$

tan θ = $\frac{12}{30}$

Take tan^-1 of both sides.

θ = tan^-1( $\frac{12}{30}$ )

θ = 21.80° ≈ 22°(nearest degree)

A point on the ground is 5 m away from the foot of a vertical wall 12 m high.

Calculate, correct to the nearest degree, the angle of depression of the point from the top of the wall.

25^o

23^o

67^o

65^o

The angle of depression is the angle between the horizontal line and the observation of the object from the horizontal line.

Let θ = angle of depression.

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

The opposite and the adjacent sides of the angle of depression are known.

tan(θ) = $\frac{Opposite}{Adjacent}$

tan(θ) = $\frac{12 cm}{5 cm}$

θ = tan^-1 $(\frac{12}{5})$

θ = 67.380135052^o ≈ 67^o (nearest degree)

The angle of elevation of the top of a vertical pole from a point, P on a level ground is 60°. The distance from P to the foot of the pole is 55 m. Find the height of the pole.

$\frac{55}{2}$ m

55 $\sqrt{3}$ m

$\frac{55 \sqrt{3}}{3}$ m

$\frac{55 \sqrt{3}}{2}$ m

Let the height of the pole = h

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

Since we know both the opposite and adjacent to the angle 60°, tan is applied.

tan(θ) = $\frac{Opposite}{Adjacent}$

tan 60° = $\frac{h}{55 m}$

But tan 60° = $\sqrt{3}$

$\sqrt{3}$ = $\frac{h}{55 m}$

Multiply both sides by 55 m

h = $\sqrt{3}$ x 55 m

h = 55 $\sqrt{3}$ m

The shadow of a vertical pole is 50 cm. If the height of the pole is 2 m, find, correct to the nearest degree, the angle of elevation of the sun from the ground level.

45°

56°

60°

76°

Tan(θ) = $\frac{Opposite}{Adjacent}$

Tan(θ) = $\frac{2}{0.5}$

θ = Tan^-1( $\frac{2}{0.5}$ )

θ = 76° (nearest degree)

A fence 2.4 m tall, is 10 m away from a tree of height 16 m. Calculate the angle of elevation of the top of the tree from the top of the fence.

51.32°

52.40°

53.67°

76.11°

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) = $\frac{Opposite}{Adjacent}$

Opposite side to θ = 13.6 m

Adjacent side to θ = 10 m

tan(θ) = $\frac{13.6 m}{10 m}$

Take tan inverse.

θ = tan^-1( $\frac{13.6 m}{10 m}$ )

θ = 53.6731740479° ≈ 53.67°

From the top of a vertical cliff 20 m high, a boat at sea can be sighted 75 m away and on the same horizontal position as the foot of the cliff.

Calculate, correct to the nearest degree, the angle of depression of the boat from the top of the cliff.

15°

16°

75°

56°

Angle of depression

The angle of depression is defined as an angle constructed by a horizontal line and the line joining the object and observer's eye.

Let θ = the angle of depression.

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 = $\frac{Opposite}{Adjacent}$

The opposite side of θ = 20 m

The adjacent side of θ = 75 m

tan θ = $\frac{20 m}{75 m}$

tan θ = $\frac{20}{75}$

Take tan^-1 of both sides.

θ = tan^-1( $\frac{20}{75}$ ) = 14.93° ≈ 15° (nearest degree)

From a height of 2 m above the ground and at a horizontal distance of 12 $\sqrt{3}$ m from a tree, the angle of elevation of the top of the tree is 30°.

How tall is the tree?

8 m

8 $\sqrt{3}$ m

14 m

18 m

Let h = height of the tree

h = y + 2 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 30° = $\frac{y}{12 \sqrt{3}}$

Get rid of the fraction by multiplying both sides by the denominator 12 $\sqrt{3}$

y = 12 $\sqrt{3}$ x tan 30° = 12 m

h = 12 m + 2 m = 14 m

A tree 8 m tall casts a shadow 10 m long.

How tall is a tree on the same ground level that casts a shadow 40 m long at the same time of the day?

32 m

50 m

80 m

96 m

Let h = height of tree.

At the same time of the day, the angle of elevation from any point on the ground to the sun is considered equal.

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) = $\frac{Opposite}{Adjacent}$

8 m tall tree

tan(θ) = $\frac{8 m}{10 m}$ = $\frac{4}{5}$

h m tall tree

tan(θ) = $\frac{h m}{40 m}$ = $\frac{h}{40}$

Since the angle, θ is the same for both trees, the tan(θ) is the same for both.

$\frac{h}{40}$ = $\frac{4}{5}$

Cross multiply.

h x 5 = 40 x 4

Divide both sides by 5.

h = $\frac{40 x 4}{5}$ = 32

A point on the ground is 5 m away from the foot of a vertical wall 12 m high.

Calculate, correct to the nearest degree, the angle of depression of the point from the top of the wall.

25^o

23^o

67^o

65^o

The angle of depression is the angle between the horizontal line and the observation of the object from the horizontal line.

Let θ = angle of depression.

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

The opposite and the adjacent sides of the angle of depression are known.

tan(θ) = $\frac{Opposite}{Adjacent}$

tan(θ) = $\frac{12 cm}{5 cm}$

θ = tan^-1 $(\frac{12}{5})$

θ = 67.380135052^o ≈ 67^o (nearest degree)

10.

The angle of elevation of the top of a vertical pole from a point, P on a level ground is 60°. The distance from P to the foot of the pole is 55 m. Find the height of the pole.

$\frac{55}{2}$ m

55 $\sqrt{3}$ m

$\frac{55 \sqrt{3}}{3}$ m

$\frac{55 \sqrt{3}}{2}$ m

Let the height of the pole = h

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

Since we know both the opposite and adjacent to the angle 60°, tan is applied.

tan(θ) = $\frac{Opposite}{Adjacent}$

tan 60° = $\frac{h}{55 m}$

But tan 60° = $\sqrt{3}$

$\sqrt{3}$ = $\frac{h}{55 m}$

Multiply both sides by 55 m

h = $\sqrt{3}$ x 55 m

h = 55 $\sqrt{3}$ m

11.

The shadow of a vertical pole is 50 cm. If the height of the pole is 2 m, find, correct to the nearest degree, the angle of elevation of the sun from the ground level.

45°

56°

60°

76°

Tan(θ) = $\frac{Opposite}{Adjacent}$

Tan(θ) = $\frac{2}{0.5}$

θ = Tan^-1( $\frac{2}{0.5}$ )

θ = 76° (nearest degree)

12.

A fence 2.4 m tall, is 10 m away from a tree of height 16 m. Calculate the angle of elevation of the top of the tree from the top of the fence.

51.32°

52.40°

53.67°

76.11°

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) = $\frac{Opposite}{Adjacent}$

Opposite side to θ = 13.6 m

Adjacent side to θ = 10 m

tan(θ) = $\frac{13.6 m}{10 m}$

Take tan inverse.

θ = tan^-1( $\frac{13.6 m}{10 m}$ )

θ = 53.6731740479° ≈ 53.67°

13.

From the top of a vertical cliff 20 m high, a boat at sea can be sighted 75 m away and on the same horizontal position as the foot of the cliff.

Calculate, correct to the nearest degree, the angle of depression of the boat from the top of the cliff.

15°

16°

75°

56°

Angle of depression

The angle of depression is defined as an angle constructed by a horizontal line and the line joining the object and observer's eye.

Let θ = the angle of depression.

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 = $\frac{Opposite}{Adjacent}$

The opposite side of θ = 20 m

The adjacent side of θ = 75 m

tan θ = $\frac{20 m}{75 m}$

tan θ = $\frac{20}{75}$

Take tan^-1 of both sides.

θ = tan^-1( $\frac{20}{75}$ ) = 14.93° ≈ 15° (nearest degree)

14.

From a height of 2 m above the ground and at a horizontal distance of 12 $\sqrt{3}$ m from a tree, the angle of elevation of the top of the tree is 30°.

How tall is the tree?

8 m

8 $\sqrt{3}$ m

14 m

18 m

Let h = height of the tree

h = y + 2 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 30° = $\frac{y}{12 \sqrt{3}}$

Get rid of the fraction by multiplying both sides by the denominator 12 $\sqrt{3}$

y = 12 $\sqrt{3}$ x tan 30° = 12 m

h = 12 m + 2 m = 14 m

SponsoredAdvertise

Kuulchat Media

Get a professional & affordable website

kodevibe.com

Master coding: The Ultimate J.H.S & S.H.S Guide

kuulpay.com

Buy B.E.C.E/W.A.S.S.C.E result checker @ kuulpay.com

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