Three-Dimensional Trigonometry (Grade 12) — Worksheet & Memo

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Grade 12 · Trigonometry

Three-Dimensional Trigonometry (Grade 12)

MARKING GUIDELINE

Marks

Duration

35 minutes

Questions

Name:

Class:

Date:

Mark

/ 23

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
In a three-dimensional problem, the usual first step is to …
(1)
A)find a triangle lying in a single plane and solve that one first✓
B)use the area rule immediately
C)assume every triangle is right-angled
D)work only in the vertical triangle
Answer: A — A 3D figure is solved by chaining 2D triangles together, starting with the one carrying the most information.
- B — the area rule is used only when an area is asked for
- C — most of the triangles in a 3D problem are NOT right-angled
- D — the horizontal triangle usually carries the given angles
1.2
A vertical pole PQ stands at Q on level ground. From R on the ground the angle of elevation of P is 32°, and QR = 40 m. The pole is … tall.
(1)
A)24,99 m✓
B)21,20 m
C)64,01 m
D)47,17 m
Answer: A — PQ = QR tan 32° = 40 tan 32° = 24,99 m.
- B — used 40 sin 32°, but 40 m is the horizontal distance, not the slope length
- C — DIVIDED 40 by tan 32° instead of multiplying
- D — used 40 ÷ cos 32°, which gives the line of sight RP
1.3
A vertical pole PQ stands at Q on level ground, with R another point on the ground. The angle of ELEVATION of P from R is …
(1)
A)∠PRQ✓
B)∠PQR
C)∠QPR
D)the angle between PQ and the vertical
Answer: A — The angle of elevation is measured at the OBSERVER, R, between the horizontal RQ and the line of sight RP.
- B — ∠PQR is the right angle at the foot of the pole
- C — ∠QPR is at the TOP of the pole, not at the observer
- D — PQ is itself vertical, so that angle is 0°
1.4
In △ABC, AB = 12, ∠A = 50° and ∠B = 70°. Then BC = …
(1)
A)10,61✓
B)13,02
C)9,19
D)13,57
Answer: A — ∠C = 60°, and BC ÷ sin A = AB ÷ sin C gives BC = 12 sin 50° ÷ sin 60°.
- B — used sin 70° — but BC lies opposite ∠A, which is 50°
- C — used 12 sin 50° without dividing by sin 60°
- D — inverted the ratio
1.5
To find the DISTANCE between the feet of two vertical poles, you solve …
(1)
A)the horizontal triangle joining the two bases✓
B)either vertical triangle on its own
C)the area rule
D)Pythagoras in a vertical plane
Answer: A — The two bases and the observer all lie in the horizontal plane, so that is the triangle to solve.
- B — a vertical triangle gives a height, not the distance between the bases
- C — the area rule finds an area, not a distance
- D — Pythagoras in a vertical plane gives a slope length
1.6
The COSINE rule is needed in a 3D problem when …
(1)
A)two sides and the angle between them are known✓
B)two angles and a side are known
C)the triangle is right-angled
D)only the angles are known
Answer: A — That is the cosine rule's standard case, in three dimensions just as in two.
- B — that is the sine rule's case
- C — a right angle needs only Pythagoras and the basic ratios
- D — angles alone never fix a length
1.7
Two vertical poles of EQUAL height stand at A and B. From a point P on the ground the angles of elevation of both tops are equal. This means …
(1)
A)PA = PB✓
B)AB equals the poles' height
C)PA equals the poles' height
D)the poles lean towards each other
Answer: A — With equal heights, equal angles of elevation force equal horizontal distances.
- B — the distance AB is not fixed by the equal angles
- C — PA equals the height only if the angle happens to be 45°
- D — the poles are vertical by definition
1.8
From a point A on level ground, 30 m from the base of a tower, the angle of elevation of the top is 55°. The tower is … tall.
(1)
A)42,84 m✓
B)24,57 m
C)21,01 m
D)52,30 m
Answer: A — Height = 30 tan 55° = 42,84 m.
- B — used 30 sin 55°, but 30 m is the horizontal distance
- C — DIVIDED 30 by tan 55° instead of multiplying
- D — used 30 ÷ cos 55°, which gives the line of sight
1.9
The angle between a line and a horizontal PLANE is the angle between that line and …
(1)
A)its projection onto the plane✓
B)the vertical
C)any line lying in the plane
D)the normal to the plane
Answer: A — The angle is measured to the shadow the line casts on the plane.
- B — that gives the COMPLEMENT of the angle wanted
- C — different lines in the plane give different angles; only the projection is correct
- D — the normal is perpendicular to the plane, so it gives the complement
1.10
Why is the SINE rule used more often than SOH-CAH-TOA in 3D problems?
(1)
A)the triangles involved are usually not right-angled✓
B)the sine rule is quicker to write out
C)SOH-CAH-TOA does not work in three dimensions
D)the sine rule needs no angles at all
Answer: A — Only the vertical triangle at the foot of a pole is right-angled; the rest are general triangles.
- B — convenience of writing is not a mathematical reason
- C — SOH-CAH-TOA works in any right-angled triangle, in 3D just as in 2D
- D — the sine rule needs at least one complete side-and-angle pair

Answer grid — marking guideline

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[13 MARKS]

In the diagram, PQ is a vertical tower. Q, R and S are points in the same horizontal plane. The angle of elevation from R to P is θ, Q̂R̂S = 120° and PQ = RS = x.

2.1
Determine QR in terms of θ and x.
(3)
In △PQR, tan θ = PQQR (2)
QR = xtan θ (1)
2.2
Explain which rule must be used to find QS, and why.
(2)
The cosine rule (1) — two sides and the INCLUDED angle (120°) are known (reason 1)
2.3
If x = 15 cm and θ = 22°, calculate the length of QS, correct to TWO decimal places.
(5)
QR = 15tan 22° = 37,13 cm (2)
QS² = QR² + RS² − 2(QR)(RS)cos 120° (2)
QS = 46,48 cm (1)
2.4
Hence calculate the size of Q̂P̂S, correct to TWO decimal places.
(3)
PQ ⊥ the horizontal plane, so △PQS is right-angled at Q (1)
tan Q̂P̂S = 46,4815 (1) = 72,11° (1)

TOTAL: 23 marks

This question paper consists of 2 questions.