Trigonometry; Sine Cosine and Area Rules Questions and Answers Mathematics Grade 12. The solution for an oblique triangle can be done with the application of the Law of Sine and Law of Cosine, simply called the Sine and Cosine Rules. An oblique triangle, as we all know, is a triangle with no right angle. It is a triangle whose angles are all acute or a triangle with one obtuse angle.
The two general forms of an oblique triangle are as shown:
Sine Rule (The Law of Sine)
The Sine Rule is used in the following cases:
CASE 1: Given two angles and one side (AAS or ASA)
CASE 2: Given two sides and a non-included angle (SSA)
The Sine Rule states that the sides of a triangle are proportional to the sines of the opposite angles. In symbols,
Case 2: SSA or The Ambiguous Case
In this case, there may be two triangles, one triangle, or no triangle with the given properties. For this reason, it is sometimes called the ambiguous case. Thus, we need to examine the possibility of no solution, one or two solutions.
Cosine Rule (The Law of Cosine)
The Cosine Rule is used in the following cases:
1. Given two sides and an included angle (SAS)
2. Given three sides (SSS)
The Cosine Rule states that the square of the length of any side of a triangle equals the sum of the squares of the length of the other sides minus twice their product multiplied by the cosine of their included angle. In symbols:
TRIGONOMETRY:SINE, COSINE AND AREA RULES QUESTIONS AND ANSWERS GRADE 12
Activity 1
In triangle ABC, ^B = 37° and AC = 16 cm. ^C = 90°. Calculate the length of AB and BC (correct to one decimal place). (3)
| Solution To calculate the length of AB, use 37° as the reference angle, then AC = 16 cm is the opposite side and AB is the hypotenuse. Use the sine ratio. sin 37° = opp =16 hyp AB AB sin 37° = 16 AB = 16 = 26,6 cm sin37° To find the length of BC, you can use cos 37° = adj = BC/26.6 hyp 26,6 cos 37° = BC BC = 21,2 cm (to one decimal place) You can also use Pythagoras’ theorem: AB2 = AC2 + BC2 [3] [3] |
Activity 2
In triangle PQR, PQ = 12,3 m and PR = 13 m. Calculate the size of ^Q . (2)
[2]
| Solution Use PQ and PR. tan θ = opp = 13 12.3 θ = tan–1 (13/12.3) = 46,58° [2] |
Activity 3
Solve ∆XYZ in which z = 7,3 m, ^X = 43° and ^Y = 96°. Give your solutions correct to 3 decimal places. (4)
[4]
| Solution The angle opposite the known side is not given, but you can work it out. ^Z = 180° – (43° + 96°) (sum angles of ∆) ^Z = 41° To find y: y = 7,3 sin 96° sin41° y = 7,3 sin 96° sin41° y = 11,066 m Using the sine rule again to find x: x = 7,3 sin43° sin41° x = 7,3 sin 43° sin41° x = 7,589 m [4] |
Activity 4
1. PQRS is a trapezium with PQ // SR, PQ = PS, SR = 10 cm,
QR = 7 cm, ^R = 63°.
Calculate:
- SQ (2)
- PS (6)
- area of quadrilateral PQRS. (correct to 2 decimal places) (5)
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Solutions
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When solving triangles, start with the triangle which has most information (i.e. triangle with three sides or two sides and an angle or two angles and a side given)
Activity 5
In the diagram alongside, AC = 7 cm,
DC = 3 cm, AB = AD, DCA = 60°,
DAB = β and ABD = θ.
Show that BD = √37 sin β
sinθ
[3]
| Solution AD2 = AC2 + CD2 − 2AC.CD cos 60 ° = (7)2 + (3)2 – 2 × 7 × 3 × 0,5 AD2 =58 – 21 AD2 = 37 AD = √37 P Applying sine rule: BD = AD ⇒ BD = AD sin β but AD = √37 sinβ sinθ sinθ ∴BD = √37 sin β sinθ [3] |
Activity 6
- In the diagram alongside, ABC is a right angled triangle. KC is the bisector of ACB. AC = r units and BCK = x
1.1 Write down AB in terms of x (2)
1.2 Give the size of AKC in terms of x (2)
1.3 If it is given that AK = 2 , calculate the value of x (7)
AB 3
- A, B and L are points in the same horizontal plane, HL is a vertical pole of length 3 metres,
AL = 5,2 m, the angle ALB = 113° and the angle of elevation of H from B is 40°.
2.1 Calculate the length of LB. (3)
2.2 Hence, or otherwise, calculate the length of AB. (3)
2.3 Determine the area of ∆ABL. (3) - The angle of elevation from a point C on the ground, at the centre of the goalpost, to the highest point A of the arc, directly above the centre of the Moses Mabhida soccer stadium, is 64,75°. The soccer pitch is 100 metres long and 64 metres wide as prescribed by FIFA for world cup stadiums. Also AC ⊥ PC.
In the figure below PQ = 100 metres and PC = 32 metres
3.1 Determine AC (2)
3.2 Calculate PAC (2)
3.3 A camera is placed at D, 40 m directly below point A, calculate the distance from D to C (4)
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| Solutions 1.1 sin 2x = AB ∴ AB = r sin 2x (2) r 1.2 AKC = 90° + x [ext. angle of ∆CBK] (2) 1.3 AK = r ∴ AK = r sinx sin x sin(90° + x) cos x r sin x AK = cos x = r sin x = 1 = 2 AB r sin 2x r cos x.2 cos x sin x 2 cos2x 3 ∴ cos2x = 3/4 cos x = √3/2 Hence x = 30º (7) 2.1 In ∆HLB, tan 40° = 3 LB [∆HLB is right-angled, so use a trig ratio] LB = 3 tan40º LB = 3,5752… ≈ 3,58 metres (3) 2.2 In ∆ABL, [ΔABL not right-angled. You have two sides and included angle, so use the Cosine Rule] AB2 = AL2 + BL2 – 2(AL)(BL).cos L AB2 = (5,2)2 + (3,58)2 – 2(5,2)(3,58).cos 113° AB2 = 54,40410… m2 AB = 7,38 m (3) 2.3 Area ∆ABL = ½ AL × BL × sin ALB = ½ (5,2) × (3,58) × sin 113° = 8,56805… ≈ 8,57 m2 (3) 3.1 cos 64,750° = CM ∴ AC = CM = 50m = 117,21 (2) AC cos64.75° 0.426569 3.2 tanPAC = PC AC PAC = tan−1 (32/AC) = 15,27° (2) 3.3 DC2 = AC2 + AD2 − 2AC.ADcos(90° − 64,75°) DC2 = (117,21 )2 + (40)2 − 2(117,21).40cos(25,25°) = 6857,289 DC = 82,81m (4) [28] |