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Secondary 3 Elementary Mathematics Practice Paper 1

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TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 3

TuitionGoWhere Practice Paper (AI)
Version: 1 of 5
Subject: Elementary Mathematics
Level: Secondary 3
Paper: Practice Paper 1 (Geometry & Trigonometry Focus)
Duration: 1 hour 30 minutes
Total Marks: 80

Name: ________________________
Class: ________________________
Date: ________________________

Instructions to Candidates

Write your name, class, and date in the spaces above.
Answer all questions.
Write your answers in the spaces provided in this booklet.
If working is needed for any question, it must be shown below that question.
The number of marks is given in brackets [ ] at the end of each question or part question.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
Take $\pi$ to be $3.142$ or use the $\pi$ button on your calculator.

Section A: Short Answer Questions (25 Marks)

Answer all questions in this section. Each question carries 1–3 marks.

1. In triangle $ABC$ , $\angle B = 90^\circ$ , $AB = 7$ cm, and $BC = 10$ cm.
Calculate the length of $AC$ .
 Answer: ________________________ cm [2]

2. Given that $\sin \theta = 0.6$ and $\theta$ is an obtuse angle ( $90^\circ < \theta < 180^\circ$ ), find the value of $\cos \theta$ .
 Answer: ________________________ [2]

3. Convert $2.5$ radians into degrees.
 Answer: ________________________ $^\circ$ [1]

4. The bearing of point $A$ from point $B$ is $135^\circ$ .
Find the bearing of point $B$ from point $A$ .
 Answer: ________________________ $^\circ$ [1]

5. In the diagram, $O$ is the centre of the circle. Points $A, B,$ and $C$ lie on the circumference.
$\angle AOC = 110^\circ$ .
Calculate $\angle ABC$ .
 Answer: ________________________ $^\circ$ [2]

6. A sector of a circle has radius $12$ cm and an angle of $1.2$ radians.
Calculate the area of the sector.
 Answer: ________________________ cm $^2$ [2]

7. Solve the equation $2\sin x = 1$ for $0^\circ \le x \le 360^\circ$ .
 Answer: $x =$ ________________________ $^\circ$ [2]

8. In triangle $PQR$ , $PQ = 8$ cm, $PR = 10$ cm, and $\angle QPR = 60^\circ$ .
Calculate the length of $QR$ .
 Answer: ________________________ cm [3]

9. The diagram shows a cuboid $ABCDEFGH$ .
$AB = 6$ cm, $BC = 4$ cm, and $CG = 3$ cm.
Calculate the length of the diagonal $AG$ .
 Answer: ________________________ cm [2]

10. Points $A(2, 5)$ and $B(8, 1)$ are given.
Find the gradient of the line perpendicular to $AB$ .
 Answer: ________________________ [2]

Section B: Structured Questions (35 Marks)

Answer all questions in this section. Show your working clearly.

11. The diagram shows a triangle $ABC$ with $AB = 12$ cm, $AC = 9$ cm, and $\angle BAC = 45^\circ$ .

(a) Calculate the area of triangle $ABC$ .
 Answer: ________________________ cm $^2$ [2]

(b) Calculate the length of $BC$ .
 Answer: ________________________ cm [3]

12. The diagram shows a circle with centre $O$ . $TA$ and $TB$ are tangents to the circle from an external point $T$ . $\angle AOB = 100^\circ$ .

(a) State the value of $\angle OAT$ .
 Answer: ________________________ $^\circ$ [1]

(b) Calculate $\angle ATB$ .
 Answer: ________________________ $^\circ$ [2]

(c) Point $C$ lies on the major arc $AB$ . Calculate $\angle ACB$ .
 Answer: ________________________ $^\circ$ [2]

13. A vertical tower $ST$ stands on horizontal ground. Point $A$ is due North of the tower, and point $B$ is due East of the tower.
The angle of elevation of the top of the tower $T$ from $A$ is $30^\circ$ .
The angle of elevation of $T$ from $B$ is $45^\circ$ .
The height of the tower $ST$ is $20$ m.

(a) Calculate the distance $SA$ .
 Answer: ________________________ m [2]

(b) Calculate the distance $SB$ .
 Answer: ________________________ m [2]

(d) Find the bearing of $B$ from $A$ .
 Answer: ________________________ $^\circ$ [2]

14. Consider the function $y = 2\sin(3x)$ for $0^\circ \le x \le 360^\circ$ .

(a) State the amplitude of the graph.
 Answer: ________________________ [1]

(b) State the period of the graph in degrees.
 Answer: ________________________ $^\circ$ [1]

[4]

15. In the diagram, $ABCD$ is a cyclic quadrilateral. $AB$ is parallel to $DC$ . $\angle DAB = 70^\circ$ and $\angle ABD = 30^\circ$ .

(a) Calculate $\angle ADB$ .
 Answer: ________________________ $^\circ$ [2]

(b) Calculate $\angle BCD$ .
 Answer: ________________________ $^\circ$ [2]

Section C: Problem Solving (20 Marks)

Answer all questions in this section. These questions require multi-step reasoning.

16. A ship leaves port $P$ and sails on a bearing of $050^\circ$ for $40$ km to reach point $Q$ .
From $Q$ , it changes course and sails on a bearing of $110^\circ$ for $30$ km to reach point $R$ .

(a) Calculate the size of $\angle PQR$ .
 Answer: ________________________ $^\circ$ [2]

(b) Calculate the distance $PR$ .
 Answer: ________________________ km [3]

17. The diagram shows a right pyramid with a square base $ABCD$ of side $10$ cm. The vertex $V$ is vertically above the centre $O$ of the base. The slant edge $VA = 13$ cm.

(a) Calculate the length of the diagonal $AC$ of the base.
 Answer: ________________________ cm [2]

(b) Calculate the height $VO$ of the pyramid.
 Answer: ________________________ cm [3]

(c) Calculate the angle between the slant edge $VA$ and the base $ABCD$ .
 Answer: ________________________ $^\circ$ [2]

(d) Calculate the total surface area of the pyramid.
 Answer: ________________________ cm $^2$ [3]

18. Points $A, B,$ and $C$ lie on a circle with centre $O$ and radius $8$ cm. The chord $AB$ has length $10$ cm.

(a) Calculate $\angle AOB$ .
 Answer: ________________________ $^\circ$ [3]

(b) Calculate the area of the minor segment bounded by chord $AB$ and the arc $AB$ .
 Answer: ________________________ cm $^2$ [4]

19. In triangle $XYZ$ , $XY = 15$ cm, $YZ = 12$ cm, and $\angle XYZ = 120^\circ$ .

(a) Calculate the area of triangle $XYZ$ .
 Answer: ________________________ cm $^2$ [2]

(b) Calculate the length of $XZ$ .
 Answer: ________________________ cm [3]

(c) Point $W$ lies on $XZ$ such that $YW$ is perpendicular to $XZ$ . Calculate the length of $YW$ .
 Answer: ________________________ cm [2]

20. The diagram shows two triangles, $\triangle ABC$ and $\triangle ADE$ , sharing vertex $A$ . $BC$ is parallel to $DE$ .
$AB = 6$ cm, $AC = 8$ cm, $AD = 9$ cm, and $AE = 12$ cm.

(a) Show that $\triangle ABC$ is similar to $\triangle ADE$ .
 [2]

(b) If the area of $\triangle ABC$ is $24$ cm $^2$ , calculate the area of $\triangle ADE$ .
 Answer: ________________________ cm $^2$ [3]

(c) Given that $\angle BAC = 50^\circ$ , calculate the length of $BC$ .
 Answer: ________________________ cm [3]

End of Paper

Answers

TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 3

Answer Key and Marking Scheme

Version: 1 of 5
Subject: Elementary Mathematics
Level: Secondary 3

Section A: Short Answer Questions

1.
Using Pythagoras' Theorem:
$AC^2 = AB^2 + BC^2 = 7^2 + 10^2 = 49 + 100 = 149$
$AC = \sqrt{149} \approx 12.206$
Answer: $12.2$ cm [2]
(1 mark for substitution, 1 mark for correct answer)

2.
$\sin \theta = 0.6$ . Since $\theta$ is obtuse, it is in the 2nd quadrant where cosine is negative.
$\cos^2 \theta + \sin^2 \theta = 1$
$\cos^2 \theta = 1 - 0.6^2 = 1 - 0.36 = 0.64$
$\cos \theta = -\sqrt{0.64} = -0.8$
Answer: $-0.8$ [2]
(1 mark for magnitude, 1 mark for negative sign)

3.
Degrees $= 2.5 \times \frac{180}{\pi} \approx 143.239$
Answer: $143^\circ$ (or $143.2^\circ$ ) [1]

4.
Back bearing $= 135^\circ + 180^\circ = 315^\circ$
Answer: $315^\circ$ [1]

5.
Reflex $\angle AOC = 360^\circ - 110^\circ = 250^\circ$
Angle at circumference $= \frac{1}{2} \times$ Angle at centre
$\angle ABC = \frac{1}{2} \times 250^\circ = 125^\circ$
Answer: $125^\circ$ [2]
(1 mark for reflex angle, 1 mark for division)

6.
Area $= \frac{1}{2} r^2 \theta = \frac{1}{2} (12)^2 (1.2) = \frac{1}{2} (144)(1.2) = 72 \times 1.2 = 86.4$
Answer: $86.4$ cm $^2$ [2]

7.
$\sin x = 0.5$
Basic angle $= 30^\circ$
Sine is positive in 1st and 2nd quadrants.
$x = 30^\circ$ or $180^\circ - 30^\circ = 150^\circ$
Answer: $30^\circ, 150^\circ$ [2]
(1 mark for each correct angle)

8.
Using Cosine Rule:
$QR^2 = PQ^2 + PR^2 - 2(PQ)(PR)\cos(60^\circ)$
$QR^2 = 8^2 + 10^2 - 2(8)(10)(0.5)$
$QR^2 = 64 + 100 - 80 = 84$
$QR = \sqrt{84} \approx 9.165$
Answer: $9.17$ cm [3]
(1 mark for formula/substitution, 1 mark for intermediate value, 1 mark for answer)

9.
Base diagonal $AC = \sqrt{6^2 + 4^2} = \sqrt{36+16} = \sqrt{52}$
Space diagonal $AG = \sqrt{AC^2 + CG^2} = \sqrt{52 + 3^2} = \sqrt{52+9} = \sqrt{61}$
$AG \approx 7.81$
Answer: $7.81$ cm [2]

10.
Gradient of $AB = \frac{1-5}{8-2} = \frac{-4}{6} = -\frac{2}{3}$
Gradient of perpendicular line $= -\frac{1}{m} = -\frac{1}{-2/3} = \frac{3}{2}$
Answer: $1.5$ (or $\frac{3}{2}$ ) [2]

Section B: Structured Questions

11.
(a) Area $= \frac{1}{2} ab \sin C = \frac{1}{2}(12)(9)\sin(45^\circ) = 54 \times 0.7071 \approx 38.18$
Answer: $38.2$ cm $^2$ [2]

(b) Cosine Rule:
$BC^2 = 12^2 + 9^2 - 2(12)(9)\cos(45^\circ)$
$BC^2 = 144 + 81 - 216(0.7071) = 225 - 152.73 = 72.27$
$BC = \sqrt{72.27} \approx 8.50$
Answer: $8.50$ cm [3]

(c) Sine Rule:
$\frac{\sin C}{12} = \frac{\sin 45^\circ}{8.50}$
$\sin C = \frac{12 \sin 45^\circ}{8.50} \approx 0.998$
$C = \sin^{-1}(0.998) \approx 86.4^\circ$
(Check: $180 - 45 - 86.4 = 48.6$ , valid triangle)
Answer: $86.4^\circ$ [2]

12.
(a) Tangent is perpendicular to radius.
Answer: $90^\circ$ [1]

(b) Quadrilateral $OATB$ : Sum of angles $= 360^\circ$ .
$\angle ATB = 360 - 90 - 90 - 100 = 80^\circ$
Answer: $80^\circ$ [2]

(c) Angle at circumference is half angle at centre.
$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2}(100^\circ) = 50^\circ$
Answer: $50^\circ$ [2]

13.
(a) In $\triangle STA$ (right-angled at $S$ ):
$\tan 30^\circ = \frac{ST}{SA} \Rightarrow SA = \frac{20}{\tan 30^\circ} = 20\sqrt{3} \approx 34.64$
Answer: $34.6$ m [2]

(b) In $\triangle STB$ (right-angled at $S$ ):
$\tan 45^\circ = \frac{ST}{SB} \Rightarrow SB = \frac{20}{1} = 20$
Answer: $20$ m [2]

(c) $\triangle SAB$ is right-angled at $S$ (North vs East).
$AB^2 = SA^2 + SB^2 = (34.64)^2 + 20^2 = 1200 + 400 = 1600$
$AB = \sqrt{1600} = 40$
Answer: $40$ m [2]

(d) Bearing of $B$ from $A$ :
In $\triangle SAB$ , $\tan(\angle SAB) = \frac{SB}{SA} = \frac{20}{34.64} \approx 0.577$
$\angle SAB = \tan^{-1}(0.577) \approx 30^\circ$
Since $B$ is East of $S$ and $A$ is North of $S$ , the bearing from $A$ involves turning clockwise from North.
Line $AS$ is South ( $180^\circ$ ). Line $AB$ is $30^\circ$ East of South.
Bearing $= 180^\circ - 30^\circ$ ? No.
Let's visualize: $A$ is North of $S$ . $B$ is East of $S$ .
Vector $A \to S$ is South ( $180^\circ$ ). Vector $S \to B$ is East ( $90^\circ$ ).
Angle $SAB = 30^\circ$ .
The bearing of $B$ from $A$ is $180^\circ - 30^\circ$ ? No, $B$ is to the right (East) of the vertical line $AS$ .
Wait, $A$ is North of $S$ . So $S$ is South of $A$ .
$B$ is East of $S$ .
Triangle $SAB$ : Angle at $S$ is $90^\circ$ .
Angle at $A$ is $30^\circ$ .
Bearing of $S$ from $A$ is $180^\circ$ .
$B$ is to the East (left/counter-clockwise from South? No, East is left if facing South? No. Facing South, East is to your Left. Bearing decreases? No.
Standard Bearing: North is $0^\circ$ . East is $90^\circ$ . South is $180^\circ$ .
$A$ is at $(0, y)$ . $S$ is at $(0,0)$ . $B$ is at $(x, 0)$ .
Vector $AB = (x, -y)$ .
Angle from North ( $0,1$ ) to $AB$ .
$\tan \alpha = \frac{x}{y} = \frac{20}{34.64}$ . $\alpha = 30^\circ$ from the South line towards East.
So Bearing $= 180^\circ - 30^\circ$ ? No. East is $90^\circ$ . South is $180^\circ$ .
From $A$ , looking South is $180^\circ$ . $B$ is to the East.
So we subtract $30^\circ$ from $180^\circ$ ?
Let's check coordinates. $A=(0, 34.64)$ , $B=(20, 0)$ .
$\Delta x = 20, \Delta y = -34.64$ .
Angle $\theta$ with vertical: $\tan \theta = 20/34.64$ . $\theta = 30^\circ$ .
Since $\Delta x > 0$ (East) and $\Delta y < 0$ (South), it is in SE quadrant.
Bearing $= 180^\circ - 30^\circ = 150^\circ$ ?
Wait. Bearing is clockwise from North.
North is Up. East is Right.
Vector is Down and Right.
Angle from South (Down) to Vector is $30^\circ$ towards East (Right).
Clockwise from North: $180^\circ$ (South) minus $30^\circ$ ? No.
Clockwise from North to East is $90^\circ$ . To South is $180^\circ$ .
The vector is between East and South.
Angle from East: $90 - 30 = 60^\circ$ ? No.
Angle from South is $30^\circ$ towards East.
So Bearing $= 180^\circ - 30^\circ = 150^\circ$ ?
Let's re-verify.
Tan(angle from South) = Opp/Adj = $SB/SA = 20/34.64$ .
Angle is $30^\circ$ .
Direction is South-East.
Bearing of South is $180^\circ$ . East is $90^\circ$ .
SE is between $90$ and $180$ .
So Bearing $= 180 - 30 = 150^\circ$ .
Answer: $150^\circ$ [2]

14.
(a) Amplitude $= 2$ [1]
(b) Period $= \frac{360}{3} = 120^\circ$ [1]
(c) Sketch:

Starts at $(0,0)$ .
Max at $(30, 2)$ .
Zero at $(60, 0)$ .
Min at $(90, -2)$ .
Zero at $(120, 0)$ .
Repeats 3 times up to $360^\circ$ .
[4] (1 mark for shape, 1 for amplitude, 1 for period/frequency, 1 for labels)

15.
(a) $\triangle ABD$ : Sum of angles $= 180^\circ$ .
$\angle ADB = 180 - 70 - 30 = 80^\circ$
Answer: $80^\circ$ [2]

(b) Cyclic Quad: Opposite angles sum to $180^\circ$ .
$\angle BCD + \angle DAB = 180^\circ$
$\angle BCD = 180 - 70 = 110^\circ$
Answer: $110^\circ$ [2]

(c) $AB \parallel DC$ . Alternate interior angles are equal.
$\angle BDC = \angle ABD = 30^\circ$ .
In $\triangle BCD$ : Sum $= 180^\circ$ .
$\angle CBD = 180 - \angle BCD - \angle BDC = 180 - 110 - 30 = 40^\circ$
Answer: $40^\circ$ [3]

Section C: Problem Solving

16.
(a) Bearing $P \to Q$ is $050^\circ$ . North line at $Q$ is parallel.
Interior angle at $Q$ (from North back to $P$ ) is $180 - 50 = 130^\circ$ ? No.
Alternate angle: Angle between South at $Q$ and $QP$ is $50^\circ$ .
So Angle between North at $Q$ and $QP$ is $180 + 50 = 230^\circ$ (Back bearing).
Or simpler:
Draw North at $Q$ . Angle from North clockwise to $QP$ is $180+50 = 230^\circ$ .
Bearing $Q \to R$ is $110^\circ$ .
$\angle PQR = 230^\circ - 110^\circ = 120^\circ$ ?
Let's use geometry.
North at $Q$ . Line $QP$ goes South-West. Angle with South is $50^\circ$ (alt interior).
Line $QR$ goes South-East. Bearing $110^\circ$ means $110^\circ$ from North.
Angle between North and $QR$ is $110^\circ$ .
Angle between North and $QP$ is $180+50 = 230^\circ$ .
$\angle PQR = 230 - 110 = 120^\circ$ .
Answer: $120^\circ$ [2]

(b) Cosine Rule on $\triangle PQR$ :
$PR^2 = 40^2 + 30^2 - 2(40)(30)\cos(120^\circ)$
$\cos(120^\circ) = -0.5$
$PR^2 = 1600 + 900 - 2400(-0.5) = 2500 + 1200 = 3700$
$PR = \sqrt{3700} \approx 60.82$
Answer: $60.8$ km [3]

(c) Sine Rule to find $\angle QPR$ :
$\frac{\sin(\angle QPR)}{30} = \frac{\sin(120^\circ)}{60.82}$
$\sin(\angle QPR) = \frac{30 \sin(120^\circ)}{60.82} \approx 0.426$
$\angle QPR = \sin^{-1}(0.426) \approx 25.2^\circ$
Bearing of $Q$ from $P$ is $050^\circ$ .
$R$ is to the "right" of $PQ$ ?
Check geometry: $Q$ is NE of $P$ . $R$ is SE of $Q$ .
Triangle $PQR$ . Angle at $P$ is $25.2^\circ$ .
Is $R$ clockwise or anti-clockwise from $Q$ relative to $P$ ?
Bearing $P \to Q$ is $50^\circ$ .
$\angle QPR$ is inside the triangle.
We need Bearing $P \to R$ .
Since $R$ is generally East/South of $Q$ , and $Q$ is NE of $P$ , the line $PR$ will have a bearing greater than $50^\circ$ .
Bearing $P \to R = 50^\circ + 25.2^\circ = 75.2^\circ$ .
Bearing of $P$ from $R$ is Back Bearing of $R$ from $P$ .
Back Bearing $= 75.2^\circ + 180^\circ = 255.2^\circ$ .
Answer: $255^\circ$ [4]

17.
(a) Diagonal of square base $AC = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \approx 14.14$
Answer: $14.1$ cm [2]

(b) $O$ is midpoint of $AC$ . $AO = \frac{14.14}{2} = 7.07$ cm.
In $\triangle VOA$ (right-angled at $O$ ):
$VO^2 + AO^2 = VA^2$
$VO^2 + 7.07^2 = 13^2$
$VO^2 + 50 = 169$
$VO^2 = 119 \Rightarrow VO = \sqrt{119} \approx 10.91$
Answer: $10.9$ cm [3]

(c) Angle between $VA$ and base is $\angle VAO$ .
$\cos(\angle VAO) = \frac{AO}{VA} = \frac{7.07}{13} \approx 0.544$
$\angle VAO = \cos^{-1}(0.544) \approx 57.0^\circ$
Answer: $57.0^\circ$ [2]

(d) Surface Area = Base Area + 4 $\times$ Area of Triangular Face.
Base Area $= 10 \times 10 = 100$ cm $^2$ .
Triangular Face (e.g., $\triangle VAB$ ): Base $AB=10$ . Need slant height $VM$ (midpoint of $AB$ ).
In $\triangle VOM$ : $VO = 10.91$ , $OM = 5$ (half side).
$VM = \sqrt{10.91^2 + 5^2} = \sqrt{119 + 25} = \sqrt{144} = 12$ cm.
Area of one face $= \frac{1}{2} \times 10 \times 12 = 60$ cm $^2$ .
Total SA $= 100 + 4(60) = 340$ cm $^2$ .
Answer: $340$ cm $^2$ [3]

18.
(a) $\triangle AOB$ is isosceles with $OA=OB=8$ , $AB=10$ .
Use Cosine Rule on $\triangle AOB$ :
$10^2 = 8^2 + 8^2 - 2(8)(8)\cos(\angle AOB)$
$100 = 128 - 128\cos(\angle AOB)$
$128\cos(\angle AOB) = 28$
$\cos(\angle AOB) = \frac{28}{128} = 0.21875$
$\angle AOB = \cos^{-1}(0.21875) \approx 77.36^\circ$
Answer: $77.4^\circ$ [3]

(b) Area of Sector $= \frac{77.36}{360} \times \pi (8)^2 \approx 43.01$ cm $^2$ .
Area of $\triangle AOB = \frac{1}{2}(8)(8)\sin(77.36^\circ) \approx 31.22$ cm $^2$ .
Area of Segment $= 43.01 - 31.22 = 11.79$ cm $^2$ .
Answer: $11.8$ cm $^2$ [4]

19.
(a) Area $= \frac{1}{2}(15)(12)\sin(120^\circ) = 90 \times \frac{\sqrt{3}}{2} = 45\sqrt{3} \approx 77.94$
Answer: $77.9$ cm $^2$ [2]

(b) Cosine Rule:
$XZ^2 = 15^2 + 12^2 - 2(15)(12)\cos(120^\circ)$
$XZ^2 = 225 + 144 - 360(-0.5) = 369 + 180 = 549$
$XZ = \sqrt{549} \approx 23.43$
Answer: $23.4$ cm [3]

(c) Area $= \frac{1}{2} \times \text{Base} \times \text{Height}$
$77.94 = \frac{1}{2} (23.43) (YW)$
$YW = \frac{2 \times 77.94}{23.43} \approx 6.65$
Answer: $6.65$ cm [2]

20.
(a) $\frac{AB}{AD} = \frac{6}{9} = \frac{2}{3}$ .
$\frac{AC}{AE} = \frac{8}{12} = \frac{2}{3}$ .
$\angle BAC = \angle DAE$ (Common angle).
Therefore, $\triangle ABC \sim \triangle ADE$ by SAS similarity. [2]

(b) Ratio of areas $= (\text{Ratio of lengths})^2 = (\frac{2}{3})^2 = \frac{4}{9}$ .
$\frac{\text{Area } ABC}{\text{Area } ADE} = \frac{4}{9}$
$\text{Area } ADE = \text{Area } ABC \times \frac{9}{4} = 24 \times 2.25 = 54$
Answer: $54$ cm $^2$ [3]

(c) In $\triangle ABC$ :
$BC^2 = 6^2 + 8^2 - 2(6)(8)\cos(50^\circ)$
$BC^2 = 36 + 64 - 96(0.6428) = 100 - 61.71 = 38.29$
$BC = \sqrt{38.29} \approx 6.19$
Answer: $6.19$ cm [3]