Secondary 3 Elementary Mathematics Geometry Trigonometry Quiz

Secondary 3 Elementary Mathematics Quiz - Geometry Trigonometry (Answer Key)

1.
Using Pythagoras' Theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = 12^2 + 5^2 = 144 + 25 = 169$
$AC = \sqrt{169} = 13$ cm
Answer: 13 cm [2]

2.
$\cos(\angle QPR) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{8}{17}$
$\angle QPR = \cos^{-1}\left(\frac{8}{17}\right)$
$\angle QPR \approx 61.9^\circ$
Answer: $61.9^\circ$ [2]

3.
Let $\theta$ be the angle with the ground.
$\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2.5}{6}$
$\theta = \cos^{-1}\left(\frac{2.5}{6}\right)$
$\theta \approx 65.4^\circ$
Answer: $65.4^\circ$ [2]

4.
$M$ is midpoint of $CD$, so $DM = MC = 5$ cm.
In $\triangle ADM$ (right-angled at $D$):
$\tan(\angle AMD) = \frac{AD}{DM} = \frac{6}{5} = 1.2 \Rightarrow \angle AMD = \tan^{-1}(1.2) \approx 50.19^\circ$
In $\triangle BCM$ (right-angled at $C$):
$\tan(\angle BMC) = \frac{BC}{MC} = \frac{6}{5} = 1.2 \Rightarrow \angle BMC = \tan^{-1}(1.2) \approx 50.19^\circ$
Angles on a straight line $CD$:
$\angle AMB = 180^\circ - (\angle AMD + \angle BMC)$
$\angle AMB = 180^\circ - (50.19^\circ + 50.19^\circ) = 180^\circ - 100.38^\circ = 79.62^\circ$
Answer: $79.6^\circ$ [4]

5.
Area $= \frac{1}{2} ab \sin C$
Area $= \frac{1}{2} (12)(9) \sin 40^\circ$
Area $= 54 \sin 40^\circ \approx 34.71$
Answer: $34.7$ cm$^2$ [2]

6.
Using Cosine Rule:
$XZ^2 = XY^2 + YZ^2 - 2(XY)(YZ) \cos(\angle XYZ)$
$XZ^2 = 10^2 + 14^2 - 2(10)(14) \cos 110^\circ$
$XZ^2 = 100 + 196 - 280 \cos 110^\circ$
$XZ^2 = 296 - 280(-0.3420) \approx 296 + 95.76 = 391.76$
$XZ = \sqrt{391.76} \approx 19.79$
Answer: $19.8$ cm [3]

7.
Using Cosine Rule for angle:
$\cos(\angle PQR) = \frac{PQ^2 + QR^2 - PR^2}{2(PQ)(QR)}$
$\cos(\angle PQR) = \frac{8^2 + 11^2 - 15^2}{2(8)(11)}$
$\cos(\angle PQR) = \frac{64 + 121 - 225}{176} = \frac{-40}{176}$
$\angle PQR = \cos^{-1}\left(-\frac{40}{176}\right) \approx 103.14^\circ$
Answer: $103^\circ$ [3]

8.
Using Sine Rule:
$\frac{\sin(\angle ACB)}{AB} = \frac{\sin(\angle BAC)}{BC}$
$\frac{\sin C}{7} = \frac{\sin 35^\circ}{9}$
$\sin C = \frac{7 \sin 35^\circ}{9} \approx 0.4456$
$C = \sin^{-1}(0.4456) \approx 26.46^\circ$
Since $\angle ACB$ is acute, we take the acute value.
Answer: $26.5^\circ$ [3]

9.
First, find area of $\triangle ABC$:
Area $= \frac{1}{2}(20)(25) \sin 130^\circ = 250 \sin 130^\circ \approx 191.51$ m$^2$
Next, find side $AC$ using Cosine Rule:
$AC^2 = 20^2 + 25^2 - 2(20)(25) \cos 130^\circ$
$AC^2 = 400 + 625 - 1000(-0.6428) = 1025 + 642.8 = 1667.8$
$AC = \sqrt{1667.8} \approx 40.84$ m
Let $h$ be the shortest distance from $B$ to $AC$ (height).
Area $= \frac{1}{2} (\text{base}) (\text{height}) = \frac{1}{2} (AC) (h)$
$191.51 = \frac{1}{2} (40.84) (h)$
$h = \frac{383.02}{40.84} \approx 9.38$
Answer: $9.38$ m [4]

10.
Let $\theta$ be the angle between $AG$ and base $ABCD$. This is $\angle GAC$.
First, find diagonal of base $AC$:
$AC = \sqrt{10^2 + 6^2} = \sqrt{100 + 36} = \sqrt{136} \approx 11.66$ cm
In right-angled $\triangle ACG$ (vertical height $CG = 8$):
$\tan \theta = \frac{CG}{AC} = \frac{8}{\sqrt{136}}$
$\theta = \tan^{-1}\left(\frac{8}{11.66}\right) \approx 34.5^\circ$
Answer: $34.5^\circ$ [4]

11.
Let $h$ be height $AB$.
In $\triangle ABD$ (right-angled at $B$):
$\tan 45^\circ = \frac{h}{BD} \Rightarrow 1 = \frac{h}{BD} \Rightarrow BD = h$
In $\triangle ABC$ (right-angled at $B$):
$\tan 30^\circ = \frac{h}{BC} = \frac{h}{BD + 20} = \frac{h}{h + 20}$
$\frac{1}{\sqrt{3}} = \frac{h}{h + 20}$
$h + 20 = h\sqrt{3}$
$20 = h\sqrt{3} - h = h(\sqrt{3} - 1)$
$h = \frac{20}{\sqrt{3} - 1} \approx \frac{20}{0.732} \approx 27.32$
Answer: $27.3$ m [4]

12.
Draw North lines at $A, B, C$.
Bearing $B$ from $A$ is $050^\circ$. Interior angle at $A$ relative to North is $50^\circ$.
At $B$, North line is parallel. Angle between North (down) and $BA$ is $50^\circ$ (alternate).
Bearing $C$ from $B$ is $140^\circ$. Angle between North (up) and $BC$ is $140^\circ$.
Angle $\angle ABC = 180^\circ - 50^\circ + (180^\circ - 140^\circ)$? No, simpler:
Angle of $BA$ from South is $50^\circ$. Angle of $BC$ from North is $140^\circ$.
$\angle ABC = 180^\circ - (140^\circ - 50^\circ)$? Let's use coordinates or geometry.
North at $B$. $BA$ is bearing $230^\circ$ (reverse of 050). $BC$ is bearing $140^\circ$.
$\angle ABC = 230^\circ - 140^\circ = 90^\circ$.
So $\triangle ABC$ is right-angled at $B$.
$AB = 12, BC = 15$.
$\tan(\angle BCA) = \frac{AB}{BC} = \frac{12}{15} = 0.8$.
$\angle BCA = \tan^{-1}(0.8) \approx 38.66^\circ$.
Bearing of $C$ from $B$ is $140^\circ$.
North line at $C$. The line $CB$ has bearing $140^\circ + 180^\circ = 320^\circ$ from $C$? No.
Bearing $B$ from $C$ is $140^\circ + 180^\circ = 320^\circ$.
Angle $\angle BCA$ is "inside" the triangle to the left of $CB$?
Let's check orientation. $A$ is SW of $B$? No, $B$ is NE of $A$. $C$ is SE of $B$.
Triangle $ABC$: $B$ is the right angle.
Bearing $C$ from $B$ is $140^\circ$.
Bearing $A$ from $B$ is $230^\circ$.
Vector $BC$ is at $140^\circ$. Vector $BA$ is at $230^\circ$.
To find bearing of $A$ from $C$:
Draw North at $C$. Bearing of $B$ from $C$ is $320^\circ$ ($140+180$).
In $\triangle ABC$, angle at $C$ is $38.66^\circ$.
Since $A$ is to the "left" of line $CB$ when looking from $C$ to $B$?
Let's use coordinates. $B=(0,0)$. $A = (12 \sin 230^\circ, 12 \cos 230^\circ)$. $C = (15 \sin 140^\circ, 15 \cos 140^\circ)$.
Actually, simpler geometry:
North at $C$. Line $CB$ is bearing $320^\circ$.
Angle $\angle BCA = 38.7^\circ$.
Is $A$ at bearing $320^\circ - 38.7^\circ$ or $320^\circ + 38.7^\circ$?
$A$ is West of $B$. $C$ is East of $B$ (roughly). So $A$ is further West.
Bearing should be larger (more West/South).
Wait, $B$ is origin. $A$ is roughly SW. $C$ is roughly SE.
From $C$, looking at $B$ (NW, bearing 320). $A$ is further left (SW).
So Bearing $A$ from $C = 320^\circ + 38.66^\circ = 358.66^\circ$?
Let's re-verify $\angle ABC$.
Bearing $A \to B = 050$. Bearing $B \to C = 140$.
Angle $ABC = 180 - (140-50) = 90$?
North at $B$. $BA$ is $50^\circ$ from South (West side). $BC$ is $140^\circ$ from North (East side).
Angle $S-B-A = 50^\circ$. Angle $N-B-C = 140^\circ$.
Angle $S-B-C = 180-140=40^\circ$.
Angle $ABC = 50+40=90^\circ$. Correct.
Triangle is Right Angled.
$\tan C = 12/15$. $C = 38.66^\circ$.
Bearing $B$ from $C$: Reverse of $140^\circ$ is $320^\circ$.
$A$ is to the "right" of $CB$?
Coordinates: $B(0,0)$. $C(15 \sin 140, 15 \cos 140) \approx (9.64, -11.49)$.
$A(12 \sin 230, 12 \cos 230) \approx (-9.19, -7.71)$.
Vector $CA = A - C = (-9.19 - 9.64, -7.71 - (-11.49)) = (-18.83, 3.78)$.
$\tan \alpha = \frac{-18.83}{3.78}$. $x$ negative, $y$ positive $\rightarrow$ 2nd Quadrant (NW).
Angle from North: $\tan^{-1}(\frac{18.83}{3.78}) \approx 78.6^\circ$ West of North.
Bearing $= 360 - 78.6 = 281.4^\circ$.
Let's re-evaluate geometric addition.
Bearing $B$ from $C$ is $320^\circ$.
Angle $BCA = 38.7^\circ$.
Vector $CA$ is roughly West. Vector $CB$ is NW.
So $A$ is "left" of $B$ from $C$'s perspective?
$320 - 38.7 = 281.3^\circ$.
Answer: $281^\circ$ [4]

13.
$O$ is centre of square base. Diagonal $AC = \sqrt{10^2+10^2} = 10\sqrt{2}$.
$AO = \frac{1}{2} AC = 5\sqrt{2} \approx 7.07$ cm.
In right-angled $\triangle VOA$:
$\tan(\angle VAO) = \frac{VO}{AO} = \frac{12}{5\sqrt{2}}$.
$\angle VAO = \tan^{-1}\left(\frac{12}{7.07}\right) \approx 59.5^\circ$.
Answer: $59.5^\circ$ [3]

14.
Arc Length $s = r\theta$
$s = 8 \times 1.2 = 9.6$ cm
Answer: $9.6$ cm [2]

15.
Area of Sector $= \frac{1}{2} r^2 \theta = \frac{1}{2} (8^2) (1.2) = 32(1.2) = 38.4$ cm$^2$.
Area of Triangle $AOB = \frac{1}{2} r^2 \sin \theta = \frac{1}{2} (64) \sin(1.2)$.
$\sin(1.2 \text{ rad}) \approx 0.9320$.
Area $\triangle AOB = 32 \times 0.9320 \approx 29.82$ cm$^2$.
Area of Segment = Area of Sector - Area of Triangle
Area $= 38.4 - 29.82 = 8.58$ cm$^2$.
Answer: $8.58$ cm$^2$ [3]

16.
$\angle BDC$ and $\angle BAC$ subtend the same arc $BC$.
Therefore, $\angle BDC = \angle BAC$.
$\angle BDC = 25^\circ$.
Answer: $25^\circ$ [2]

17.
In quadrilateral $OATB$, $\angle OAT = 90^\circ$ and $\angle OBT = 90^\circ$ (tangents are perpendicular to radius).
Sum of angles in quadrilateral $= 360^\circ$.
$\angle ATB + \angle AOB + 90^\circ + 90^\circ = 360^\circ$
$\angle ATB + 110^\circ + 180^\circ = 360^\circ$
$\angle ATB = 360^\circ - 290^\circ = 70^\circ$.
Answer: $70^\circ$ [3]

18.
Area of Sector $= \frac{1}{2} r^2 \theta$
$15 = \frac{1}{2} (5^2) \theta$
$15 = \frac{25}{2} \theta$
$30 = 25 \theta$
$\theta = \frac{30}{25} = 1.2$ radians.
Answer: $1.2$ rad [2]

19.
By the Alternate Segment Theorem, the angle between the tangent and chord ($40^\circ$) is equal to the angle in the alternate segment ($\angle PRQ$ or $\angle PQR$?).
The chord is $PQ$. The angle in the alternate segment is $\angle PRQ$. So $\angle PRQ = 40^\circ$.
Since $PQ = PR$, $\triangle PQR$ is isosceles with base $QR$.
Therefore, base angles are equal: $\angle PQR = \angle PRQ$.
$\angle PQR = 40^\circ$.
Answer: $40^\circ$ [3]

20.
Opposite angles in a cyclic quadrilateral sum to $180^\circ$.
$\angle BCD + \angle DAB = 180^\circ$
$\angle BCD + 85^\circ = 180^\circ \Rightarrow \angle BCD = 95^\circ$.
$\angle CDA + \angle ABC = 180^\circ$
$\angle CDA + 100^\circ = 180^\circ \Rightarrow \angle CDA = 80^\circ$.
Answer: $\angle BCD = 95^\circ, \angle CDA = 80^\circ$ [2]