O-Level Additional Mathematics Quiz - Geometry Trigonometry
Answer Key and Marking Scheme
Total Marks: 60
Section A: Trigonometric Functions and Graphs (Questions 1–5)
1. Given sin θ = 5 13 \sin \theta = \frac{5}{13} sin θ = 13 5 θ \theta θ 90 ∘ < θ < 180 ∘ 90^\circ < \theta < 180^\circ 9 0 ∘ < θ < 18 0 ∘
In second quadrant, cos θ < 0 \cos \theta < 0 cos θ < 0 tan θ < 0 \tan \theta < 0 tan θ < 0
cos 2 θ = 1 − sin 2 θ = 1 − 25 169 = 144 169 \cos^2\theta = 1 - \sin^2\theta = 1 - \frac{25}{169} = \frac{144}{169} cos 2 θ = 1 − sin 2 θ = 1 − 169 25 = 169 144
cos θ = − 12 13 \cos\theta = -\frac{12}{13} cos θ = − 13 12
tan θ = sin θ cos θ = 5 / 13 − 12 / 13 = − 5 12 \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{5/13}{-12/13} = -\frac{5}{12} tan θ = c o s θ s i n θ = − 12/13 5/13 = − 12 5
Answer: cos θ = − 12 13 \cos\theta = -\frac{12}{13} cos θ = − 13 12 tan θ = − 5 12 \tan\theta = -\frac{5}{12} tan θ = − 12 5
2. f ( x ) = 2 sin ( 3 x ) − 1 f(x) = 2\sin(3x) - 1 f ( x ) = 2 sin ( 3 x ) − 1
(a) Amplitude = ∣ 2 ∣ = 2 |2| = 2 ∣2∣ = 2
(b) Period = 2 π 3 \frac{2\pi}{3} 3 2 π
(c) Since − 1 ≤ sin ( 3 x ) ≤ 1 -1 \leq \sin(3x) \leq 1 − 1 ≤ sin ( 3 x ) ≤ 1 − 2 ≤ 2 sin ( 3 x ) ≤ 2 -2 \leq 2\sin(3x) \leq 2 − 2 ≤ 2 sin ( 3 x ) ≤ 2 − 3 ≤ f ( x ) ≤ 1 -3 \leq f(x) \leq 1 − 3 ≤ f ( x ) ≤ 1 [ − 3 , 1 ] [-3, 1] [ − 3 , 1 ]
3. cos A = − 3 5 \cos A = -\frac{3}{5} cos A = − 5 3 180 ∘ < A < 270 ∘ 180^\circ < A < 270^\circ 18 0 ∘ < A < 27 0 ∘
In third quadrant, sin A < 0 \sin A < 0 sin A < 0
sin 2 A = 1 − cos 2 A = 1 − 9 25 = 16 25 \sin^2A = 1 - \cos^2A = 1 - \frac{9}{25} = \frac{16}{25} sin 2 A = 1 − cos 2 A = 1 − 25 9 = 25 16
sin A = − 4 5 \sin A = -\frac{4}{5} sin A = − 5 4
sin 2 A = 2 sin A cos A = 2 ( − 4 5 ) ( − 3 5 ) = 24 25 \sin 2A = 2\sin A\cos A = 2\left(-\frac{4}{5}\right)\left(-\frac{3}{5}\right) = \frac{24}{25} sin 2 A = 2 sin A cos A = 2 ( − 5 4 ) ( − 5 3 ) = 25 24
Answer: sin 2 A = 24 25 \sin 2A = \frac{24}{25} sin 2 A = 25 24
4. y = 3 cos ( x 2 ) y = 3\cos\left(\frac{x}{2}\right) y = 3 cos ( 2 x ) 0 ∘ ≤ x ≤ 720 ∘ 0^\circ \leq x \leq 720^\circ 0 ∘ ≤ x ≤ 72 0 ∘
Amplitude = 3
Period = 360 ∘ 1 / 2 = 720 ∘ \frac{360^\circ}{1/2} = 720^\circ 1/2 36 0 ∘ = 72 0 ∘
Maximum points: ( 0 ∘ , 3 ) (0^\circ, 3) ( 0 ∘ , 3 ) ( 720 ∘ , 3 ) (720^\circ, 3) ( 72 0 ∘ , 3 )
Minimum point: ( 360 ∘ , − 3 ) (360^\circ, -3) ( 36 0 ∘ , − 3 )
Crosses x x x cos ( x / 2 ) = 0 \cos(x/2) = 0 cos ( x /2 ) = 0 x / 2 = 90 ∘ , 270 ∘ , 450 ∘ , 630 ∘ x/2 = 90^\circ, 270^\circ, 450^\circ, 630^\circ x /2 = 9 0 ∘ , 27 0 ∘ , 45 0 ∘ , 63 0 ∘ x = 180 ∘ , 540 ∘ x = 180^\circ, 540^\circ x = 18 0 ∘ , 54 0 ∘
Marking: [1 mark] for correct shape (cosine curve), [1 mark] for correct amplitude and period, [1 mark] for correctly labelled key points.
5. sin − 1 ( sin 5 π 6 ) \sin^{-1}\left(\sin\frac{5\pi}{6}\right) sin − 1 ( sin 6 5 π )
sin 5 π 6 = sin ( π − π 6 ) = sin π 6 = 1 2 \sin\frac{5\pi}{6} = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\frac{\pi}{6} = \frac{1}{2} sin 6 5 π = sin ( π − 6 π ) = sin 6 π = 2 1
Since 1 2 \frac{1}{2} 2 1 sin − 1 \sin^{-1} sin − 1 π 6 \frac{\pi}{6} 6 π [ − π 2 , π 2 ] [-\frac{\pi}{2}, \frac{\pi}{2}] [ − 2 π , 2 π ]
sin − 1 ( 1 2 ) = π 6 \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} sin − 1 ( 2 1 ) = 6 π
Answer: π 6 \frac{\pi}{6} 6 π
Section B: Trigonometric Identities and Equations (Questions 6–10)
6. Prove sin θ 1 + cos θ + 1 + cos θ sin θ = 2 csc θ \frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} = 2\csc\theta 1 + c o s θ s i n θ + s i n θ 1 + c o s θ = 2 csc θ
LHS = sin 2 θ + ( 1 + cos θ ) 2 sin θ ( 1 + cos θ ) = \frac{\sin^2\theta + (1 + \cos\theta)^2}{\sin\theta(1 + \cos\theta)} = s i n θ ( 1 + c o s θ ) s i n 2 θ + ( 1 + c o s θ ) 2
= sin 2 θ + 1 + 2 cos θ + cos 2 θ sin θ ( 1 + cos θ ) = \frac{\sin^2\theta + 1 + 2\cos\theta + \cos^2\theta}{\sin\theta(1 + \cos\theta)} = s i n θ ( 1 + c o s θ ) s i n 2 θ + 1 + 2 c o s θ + c o s 2 θ
= ( sin 2 θ + cos 2 θ ) + 1 + 2 cos θ sin θ ( 1 + cos θ ) = \frac{(\sin^2\theta + \cos^2\theta) + 1 + 2\cos\theta}{\sin\theta(1 + \cos\theta)} = s i n θ ( 1 + c o s θ ) ( s i n 2 θ + c o s 2 θ ) + 1 + 2 c o s θ
= 1 + 1 + 2 cos θ sin θ ( 1 + cos θ ) = 2 ( 1 + cos θ ) sin θ ( 1 + cos θ ) = \frac{1 + 1 + 2\cos\theta}{\sin\theta(1 + \cos\theta)} = \frac{2(1 + \cos\theta)}{\sin\theta(1 + \cos\theta)} = s i n θ ( 1 + c o s θ ) 1 + 1 + 2 c o s θ = s i n θ ( 1 + c o s θ ) 2 ( 1 + c o s θ )
= 2 sin θ = 2 csc θ = = \frac{2}{\sin\theta} = 2\csc\theta = = s i n θ 2 = 2 csc θ =
7. 2 cos 2 θ + 3 sin θ = 3 2\cos^2\theta + 3\sin\theta = 3 2 cos 2 θ + 3 sin θ = 3 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ \leq \theta \leq 360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
Using cos 2 θ = 1 − sin 2 θ \cos^2\theta = 1 - \sin^2\theta cos 2 θ = 1 − sin 2 θ
2 ( 1 − sin 2 θ ) + 3 sin θ = 3 2(1 - \sin^2\theta) + 3\sin\theta = 3 2 ( 1 − sin 2 θ ) + 3 sin θ = 3
2 − 2 sin 2 θ + 3 sin θ = 3 2 - 2\sin^2\theta + 3\sin\theta = 3 2 − 2 sin 2 θ + 3 sin θ = 3
− 2 sin 2 θ + 3 sin θ − 1 = 0 -2\sin^2\theta + 3\sin\theta - 1 = 0 − 2 sin 2 θ + 3 sin θ − 1 = 0
2 sin 2 θ − 3 sin θ + 1 = 0 2\sin^2\theta - 3\sin\theta + 1 = 0 2 sin 2 θ − 3 sin θ + 1 = 0
( 2 sin θ − 1 ) ( sin θ − 1 ) = 0 (2\sin\theta - 1)(\sin\theta - 1) = 0 ( 2 sin θ − 1 ) ( sin θ − 1 ) = 0
sin θ = 1 2 \sin\theta = \frac{1}{2} sin θ = 2 1 sin θ = 1 \sin\theta = 1 sin θ = 1
sin θ = 1 2 ⟹ θ = 30 ∘ , 150 ∘ \sin\theta = \frac{1}{2} \implies \theta = 30^\circ, 150^\circ sin θ = 2 1 ⟹ θ = 3 0 ∘ , 15 0 ∘
sin θ = 1 ⟹ θ = 90 ∘ \sin\theta = 1 \implies \theta = 90^\circ sin θ = 1 ⟹ θ = 9 0 ∘
Answer: θ = 30 ∘ , 90 ∘ , 150 ∘ \theta = 30^\circ, 90^\circ, 150^\circ θ = 3 0 ∘ , 9 0 ∘ , 15 0 ∘
8. tan A = 3 4 \tan A = \frac{3}{4} tan A = 4 3 tan B = 5 12 \tan B = \frac{5}{12} tan B = 12 5 A A A B B B
tan ( A + B ) = tan A + tan B 1 − tan A tan B = 3 4 + 5 12 1 − 3 4 ⋅ 5 12 \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B} = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \cdot \frac{5}{12}} tan ( A + B ) = 1 − t a n A t a n B t a n A + t a n B = 1 − 4 3 ⋅ 12 5 4 3 + 12 5
= 9 12 + 5 12 1 − 15 48 = 14 12 33 48 = 14 12 × 48 33 = 14 × 4 33 = 56 33 = \frac{\frac{9}{12} + \frac{5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{\frac{33}{48}} = \frac{14}{12} \times \frac{48}{33} = \frac{14 \times 4}{33} = \frac{56}{33} = 1 − 48 15 12 9 + 12 5 = 48 33 12 14 = 12 14 × 33 48 = 33 14 × 4 = 33 56
Since tan ( A + B ) > 0 \tan(A + B) > 0 tan ( A + B ) > 0 A + B A + B A + B A , B < 90 ∘ A, B < 90^\circ A , B < 9 0 ∘ tan ( A + B ) > 0 \tan(A+B) > 0 tan ( A + B ) > 0 A + B = tan − 1 ( 56 33 ) A + B = \tan^{-1}\left(\frac{56}{33}\right) A + B = tan − 1 ( 33 56 ) 56 33 ≈ 1.697 \frac{56}{33} \approx 1.697 33 56 ≈ 1.697 A + B ≈ 59.5 ∘ A + B \approx 59.5^\circ A + B ≈ 59. 5 ∘ A + B = tan − 1 ( 56 33 ) A + B = \tan^{-1}\left(\frac{56}{33}\right) A + B = tan − 1 ( 33 56 ) 59.5 ∘ 59.5^\circ 59. 5 ∘
Answer: tan ( A + B ) = 56 33 \tan(A + B) = \frac{56}{33} tan ( A + B ) = 33 56 A + B ≈ 59.5 ∘ A + B \approx 59.5^\circ A + B ≈ 59. 5 ∘ tan − 1 ( 56 33 ) \tan^{-1}(\frac{56}{33}) tan − 1 ( 33 56 )
9. 5 sin θ + 12 cos θ = R sin ( θ + α ) 5\sin\theta + 12\cos\theta = R\sin(\theta + \alpha) 5 sin θ + 12 cos θ = R sin ( θ + α )
R = 5 2 + 12 2 = 25 + 144 = 169 = 13 R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 R = 5 2 + 1 2 2 = 25 + 144 = 169 = 13
α = tan − 1 ( 12 5 ) ≈ 67.38 ∘ \alpha = \tan^{-1}\left(\frac{12}{5}\right) \approx 67.38^\circ α = tan − 1 ( 5 12 ) ≈ 67.3 8 ∘
Maximum value of 5 sin θ + 12 cos θ 5\sin\theta + 12\cos\theta 5 sin θ + 12 cos θ R = 13 R = 13 R = 13
Maximum value of 5 sin θ + 12 cos θ + 7 = 13 + 7 = 20 5\sin\theta + 12\cos\theta + 7 = 13 + 7 = 20 5 sin θ + 12 cos θ + 7 = 13 + 7 = 20
Answer: R = 13 R = 13 R = 13 α ≈ 67.4 ∘ \alpha \approx 67.4^\circ α ≈ 67. 4 ∘
10. cos 2 θ = sin θ \cos 2\theta = \sin\theta cos 2 θ = sin θ 0 ≤ θ ≤ 2 π 0 \leq \theta \leq 2\pi 0 ≤ θ ≤ 2 π
Using cos 2 θ = 1 − 2 sin 2 θ \cos 2\theta = 1 - 2\sin^2\theta cos 2 θ = 1 − 2 sin 2 θ
1 − 2 sin 2 θ = sin θ 1 - 2\sin^2\theta = \sin\theta 1 − 2 sin 2 θ = sin θ
2 sin 2 θ + sin θ − 1 = 0 2\sin^2\theta + \sin\theta - 1 = 0 2 sin 2 θ + sin θ − 1 = 0
( 2 sin θ − 1 ) ( sin θ + 1 ) = 0 (2\sin\theta - 1)(\sin\theta + 1) = 0 ( 2 sin θ − 1 ) ( sin θ + 1 ) = 0
sin θ = 1 2 \sin\theta = \frac{1}{2} sin θ = 2 1 sin θ = − 1 \sin\theta = -1 sin θ = − 1
sin θ = 1 2 ⟹ θ = π 6 , 5 π 6 \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}, \frac{5\pi}{6} sin θ = 2 1 ⟹ θ = 6 π , 6 5 π
sin θ = − 1 ⟹ θ = 3 π 2 \sin\theta = -1 \implies \theta = \frac{3\pi}{2} sin θ = − 1 ⟹ θ = 2 3 π
Answer: θ = π 6 , 5 π 6 , 3 π 2 \theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2} θ = 6 π , 6 5 π , 2 3 π
Section C: Coordinate Geometry (Questions 11–15)
11. A ( 2 , − 1 ) A(2, -1) A ( 2 , − 1 ) B ( 8 , 7 ) B(8, 7) B ( 8 , 7 )
Midpoint = ( 2 + 8 2 , − 1 + 7 2 ) = ( 5 , 3 ) = \left(\frac{2+8}{2}, \frac{-1+7}{2}\right) = (5, 3) = ( 2 2 + 8 , 2 − 1 + 7 ) = ( 5 , 3 )
Length A B = ( 8 − 2 ) 2 + ( 7 − ( − 1 ) ) 2 = 6 2 + 8 2 = 36 + 64 = 100 = 10 AB = \sqrt{(8-2)^2 + (7-(-1))^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 A B = ( 8 − 2 ) 2 + ( 7 − ( − 1 ) ) 2 = 6 2 + 8 2 = 36 + 64 = 100 = 10
Answer: Midpoint = ( 5 , 3 ) (5, 3) ( 5 , 3 ) 10 10 10
12. P ( − 3 , 2 ) P(-3, 2) P ( − 3 , 2 ) Q ( 5 , − 4 ) Q(5, -4) Q ( 5 , − 4 )
Midpoint of P Q = ( − 3 + 5 2 , 2 + ( − 4 ) 2 ) = ( 1 , − 1 ) PQ = \left(\frac{-3+5}{2}, \frac{2+(-4)}{2}\right) = (1, -1) P Q = ( 2 − 3 + 5 , 2 2 + ( − 4 ) ) = ( 1 , − 1 )
Gradient of P Q = − 4 − 2 5 − ( − 3 ) = − 6 8 = − 3 4 PQ = \frac{-4-2}{5-(-3)} = \frac{-6}{8} = -\frac{3}{4} P Q = 5 − ( − 3 ) − 4 − 2 = 8 − 6 = − 4 3
Gradient of perpendicular bisector = 4 3 = \frac{4}{3} = 3 4
Equation: y − ( − 1 ) = 4 3 ( x − 1 ) y - (-1) = \frac{4}{3}(x - 1) y − ( − 1 ) = 3 4 ( x − 1 )
y + 1 = 4 3 x − 4 3 y + 1 = \frac{4}{3}x - \frac{4}{3} y + 1 = 3 4 x − 3 4
Multiply by 3: 3 y + 3 = 4 x − 4 3y + 3 = 4x - 4 3 y + 3 = 4 x − 4
4 x − 3 y − 7 = 0 4x - 3y - 7 = 0 4 x − 3 y − 7 = 0
Answer: 4 x − 3 y − 7 = 0 4x - 3y - 7 = 0 4 x − 3 y − 7 = 0
13. Centre C ( 3 , − 2 ) C(3, -2) C ( 3 , − 2 ) P ( 7 , 1 ) P(7, 1) P ( 7 , 1 )
Radius = C P = ( 7 − 3 ) 2 + ( 1 − ( − 2 ) ) 2 = 4 2 + 3 2 = 16 + 9 = 25 = 5 = CP = \sqrt{(7-3)^2 + (1-(-2))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 = C P = ( 7 − 3 ) 2 + ( 1 − ( − 2 ) ) 2 = 4 2 + 3 2 = 16 + 9 = 25 = 5
Equation: ( x − 3 ) 2 + ( y − ( − 2 ) ) 2 = 5 2 (x - 3)^2 + (y - (-2))^2 = 5^2 ( x − 3 ) 2 + ( y − ( − 2 ) ) 2 = 5 2
( x − 3 ) 2 + ( y + 2 ) 2 = 25 (x - 3)^2 + (y + 2)^2 = 25 ( x − 3 ) 2 + ( y + 2 ) 2 = 25
Answer: ( x − 3 ) 2 + ( y + 2 ) 2 = 25 (x - 3)^2 + (y + 2)^2 = 25 ( x − 3 ) 2 + ( y + 2 ) 2 = 25
14. Line: y = 2 x − 1 y = 2x - 1 y = 2 x − 1 x 2 + y 2 = 10 x^2 + y^2 = 10 x 2 + y 2 = 10
Substitute: x 2 + ( 2 x − 1 ) 2 = 10 x^2 + (2x - 1)^2 = 10 x 2 + ( 2 x − 1 ) 2 = 10
x 2 + 4 x 2 − 4 x + 1 = 10 x^2 + 4x^2 - 4x + 1 = 10 x 2 + 4 x 2 − 4 x + 1 = 10
5 x 2 − 4 x − 9 = 0 5x^2 - 4x - 9 = 0 5 x 2 − 4 x − 9 = 0
( 5 x − 9 ) ( x + 1 ) = 0 (5x - 9)(x + 1) = 0 ( 5 x − 9 ) ( x + 1 ) = 0
x = 9 5 x = \frac{9}{5} x = 5 9 x = − 1 x = -1 x = − 1
When x = 9 5 x = \frac{9}{5} x = 5 9 y = 2 ( 9 5 ) − 1 = 18 5 − 5 5 = 13 5 y = 2(\frac{9}{5}) - 1 = \frac{18}{5} - \frac{5}{5} = \frac{13}{5} y = 2 ( 5 9 ) − 1 = 5 18 − 5 5 = 5 13
When x = − 1 x = -1 x = − 1 y = 2 ( − 1 ) − 1 = − 3 y = 2(-1) - 1 = -3 y = 2 ( − 1 ) − 1 = − 3
Answer: ( 9 5 , 13 5 ) \left(\frac{9}{5}, \frac{13}{5}\right) ( 5 9 , 5 13 ) ( − 1 , − 3 ) (-1, -3) ( − 1 , − 3 )
15. x 2 + y 2 − 6 x + 4 y − 12 = 0 x^2 + y^2 - 6x + 4y - 12 = 0 x 2 + y 2 − 6 x + 4 y − 12 = 0
Complete the square:
( x 2 − 6 x ) + ( y 2 + 4 y ) = 12 (x^2 - 6x) + (y^2 + 4y) = 12 ( x 2 − 6 x ) + ( y 2 + 4 y ) = 12
( x − 3 ) 2 − 9 + ( y + 2 ) 2 − 4 = 12 (x - 3)^2 - 9 + (y + 2)^2 - 4 = 12 ( x − 3 ) 2 − 9 + ( y + 2 ) 2 − 4 = 12
( x − 3 ) 2 + ( y + 2 ) 2 = 25 (x - 3)^2 + (y + 2)^2 = 25 ( x − 3 ) 2 + ( y + 2 ) 2 = 25
Centre = ( 3 , − 2 ) = (3, -2) = ( 3 , − 2 ) = 25 = 5 = \sqrt{25} = 5 = 25 = 5
Answer: Centre = ( 3 , − 2 ) (3, -2) ( 3 , − 2 ) 5 5 5
Section D: Proofs in Plane Geometry (Questions 16–20)
16. In cyclic quadrilateral A B C D ABCD A B C D E E E
∠ B A C = 35 ∘ \angle BAC = 35^\circ ∠ B A C = 3 5 ∘ ∠ A B D = 50 ∘ \angle ABD = 50^\circ ∠ A B D = 5 0 ∘
Angles in the same segment: ∠ B D C = ∠ B A C = 35 ∘ \angle BDC = \angle BAC = 35^\circ ∠ B D C = ∠ B A C = 3 5 ∘
Similarly, ∠ A C D = ∠ A B D = 50 ∘ \angle ACD = \angle ABD = 50^\circ ∠ A C D = ∠ A B D = 5 0 ∘
Answer: ∠ B D C = 35 ∘ \angle BDC = 35^\circ ∠ B D C = 3 5 ∘ ∠ A C D = 50 ∘ \angle ACD = 50^\circ ∠ A C D = 5 0 ∘
17. S T ∥ Q R ST \parallel QR S T ∥ QR P S = 4 PS = 4 P S = 4 S Q = 6 SQ = 6 S Q = 6 Q R = 15 QR = 15 QR = 15
P Q = P S + S Q = 4 + 6 = 10 PQ = PS + SQ = 4 + 6 = 10 P Q = P S + S Q = 4 + 6 = 10
By similar triangles (△ P S T ∼ △ P Q R \triangle PST \sim \triangle PQR △ P S T ∼ △ P QR
S T Q R = P S P Q \frac{ST}{QR} = \frac{PS}{PQ} QR S T = P Q P S
S T 15 = 4 10 \frac{ST}{15} = \frac{4}{10} 15 S T = 10 4
S T = 15 × 4 10 = 6 ST = 15 \times \frac{4}{10} = 6 S T = 15 × 10 4 = 6
Answer: S T = 6 ST = 6 S T = 6
18. Given: O O O A B AB A B C C C C D ⊥ A B CD \perp AB C D ⊥ A B
To prove: △ A C D ∼ △ C B D \triangle ACD \sim \triangle CBD △ A C D ∼ △ C B D
Proof:
In △ A C D \triangle ACD △ A C D △ C B D \triangle CBD △ C B D
∠ A D C = ∠ C D B = 90 ∘ \angle ADC = \angle CDB = 90^\circ ∠ A D C = ∠ C D B = 9 0 ∘ C D ⊥ A B CD \perp AB C D ⊥ A B
∠ A C B = 90 ∘ \angle ACB = 90^\circ ∠ A C B = 9 0 ∘
In △ A C D \triangle ACD △ A C D ∠ C A D = 90 ∘ − ∠ A C D \angle CAD = 90^\circ - \angle ACD ∠ C A D = 9 0 ∘ − ∠ A C D
In △ C B D \triangle CBD △ C B D ∠ B C D = 90 ∘ − ∠ A C D \angle BCD = 90^\circ - \angle ACD ∠ B C D = 9 0 ∘ − ∠ A C D ∠ A C B = 90 ∘ \angle ACB = 90^\circ ∠ A C B = 9 0 ∘
Therefore ∠ C A D = ∠ B C D \angle CAD = \angle BCD ∠ C A D = ∠ B C D
Hence △ A C D ∼ △ C B D \triangle ACD \sim \triangle CBD △ A C D ∼ △ C B D
19. Given: A B C D ABCD A B C D E E E B C BC B C A E AE A E D C DC D C F F F
To prove: C C C D F DF D F
Proof:
In △ A B E \triangle ABE △ A B E △ F C E \triangle FCE △ F C E
∠ A B E = ∠ F C E \angle ABE = \angle FCE ∠ A B E = ∠ F C E A B ∥ D C AB \parallel DC A B ∥ D C
∠ A E B = ∠ F E C \angle AEB = \angle FEC ∠ A E B = ∠ F E C
B E = C E BE = CE B E = C E E E E B C BC B C
Therefore △ A B E ≅ △ F C E \triangle ABE \cong \triangle FCE △ A B E ≅ △ F C E
Hence A B = C F AB = CF A B = C F
But A B = D C AB = DC A B = D C
So C F = D C CF = DC C F = D C C C C D F DF D F
20. Tangent-secant theorem: P T 2 = P A × P B PT^2 = PA \times PB P T 2 = P A × P B
6 2 = 4 × P B 6^2 = 4 \times PB 6 2 = 4 × P B
36 = 4 × P B 36 = 4 \times PB 36 = 4 × P B
P B = 9 PB = 9 P B = 9
A B = P B − P A = 9 − 4 = 5 AB = PB - PA = 9 - 4 = 5 A B = P B − P A = 9 − 4 = 5
Answer: A B = 5 AB = 5 A B = 5
END OF ANSWER KEY