Mathematics Grade 12
Euclidean geometry Grade 12 Questions and Answers
Euclidean geometry Grade 12 Questions and Answers:Euclidean geometry is a mathematical system attributed to Alexandrian Greek mathematician Euclid, which he described in his textbook on geometry: the Elements. Euclid’s method consists in assuming a small set of intuitively appealing axioms, and deducing many other propositions from these.
Contents
Types of triangles
| Name | Diagram | Properties |
| Scalene |  | All sides and angles are different. |
| Isosceles |  | Two sides are equal in length. The angles opposite the equal sides are also equal. |
| Equilateral |  | All three sides are equal in length and all three angles are equal. |
| Acute-angled |  | Each of the three interior angles is less than 9090°. |
| Obtuse-angled |  | One interior angle is greater than 9090°. |
| Right-angled |  | One interior angle is 9090°. |
Congruent triangles (EMCJ2)
| Condition | Diagram |
| SSS(side, side, side) |  △ABC≡△EDF△ABC≡△EDF |
| SAS(side, incl. angle, side) |  △GHI≡△JKL△GHI≡△JKL |
| AAS(angle, angle, side) |  △MNO≡△PQR△MNO≡△PQR |
| RHS(9090°, hypotenuse, side) |  △STU≡△VWX△STU≡△VWX |
Source:https://intl.siyavula.com/read/maths/grade-12/euclidean-geometry/08-euclidean-geometry-01
Euclidean Geometry Grade 12 Questions and Answers (Downloadable pdf)
Hey, Grade 12 Learners. My Courses portal has everything you needed to ace your matric exams, tests, assessments, research tasks, and assignments. Feel free to explore all resources for grade 12 learners, such as Study Guides, Department of Basic Education Past Exam Papers with Memos, and Speech Topics.
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Mathematics Grade 12
Mathematics Grade 12 November 2020 Question Papers and Memos for study revision
Mathematics Grade 12 November 2020 Question Papers and Memos pdf download: Paper 1 and Paper 2
List Mathematics Grade 12 November 2020 Question Papers and Memos:
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Cubic Functions Grade 12 Questions and Answers pdf: In mathematics, a cubic function is a function of the form f(x)=ax^3+bx^2+cx+d where the coefficients a, b, c, and d are real numbers, and the variable x takes real values, and a ≠ 0. In other words, it is both a polynomial function of degree three, and a real function.
Downloadable Cubic Functions Grade 12 Questions and Answers pdf
Questions and Answers
My Courses has a large Questions and Answers repository for the most popular High School and Tertiary Schools subjects. This comes in handy when doing your revision or preparing for exams, tests, research tasks, and assignments.
General Solution Trigonometric Grade 12 pdf: This page contains a guide for Mathematics Grade 12 Students with the following Trigonometric notes: Trigonometric Functions, 3D Trigonometry, Identities Problems with Solutions, and Challenging Exam Practices.
Contents
Concepts and skills
- Simplify trigonometric equations.
- Find the reference angle.
- Use reduction formulae to find other angles within each quadrant.
- Find the general solution of a given trigonometric equation.
Prior knowledge:
- Trigonometric ratios.
- Trigonometric identities.
- Solve problems in the Cartesian plane.
- Co-functions.
- Factorisation.
- Revise the following trigonometric ratios for right-angled triangles, which you learnt in Grade 10.
Will start with the basic equation, sin x = 0. The principal solution for this case will be x = 0, π, 2π as these values satisfy the given equation lying in the interval [0, 2π]. But, we know that if sin x = 0, then x = 0, π, 2π, π, -2π, -6π, etc. are solutions of the given equation. Hence, the general solution for sin x = 0 will be, x = nπ, where n∈I.
Similarly, general solution for cos x = 0 will be x = (2n+1)π/2, n∈I, as cos x has a value equal to 0 at π/2, 3π/2, 5π/2, -7π/2, -11π/2 etc. Below here is the table defining the general solutions of the given trigonometric functions involved equations.
Solutions Trigonometric Equations
| Equations | Solutions |
| sin x = 0 | x = nπ |
| cos x = 0 | x = (nπ + π/2) |
| tan x = 0 | x = nπ |
| sin x = 1 | x = (2nπ + π/2) = (4n+1)π/2 |
| cos x = 1 | x = 2nπ |
| sin x = sin θ | x = nπ + (-1)nθ, where θ ∈ [-π/2, π/2] |
| cos x = cos θ | x = 2nπ ± θ, where θ ∈ (0, π] |
| tan x = tan θ | x = nπ + θ, where θ ∈ (-π/2 , π/2] |
| sin2 x = sin2 θ | x = nπ ± θ |
| cos2 x = cos2 θ | x = nπ ± θ |
| tan2 x = tan2 θ | x = nπ ± θ |
Watch: Video
How to solve general trigonometric equations formula and find solutions
The periodicity of the trigonometric functions means that there are an infinite number of positive and negative angles that satisfy an equation. If we do not restrict the solution, then we need to determine the general solution to the equation. We know that the sine and cosine functions have a period of 360360° and the tangent function has a period of 180180°.
Method for finding the solution:
- Simplify the equation using algebraic methods and trigonometric identities.
- Determine the reference angle (use a positive value).
- Use the CAST diagram to determine where the function is positive or negative (depending on the given equation/information).
- Restricted values: find the angles that lie within a specified interval by adding/subtracting multiples of the appropriate period.
- General solution: find the angles in the interval [0°;360°][0°;360°] that satisfy the equation and add multiples of the period to each answer.
- Check answers using a calculator.
Questions and Answers
My Courses has a large Questions and Answers repository for the most popular High School and Tertiary Schools subjects. This comes in handy when doing your revision or preparing for exams, tests, research tasks, and assignments.
Download General Solution Trigonometric Grade 12 pdf
Identities Problems with Solutions for exam Practice
1. (1 – sin A)/(1 + sin A) = (sec A – tan A)2
Solution:
L.H.S = (1 – sin A)/(1 + sin A)
= (1 – sin A)2/(1 – sin A) (1 + sin A),[Multiply both numerator and denominator by (1 – sin A)
= (1 – sin A)2/(1 – sin2 A)
= (1 – sin A)2/(cos2 A), [Since sin2 θ + cos2 θ = 1 ⇒ cos2 θ = 1 – sin2 θ]
= {(1 – sin A)/cos A}2
= (1/cos A – sin A/cos A)2
= (sec A – tan A)2 = R.H.S. Proved.
2. Prove that, √{(sec θ – 1)/(sec θ + 1)} = cosec θ – cot θ.
Solution:
L.H.S.= √{(sec θ – 1)/(sec θ + 1)}
= √[{(sec θ – 1) (sec θ – 1)}/{(sec θ + 1) (sec θ – 1)}]; [multiplying numerator and denominator by (sec θ – l) under radical sign]
= √{(sec θ – 1)2/(sec2 θ – 1)}
=√{(sec θ -1)2/tan2 θ}; [since, sec2 θ = 1 + tan2 θ ⇒ sec2 θ – 1 = tan2 θ]
= (sec θ – 1)/tan θ
= (sec θ/tan θ) – (1/tan θ)
= {(1/cos θ)/(sin θ/cos θ)} – cot θ
= {(1/cos θ) × (cos θ/sin θ)} – cot θ
= (1/sin θ) – cot θ
= cosec θ – cot θ = R.H.S. Proved.
3. tan4 θ + tan2 θ = sec4 θ – sec2 θ
Solution:
L.H.S = tan4 θ + tan2 θ
= tan2 θ (tan2 θ + 1)
= (sec2 θ – 1) (tan2 θ + 1) [since, tan2 θ = sec2 θ – 1]
= (sec2 θ – 1) sec2 θ [since, tan2 θ + 1 = sec2 θ]
= sec4 θ – sec2 θ = R.H.S. Proved.
More problems on trigonometric identities are shown where one side of the identity ends up with the other side.
4. . cos θ/(1 – tan θ) + sin θ/(1 – cot θ) = sin θ + cos θ
Solution:
L.H.S = cos θ/(1 – tan θ) + sin θ/(1 – cot θ)
= cos θ/{1 – (sin θ/cos θ)} + sin θ/{1 – (cos θ/sin θ)}
= cos θ/{(cos θ – sin θ)/cos θ} + sin θ/{(sin θ – cos θ/sin θ)}
= cos2 θ/(cos θ – sin θ) + sin2 θ/(cos θ – sin θ)
= (cos2 θ – sin2 θ)/(cos θ – sin θ)
= [(cos θ + sin θ)(cos θ – sin θ)]/(cos θ – sin θ)
= (cos θ + sin θ) = R.H.S. Proved.
5. Show that, 1/(csc A – cot A) – 1/sin A = 1/sin A – 1/(csc A + cot A)
Solution:
We have,
1/(csc A – cot A) + 1/(csc A + cot A)
= (csc A + cot A + csc A – cot A)/(csc2 A – cot2 A)
= (2 csc A)/1; [since, csc2 A = 1 + cot2 A ⇒ csc2A – cot2 A = 1]
= 2/sin A; [since, csc A = 1/sin A]
Therefore,
1/(csc A – cot A) + 1/(csc A + cot A) = 2/sin A
⇒ 1/(csc A – cot A) + 1/(csc A + cot A) = 1/sin A + 1/sin A
Therefore, 1/(csc A – cot A) – 1/sin A = 1/sin A – 1/(csc A + cot A) Proved.
6. (tan θ + sec θ – 1)/(tan θ – sec θ + 1) = (1 + sin θ)/cos θ
Solution:
L.H.S = (tan θ + sec θ – 1)/(tan θ – sec θ + 1)
= [(tan θ + sec θ) – (sec2 θ – tan2 θ)]/(tan θ – sec θ + 1), [Since, sec2 θ – tan2 θ = 1]
= {(tan θ + sec θ) – (sec θ + tan θ) (sec θ – tan θ)}/(tan θ – sec θ + 1)
= {(tan θ + sec θ) (1 – sec θ + tan θ)}/(tan θ – sec θ + 1)
= {(tan θ + sec θ) (tan θ – sec θ + 1)}/(tan θ – sec θ + 1)
= tan θ + sec θ
= (sin θ/cos θ) + (1/cos θ)
= (sin θ + 1)/cos θ
= (1 + sin θ)/cos θ = R.H.S. Proved.
Sources:
https://www.math-only-math.com/problems-on-trigonometric-identities.html
https://intl.siyavula.com/read/maths/grade-12/trigonometry/04-trigonometry-04
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