MPM2D · Exam review

Principles of Mathematics Exam Review

A practice exam with questions from all 7 units, then the key facts from every lesson, unit by unit. For a deeper look at one unit, open its test review.

Unit test reviews

Practice exam

35 questions, the same number from each unit. Each one shows the lesson it comes from.

Unit 1: Solving Linear Systems by Graphing
1
Which ordered pair is the solution of the system $y = x + 2$ and $y = - 2 x + 8$ ?
Unit 1: Solving by Substitution
2
In which equation is it easiest to isolate a variable without creating fractions?
Unit 1: Solving by Elimination
3
Solve $2 x + y = 9$ and $3 x - y = 11$ by elimination. What is the value of $x$ ?
Unit 1: Solving by Elimination
4
Match each system to the method that suits it best and the reason.
y = 4x − 3 and 2x + 5y = 18
5x + 2y = 9 and 3x − 2y = 7
Estimate where y = 2x + 1 and y = −x + 4 cross, and see the picture
Unit 1: Problems Using Linear Systems
5
Put the steps for solving a word problem with a linear system in order.
1. 1Write two equations, one for each fact in the problem
2. 2Verify the values in the original words and state the answer in a sentence
3. 3Solve the system by substitution, elimination or graphing
4. 4Define two variables, with units
Unit 2: Midpoint and Length of a Line Segment
6
What is the midpoint of the segment joining $(- 4, 7)$ and $(10, - 3)$ ?
Unit 2: Equation of a Circle
7
What is the radius of the circle $x^{2} + y^{2} = 64$ ?
Unit 2: Slopes, Perpendicular Bisectors and Distance to a Line
8
What is the slope of a line perpendicular to a line with slope $\frac{3}{4}$ ?
Unit 2: Slopes, Perpendicular Bisectors and Distance to a Line
9
Put the steps for finding the distance from a point to a line in order.
1. 1Use the length formula from the point to the foot
2. 2Solve the system of the two lines to find the foot of the perpendicular
3. 3Write the equation of the perpendicular line through the point
4. 4Find the slope of the given line
5. 5Take the negative reciprocal to get the perpendicular slope
Unit 2: Verifying Properties of Triangles and Quadrilaterals
10
Put the steps for verifying the midsegment theorem for a given triangle in order.
1. 1Write a concluding sentence naming the property
2. 2Compare the exact slopes and lengths
3. 3Calculate the slope and length of the midsegment and of the third side
4. 4State the goal: the midsegment and the third side have equal slopes, and the midsegment is half as long
5. 5Find the midpoints of the two sides
Unit 3: Expanding and Simplifying
11
Expand and simplify $(x + 6) (x - 2)$ .
Unit 3: Common Factors and Grouping
12
What is the greatest common factor of $24 x^{2} y$ and $36 x y^{2}$ ?
Unit 3: Factoring Trinomials
13
Factor $x^{2} + 10 x + 16$ .
Unit 3: Factoring Trinomials
14
Match each trinomial with its factored form.
1. A. $(x - 3) (x - 4)$
2. B. $(x - 4) (x + 3)$
3. C. $(x + 3) (x + 4)$
4. D. $(x + 4) (x - 3)$
$x^{2} + 7 x + 12$
$x^{2} - 7 x + 12$
$x^{2} + x - 12$
$x^{2} - x - 12$
Unit 3: Perfect Squares and Difference of Squares
15
Factor $(x + 3)^{2} - 16$ .
Unit 4: Quadratic Relations and Their Properties
16
Each table uses $x = 0, 1, 2, 3$ . Which list of $y$ -values comes from a quadratic relation?
Unit 4: Transformations and Vertex Form
17
What is the $y$ -intercept of $y = - 2 (x - 1)^{2} + 7$ ?
Unit 4: Factored Form and Zeros
18
Match each relation with its zeros.
1. A. $- 6$ and $1$
2. B. $0$ and $6$
3. C. $6$ and $- 1$
4. D. $6$ only
$y = (x - 6) (x + 1)$
$y = (x + 6) (x - 1)$
$y = x (x - 6)$
$y = (x - 6)^{2}$
Unit 4: Completing the Square
19
A classmate wrote: $2 x^{2} + 8 x + 1 = 2 (x^{2} + 4 x + 4 - 4) + 1 = 2 (x + 2)^{2} - 3 .$ What went wrong?
Unit 4: Comparing y = x squared and y = 2 to the x
20
For $y = 2^{x}$ with $x = 0, 1, 2, 3, 4$ , what happens to the second differences?
Unit 5: Solving Quadratic Equations by Factoring
21
Solve $x^{2} - 7 x + 12 = 0$ .
Unit 5: The Quadratic Formula
22
To use the quadratic formula on $3 x^{2} - x = 7$ , what are $a$ , $b$ and $c$ ?
Unit 5: Real and Non-Real Roots
23
Calculate the discriminant of $3 x^{2} - 5 x + 2 = 0$ .
Unit 5: Real and Non-Real Roots
24
Which statement about $x^{2} - 4 x + 5 = 0$ is correct?
Unit 5: Solving Problems with Quadratics
25
In the model $h = - 4.9 t^{2} + v_{0} t + h_{0}$ , what does $h_{0}$ represent? Answer in a few words.
Unit 6: Similar Triangles
26
Which information is enough, by itself, to prove that two triangles are similar?
Unit 6: Similar Triangles
27
In $△ A BC$ , point $D$ is on $A B$ and point $E$ is on $A C$ with $D E$ parallel to $BC$ . $A D = 4.0$ m, $A B = 10.0$ m and $D E = 3.2$ m. Find $BC$ in metres.
m
Unit 6: The Primary Trigonometric Ratios
28
In a right triangle, the side opposite angle A is 3 and the side adjacent to angle A is 4. What is tan A?
Unit 6: Solving Right-Triangle Problems
29
A boat is 100 m from the base of a vertical cliff that is 60 m high. Find the angle of depression from the top of the cliff to the boat, to the nearest degree.
°
Unit 6: Solving Right-Triangle Problems
30
A ramp rises 0.5 m over a horizontal distance of 6.0 m. What angle does the ramp make with the ground? Round to the nearest degree.
°
Unit 7: The Sine Law
31
In $△ A BC$ , which side is opposite $∠ B$ ?
Unit 7: The Sine Law
32
Which statement about using $sin^{- 1}$ with the sine law is correct?
Unit 7: The Cosine Law
33
What is the value of $cos 9 0^{\circ}$ ? (This is why the cosine law turns into the Pythagorean theorem for a right angle.)
Unit 7: Choosing a Method and Solving Problems
34
Two hikers leave camp on straight paths that make an angle of $7 5^{\circ}$ . One walks 4.2 km and the other 5.5 km. How far apart are they, to one decimal place?
km
Unit 7: Choosing a Method and Solving Problems
35
What is the three-letter name for the case where you know two angles and the side between them?

Key facts by unit

Unit 1: Linear Systems

Solving Linear Systems by Graphing

The solution of a linear system is the ordered pair $(x, y)$ that makes every equation true; on a graph it is the point of intersection of the lines.
To check a possible solution, substitute it into each original equation; it must work in all of them.
Different slopes: the lines cross once, so the system has exactly one solution.
Same slope, different y-intercepts: parallel lines, no solution. Same slope and same y-intercept: the same line, infinitely many solutions.
Standard form $A x + B y = C$ : find the intercepts (set $x = 0$ , then $y = 0$ ), or rearrange to $y = m x + b$ when $B \neq = 0$ .

Solving by Substitution

Substitution: isolate a variable in one equation, substitute into the other equation, solve, back-substitute, check.
Isolate a variable with coefficient $1$ or $- 1$ to avoid fractions.
Put the substituted expression in brackets: $2 x - 3 y = - 9$ with $y = 14 - 3 x$ becomes $2 x - 3 (14 - 3 x) = - 9$ .
If the variables cancel and leave a true statement (such as $4 = 4$ ), there are infinitely many solutions; a false statement (such as $12 = 5$ ) means no solution.
The solution is an ordered pair $(x, y)$ : finding one variable is only halfway.

Solving by Elimination

Opposite coefficients ( $+ 3 y$ and $- 3 y$ ): add the equations. Equal coefficients ( $2 b$ and $2 b$ ): subtract them.
Multiplying every term of an equation by the same non-zero number gives an equivalent equation (the same line).
To match coefficients, multiply to their lowest common multiple, e.g. $3 x$ and $2 x$ both become $6 x$ .
If every variable cancels: a true statement ( $0 = 0$ ) means infinitely many solutions; a false statement ( $0 = 3$ ) means no solution.
Choose elimination for two standard-form equations, substitution when a variable is already isolated, graphing for a picture or estimate.

Problems Using Linear Systems

Define each variable precisely, with its units, before writing equations.
Most problems give an amount equation (totals of quantities) and a value equation (totals of cost, value or interest).
Simple interest: $I = P r t$ ; for one year $I = P r$ , with the rate as a decimal ( $4% = 0.04$ ).
With a current $c$ and still-water speed $b$ : downstream speed $b + c$ , upstream speed $b - c$ ; distance = speed × time.
Verify the solution in the original words and state it in a sentence with units.

Unit 2: Analytic Geometry

Midpoint and Length of a Line Segment

Midpoint of $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ : $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$ , the average of the coordinates.
Length of a segment: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2},$ the Pythagorean theorem with the run and the rise as legs.
Horizontal or vertical segment: the length is the difference of the coordinates that change (larger minus smaller).
To find a missing endpoint, repeat the move from the known endpoint to the midpoint: $B = (2 x_{M} - x_{A}, 2 y_{M} - y_{A})$ .
Give exact lengths as simplified radicals, such as $52 = 213$ , and round only the final decimal.

Equation of a Circle

A circle with centre $(0, 0)$ and radius $r$ has equation $x^{2} + y^{2} = r^{2}$ ; it comes from the length formula $x^{2} + y^{2} = r$ .
The radius is the positive square root of the constant: $x^{2} + y^{2} = 36$ has $r = 6$ ; $x^{2} + y^{2} = 20$ has $r = 25 \approx 4.5$ .
A circle centred at the origin through $(a, b)$ has $r^{2} = a^{2} + b^{2}$ .
Point test: $x^{2} + y^{2} < r^{2}$ inside, $= r^{2}$ on, $> r^{2}$ outside.
The intercepts of $x^{2} + y^{2} = r^{2}$ are $(\pm r, 0)$ and $(0, \pm r)$ .

Slopes, Perpendicular Bisectors and Distance to a Line

Slope: $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ (undefined for a vertical line, where $x_{2} = x_{1}$ ).
Non-vertical parallel lines have equal slopes. Perpendicular lines that are neither horizontal nor vertical have negative reciprocal slopes, whose product is $- 1$ .
The perpendicular bisector of $A B$ passes through the midpoint of $A B$ with the negative reciprocal of the slope of $A B$ ; every point on it is equidistant from $A$ and $B$ .
The distance from a point to a line is measured along the perpendicular: build the perpendicular, find the foot by solving a system, then use the length formula.
To find $b$ in $y = m x + b$ , substitute a known point and the slope, then solve.

Verifying Properties of Triangles and Quadrilaterals

Parallel: equal slopes. Perpendicular: slopes with product $- 1$ (or a horizontal side with a vertical side). Equal sides: equal lengths. Bisecting: the same midpoint.
Midsegment theorem: the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
The diagonals of a parallelogram bisect each other; in a rectangle they are also equal, and in a rhombus they are also perpendicular.
A median joins a vertex to the midpoint of the opposite side; the three medians meet at the centroid, $(\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3}) .$
If the squares of the two shorter sides add to the square of the longest side, the triangle is right-angled.

Unit 3: Quadratic Expressions

Expanding and Simplifying

Distributive property: $a (b + c) = ab + a c$ . Every term outside multiplies every term inside.
Binomial times binomial gives four products before collecting like terms: $(a + b) (c + d) = a c + a d + b c + b d$ .
$(x + m) (x + n)$ expands to $x^{2} + (m + n) x + mn$ : the middle coefficient is the sum and the constant is the product.
Squares of binomials: $(a + b)^{2} = a^{2} + 2 ab + b^{2}$ and $(a - b)^{2} = a^{2} - 2 ab + b^{2}$ .
Product of a sum and a difference: $(a + b) (a - b) = a^{2} - b^{2}$ .
Check an expansion by substituting the same value of $x$ into the original and the answer.

Common Factors and Grouping

GCF of monomials: the GCF of the coefficients times each shared variable to its lowest power.
Common factoring: $ab + a c = a (b + c)$ . Divide every term by the GCF; a term equal to the GCF leaves 1.
A common binomial factor comes out like a single term: $3 x (x - 2) + 5 (x - 2) = (x - 2) (3 x + 5) .$
Grouping: $a x + a y + b x + b y = a (x + y) + b (x + y) = (x + y) (a + b) .$
Opposite brackets: $b - a = - (a - b)$ .
Always check a factorization by expanding it back.

Factoring Trinomials

$x^{2} + b x + c = (x + m) (x + n)$ when $m + n = b$ and $mn = c$ .
Positive $c$ : same signs (the sign of $b$ ). Negative $c$ : opposite signs, and the number farther from zero has the sign of $b$ .
Decomposition for $a x^{2} + b x + c$ : find two numbers with product $a c$ and sum $b$ , split the middle term, then factor by grouping.
Take out any common factor first, including $- 1$ when the $x^{2}$ term is negative.
If no integer pair has the right product and sum, the trinomial does not factor using integers.
Check every factorization by expanding.

Perfect Squares and Difference of Squares

Difference of squares: $a^{2} - b^{2} = (a - b) (a + b)$ .
Perfect square trinomials: $a^{2} + 2 ab + b^{2} = (a + b)^{2}$ and $a^{2} - 2 ab + b^{2} = (a - b)^{2}$ .
Perfect square test: first and last terms are squares, the last term is positive, and the middle term is $\pm 2 ab$ .
A second-degree sum of squares such as $x^{2} + 25$ has no binomial factors using real numbers.
Factoring strategy: common factor first, count the terms, factor again where possible, then check by expanding.

Unit 4: Quadratic Relations

Quadratic Relations and Their Properties

A quadratic relation can be written as $y = a x^{2} + b x + c$ with $a \neq = 0$ . Its graph is a parabola.
With equal steps in $x$ : constant first differences mean linear; constant, non-zero second differences mean quadratic.
When $x$ goes up by 1 each step, the second difference is $2 a$ . Positive opens up (minimum); negative opens down (maximum).
The axis of symmetry $x = h$ passes through the vertex and lies halfway between any two points with the same $y$ -value, such as the two zeros.
The maximum or minimum value is the $y$ -coordinate of the vertex. A parabola has 0, 1 or 2 zeros; $y = a x^{2} + b x + c$ has $y$ -intercept $c$ .

Transformations and Vertex Form

$y = a (x - h)^{2} + k$ has vertex $(h, k)$ and axis of symmetry $x = h$ .
$k$ moves the graph up ( $k > 0$ ) or down ( $k < 0$ ). $h$ moves it right ( $h > 0$ ) or left ( $h < 0$ ): $y = (x + 3)^{2}$ has $h = - 3$ .
$a < 0$ reflects in the x-axis (opens down). $a$ beyond $1$ or $- 1$ stretches vertically; $a$ between $- 1$ and $1$ (not 0) compresses vertically.
From the vertex, the step pattern 1, 3, 5 of $y = x^{2}$ becomes $a$ , $3 a$ , $5 a$ .
Equation from a graph: read $(h, k)$ , substitute one other point into $y = a (x - h)^{2} + k$ , and solve for $a$ .

Factored Form and Zeros

Zero product property: if $A \times B = 0$ , then $A = 0$ or $B = 0$ .
$y = a (x - r) (x - s)$ has zeros $x = r$ and $x = s$ ; the factor $(x + 2)$ gives the zero $x = - 2$ .
The axis of symmetry is halfway between the zeros: $x = \frac{r + s}{2}$ . Substitute it to get the vertex.
The $y$ -intercept of $y = a (x - r) (x - s)$ is $a \cdot r \cdot s$ .
To find $a$ from the zeros, substitute one more point into $y = a (x - r) (x - s)$ and solve.

Completing the Square

$(x + d)^{2} = x^{2} + 2 d x + d^{2}$ : the constant is the square of half the coefficient of $x$ .
$x^{2} + b x = (x + \frac{b}{2})^{2} - (\frac{b}{2})^{2}$ . Add and subtract the same number, so the relation is unchanged.
If $a \neq = 1$ , factor $a$ from the $x^{2}$ and $x$ terms only. When the subtracted square leaves the brackets, multiply it by $a$ .
Partial factoring: $y = x (a x + b) + c$ equals $c$ at $x = 0$ and $x = - \frac{b}{a}$ , so the axis is halfway between them.
Check every completed square by expanding it back to standard form.

Comparing y = x squared and y = 2 to the x

$y = x^{2}$ has constant second differences; $y = 2^{x}$ has a constant ratio: each $y$ -value is double the one before.
$y = 2^{x}$ has $y$ -intercept 1, no zeros, is always increasing, and has the x-axis as a horizontal asymptote.
$y = x^{2}$ and $y = 2^{x}$ meet at $(2, 4)$ and $(4, 16)$ and once between $x = - 1$ and $x = 0$ ; for $x > 4$ , $2^{x} > x^{2}$ .
$a^{0} = 1$ for any $a \neq = 0$ .
$a^{- n} = \frac{1}{a ^{n}}$ for any $a \neq = 0$ : a negative exponent gives a reciprocal, not a negative number.

Unit 5: Quadratic Equations

Solving Quadratic Equations by Factoring

Zero product property: if $A \times B = 0$ , then $A = 0$ or $B = 0$ . It works only when the product equals 0.
Standard form of a quadratic equation: $a x^{2} + b x + c = 0$ with $a \neq = 0$ . Rearrange to this form before factoring.
The roots of $a x^{2} + b x + c = 0$ are the zeros ( $x$ -intercepts) of $y = a x^{2} + b x + c$ .
If $a (x - r) (x - s) = 0$ with $a \neq = 0$ , the roots are $x = r$ and $x = s$ . A factor $(x + 2)$ gives the root $- 2$ .
A perfect square such as $(x - 3)^{2} = 0$ has one distinct root (a double root), where the vertex touches the $x$ -axis.
Verify each root by substituting it into the original equation; in context, reject roots that make no sense.

The Quadratic Formula

Quadratic formula: for $a x^{2} + b x + c = 0$ with $a \neq = 0$ , $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ .
The formula comes from completing the square on $a x^{2} + b x + c = 0$ .
The equation must be in standard form before you read off $a$ , $b$ and $c$ , with their signs.
The fraction bar is under the whole numerator $- b \pm b^{2} - 4 a c$ .
The roots are equally spaced on either side of $x = - \frac{b}{2 a}$ , the axis of symmetry.
If $a$ , $b$ , $c$ are integers and $b^{2} - 4 a c$ is a perfect square, the roots are rational and the equation can be factored.

Real and Non-Real Roots

The discriminant of $a x^{2} + b x + c = 0$ is $b^{2} - 4 a c$ , the expression under the square root in the quadratic formula.
$b^{2} - 4 a c > 0$ : two distinct real roots; the parabola has two $x$ -intercepts.
$b^{2} - 4 a c = 0$ : one real root, $x = - \frac{b}{2 a}$ ; the vertex is on the $x$ -axis.
$b^{2} - 4 a c < 0$ : no real roots; the parabola does not meet the $x$ -axis. (The roots are non-real.)
From $y = a (x - h)^{2} + k$ : if $a$ and $k$ have opposite signs there are 2 zeros, if they have the same sign there are none, and if $k = 0$ there is 1.
With integer $a$ , $b$ , $c$ : a perfect-square discriminant means rational roots, so the equation factors.

Solving Problems with Quadratics

Vertical motion: $h = - 4.9 t^{2} + v_{0} t + h_{0}$ ( $h$ in metres, $t$ in seconds, air resistance ignored).
The maximum or minimum value is the second coordinate of the vertex; it occurs at $x = - \frac{b}{2 a}$ , which is halfway between the zeros when the parabola has two.
"When does it reach height $H$ ?" means solve $h = H$ . Reject roots outside the domain, such as negative times.
For a parabola that opens down, the height is above $H$ between the two roots of $h = H$ .
For maximum area or revenue, write the quantity as a product of two factors, find the zeros, and go halfway between them.

Unit 6: Similar Triangles and Right-Triangle Trigonometry

Similar Triangles

Similar triangles: corresponding angles are equal and corresponding sides are proportional. In $△ A BC \sim △ D EF$ the vertices are listed in matching order.
Scale factor: $k = \frac{side in the new triangle}{matching side in the original} .$ Every side is multiplied by $k$ ; $k = 1$ means the triangles are congruent.
Tests for similarity: AA (two pairs of equal angles), SSS similarity (all three side ratios equal), SAS similarity (two side ratios equal and the angles between those sides equal).
Congruent triangles are similar with scale factor 1. Similar triangles are congruent only when the scale factor is 1.
To find a missing side, match sides by the angles they are opposite, write a proportion with the ratios in the same order, and solve.
If lengths are scaled by $k$ , perimeters are scaled by $k$ and areas by $k^{2}$ .

The Primary Trigonometric Ratios

The hypotenuse is the side across from the right angle and is the longest side. Opposite and adjacent are named from the reference angle and swap when the reference angle changes.
$sin θ = \frac{opp}{hyp}$ , $cos θ = \frac{adj}{hyp}$ , $tan θ = \frac{opp}{adj}$ (SOH CAH TOA).
The ratios depend only on the angle, because all right triangles with the same acute angle are similar (AA).
Finding a side: unknown on top, multiply (e.g. $x = 15.0 sin 2 8^{\circ}$ ); unknown on the bottom, divide (e.g. $x = \frac{8.0}{cos 5 2 ^{\circ}}$ ).
Finding an angle: $θ = sin^{- 1} (\frac{opp}{hyp})$ , $cos^{- 1} (\frac{adj}{hyp})$ or $tan^{- 1} (\frac{opp}{adj})$ . The two acute angles add to $9 0^{\circ}$ .
Use degree mode (check: $sin 3 0^{\circ} = 0.5$ ) and keep unrounded values until the final answer.

Solving Right-Triangle Problems

Angle of elevation: measured up from the horizontal to the line of sight. Angle of depression: measured down from the horizontal to the line of sight.
The angle of depression from P to Q equals the angle of elevation from Q to P (alternate angles between parallel horizontal lines).
In a right triangle: two sides give the third by $a^{2} + b^{2} = c^{2}$ ; a side and an acute angle give another side by SOH CAH TOA; two sides give an angle by an inverse ratio; the acute angles add to $9 0^{\circ}$ .
When an angle is measured at eye level, add the eye height to the height found from the triangle.
In two-triangle problems, solve the triangle with enough information first, find the shared side, and carry it unrounded into the second triangle. If neither triangle can be solved alone, write an equation from each and solve the system.

Unit 7: Trigonometry of Acute Triangles

The Sine Law

In $△ A BC$ , side $a$ is opposite $∠ A$ , side $b$ is opposite $∠ B$ , side $c$ is opposite $∠ C$ .
Sine law: $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$ , or flipped, $\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ .
It comes from writing the altitude two ways, $h = b sin A = a sin B$ , and dividing by $sin A sin B$ .
Use it when you know an angle and its opposite side plus one more angle or side (AAS, ASA after the angle sum, or two sides and an angle opposite one of them).
To find an angle, solve for its sine, then use $sin^{- 1}$ in degree mode. In an acute triangle the result is the angle you want; the two-triangle ambiguous case is Grade 11.
The three angles add to $18 0^{\circ}$ : use this to find a third angle before or after the sine law.

The Cosine Law

Cosine law: $a^{2} = b^{2} + c^{2} - 2 b c cos A$ (and similarly $b^{2} = a^{2} + c^{2} - 2 a c cos B$ , $c^{2} = a^{2} + b^{2} - 2 ab cos C$ ).
For an angle: $cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$ , where $a$ is the side opposite $∠ A$ .
Use it for SAS (find the third side) and SSS (find an angle): the cases with no complete opposite pair.
It is developed from an altitude using the Pythagorean theorem and $x = b cos A$ ; when $∠ A = 9 0^{\circ}$ , $cos A = 0$ and it becomes $a^{2} = b^{2} + c^{2}$ .
If the longest side squared is less than the sum of the squares of the other two, the triangle is acute.

Choosing a Method and Solving Problems

Right angle: SOH CAH TOA or the Pythagorean theorem. No right angle: the sine law or the cosine law.
An angle and its opposite side known: sine law (AAS, ASA after the angle sum, SSA).
Two sides and the angle between them (SAS) or three sides (SSS): cosine law.
Multi-step problems: solve one triangle, then use its answer as a given in the next step or the next triangle.
Check: angles add to $18 0^{\circ}$ , the longest side faces the largest angle, and sides to one decimal place and angles to the nearest degree are rounded only at the end.