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Practice questions for SohCahToa

Grade 10 Math: Master Essential Skills & Concepts

1.1 Investigating the Basic Properties of Quadratic Relations

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Characteristics of quadratic functions

Transformations of quadratic functions

Quadratic function in general form: y = ax^2 + bx + c

Converting from general to vertex form by completing the square

Shortcut: Vertex formula

Converting general form into vertex form by applying the vertex formula

Vertex Formula Application

Graphing quadratic functions: General form VS. Vertex form

Finding the quadratic functions for given parabolas

Applications of quadratic functions

2.1 Relating the Graph of y = x^2 and Its Transformations

Combining transformations of functions

Transformations of functions: Horizontal translations

Transformations of functions: Vertical translations

Reflection across the y-axis: y = f(-x)

Reflection across the x-axis: y = -f(x)

Transformations of functions: Horizontal stretches

Transformations of functions: Vertical stretches

Quadratic function in vertex form: y = a(x-p)^2 + q

3.1 Solving Quadratic Equations

Factor by taking out the greatest common factor

Factor by taking out GCF

Factor by grouping

Factoring difference of squares: \(x^2 - y^2\)

Factoring difference of squares: <katex>x^2 - y^2</katex>

Factoring difference of squares: x^2 - y^2

Factoring trinomials

Factoring Trinomials

Solving quadratic equations by factoring

Solving quadratic equations by completing the square

Using quadratic formula to solve quadratic equations

Solving a quadratic equation with TWO REAL SOLUTIONS

Solving a quadratic equation with TWO COMPLEX SOLUTIONS

Nature of roots of quadratic equations: The discriminant

Applications of quadratic equations

Completing the square

Completing the Square

1.1 Using Linear Systems to Solve Problems

Determining number of solutions to linear equations

Solving systems of linear equations by graphing

Solving systems of linear equations by elimination

Solving systems of linear equations by substitution

System of linear equations

System of linear-quadratic equations

System of quadratic-quadratic equations

2.1 Solving Problems Involving Properties of Line Segments

Distance formula: \(d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

Distance formula: <katex>d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}</katex>

Distance formula

Midpoint formula: \(M = ( \frac{x_1+x_2}2  ,\frac{y_1+y_2}2)\)

Midpoint formula

Midpoint formula: <katex>M = ( \frac{x_1+x_2}2 ,\frac{y_1+y_2}2)</katex>

Midpoint formula: <katex>M = ( \frac{x_1+x_2}2 ,\frac{y_1+y_2}2)</katex>

Slope equation: \(m = \frac{y_2-y_1}{x_2- x_1}\)

Slope equation: <katex>m = \frac{y_2-y_1}{x_2- x_1}</katex>

Slope intercept form:  y = mx + b

Slope intercept form: y = mx + b

General form:  Ax  + By  + C = 0

General form: Ax + By + C = 0

Point-slope form: y - y_1 = m(x - x_1)

Graphing linear functions using table of values

Graphing linear functions using x- and y-intercepts

Graphing from slope-intercept form y=mx+b

Graphing linear functions using a single point and slope

Word problems of graphing linear functions

3.1 Using Analytic Geometry to Verify Geometric Properties

Parallel line equation

Perpendicular line equation

Combination of both parallel and perpendicular line equations

Parallel and perpendicular lines in linear functions

1.1 Investigating Similarity and Solving Problems Involving Similar Triangles

Similar triangles

Similar polygons

Isosceles and equilateral triangles

2.1 Solving Problems Involving the Trigonometry of Right Triangles

Use sine ratio to calculate angles and sides (Sin = o / h)

Use cosine ratio to calculate angles and sides (Cos = a / h)

Combination of SohCahToa questions

Word problems relating ladder in trigonometry

Word problems relating guy wire in trigonometry

Solving expressions using 45-45-90 special right triangles

Solving expressions using 30-60-90 special right triangles

Use tangent ratio to calculate angles and sides (Tan = o / a)

3.1 Solving Problems Involving the Trigonometry of Acute Triangles

Law of sines

Ambiguous case: SSA triangles

Law of cosines

Applications of the sine law and cosine law

Get better math marks with our complete Grade 10 math help &ndash; be it <a href="http://www.edu.gov.on.ca/eng/curriculum/secondary/grades.html">Ontario math curriculum</a>, math curricula of <a href="https://curriculum.gov.bc.ca/curriculum/10-12#m">B.C.</a>, <a href="https://education.alberta.ca/mathematics-10-12/programs-of-study/">Alberta</a>, <a href="http://www.edu.gov.mb.ca/k12/cur/math/index.html">Manitoba</a>, <a href="https://www.curriculum.gov.sk.ca/webapps/moe-curriculum-BBLEARN/index.jsp?lang=en&subj=mathematics&level=foundations10">Saskatchewan</a>, <a href="http://www.ed.gov.nl.ca/edu/k12/curriculum/guides/mathematics/">Newfoundland</a>, or other provinces. We&apos;ve got you all covered!

Just like your class or textbook, our complete help for grade 10 math includes topics such as Trigonometry, Quadratic Equations, Radicals, Factoring polynomials, Linear equations, Linear systems, and more. Learn the concepts with our video tutorials that show you step-by-step solutions to even the hardest grade 10 math problems. Then, strengthen your understanding with tons of grade 10 math practice. 

All our lessons are taught by experienced grade 10 math teachers. Let&apos;s finish your homework in no time, and ACE that final.

Principles of Mathematics, Grade 10, Academic (MPM2D)

1. Quadratic Relations of the Form y = ax^2 + bx + c

2. Analytic Geometry

3. Trigonometry

math

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Ex. 1

- Hi! Welcome to this question right here. So this one here you can use whatever you like. Sohcahtoa, you can use sin, cos, tangent as long as you can find the length from b to c, which is from here to here. Okay? 
So how do we know the length from there to there? We only know the angle over here is 22 and this angle here is 15. But if you look in a different way you can actually see there's one right triangle right here. So you can actually find the length from a to b because you kind of know this side's 80 and if you know this angle right here, maybe you can find the length a to b. Then if you look at a bigger scale, if I can just go ahead and treat this big triangle right here, because I know this whole thing right here is 22 plus 15, I can also figure out the a to c. Then I can take the a to c, subtract the a to b and I'll get the b to c. Okay? Kind of weird. Let's actually do it step by step. So let's actually find the length a to b first. So here is my length a to b. So here we go. To find the length a to b, you have to visualize your triangle as actually just this part right here. So you're trying to find the opposite side from this angle 22. This is the opposite, this is the adjacent. So you're going to use tangent. So for the length a to b, I will be using tangent of 22 degrees and the opposite will be the one I'm looking for which is the length a to b divided by the 80 degrees which is the adjacent which is over here. So I'm going to go ahead and calculate that. I've got tangent 22 times 80. Tangent 22 times 80. I'm just gonna write down 32.3. And let's call that centimeters for now. So what about a to c? If you're doing a to c, then you have to look at the triangle like this. Look at it like this. These two angles right here is just 22 plus 15. Twenty-two plus 15, we get 37, right. So from this angle you're looking for the opposite side and you have the adjacent 80 again. So it will just be tangent once again. Using the tangent 37, using these two triangles together is equal to the length of a to c divided by 80 again. So I'm going to go tangent 37 and I'm going to multiply it by 80. A little bit of cross multiplying. I get myself a 60.3 centimeters. So if I know from here to here it's 63 centimeters. So from here to here the blue line is 63 in total and the a to b over here is just 32 roughly, then what will be my b to c? 
Well, just subtract the difference and that's it, right? 
So it just goes 63.2 minus 32.2. . . are they all 0.3s? My bad. All 0.3s. So three and threes are gone. So 63.3-- oh, it's actually 60. I'm really sorry all those mistakes right here. So this is actually 60.3, okay. So, 60.3 minus 32.3 I get myself 28, okay. Sorry about all those scribbles. The final answer should be just 28 centimeters if that was centimeters. Okay? Thanks for watching!

Ex. 2

SOHCAHTOA is actually a memory aid to memorize how to calculate sine, cosine and <a href="[[glossary.html|{"keyword":"Tangent (in Trigonometry)"}]]">tangent</a> of an angle. SOH: sine = opposite over hypotenuse; CAH = adjacent over hypotenuse; and TOA = opposite over adjacent. Besides using them separately, the sine, cosine, and tangent ratios can be used together to solve trigonometry problems.

Use sine ratio to calculate angles and sides (Sin = \(  \frac{o}{h}\) )

Use cosine ratio to calculate angles and sides (Cos =  \(  \frac{a}{h}\)  )

Use tangent ratio to calculate angles and sides (Tan =  \(  \frac{o}{a}\)  )

ASTC rule in trigonometry (All Students Take Calculus)