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Linear Pairs of Angles

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EEVblog:

--- Quote from: janoc on April 01, 2020, 06:11:17 pm ---I am quite surprised that an 8 years old is being asked to solve what amounts to linear equations? Isn't that like second year?
At that level kids can usually barely read and write and are learning multiplication tables! Either Sagan is very smart and is doing some extra math or Australia has some rather interesting math curriculum, because these things are usually taught from 5-6th grade up ...

--- End quote ---

Yes, this is "extension" class material for those who are good at maths.

Nominal Animal:
Sheet 1, B2: Difference in a linear pair of angles is \$42^o\$.

Using a single variable \$x\$ to denote the smaller angle:
$$\begin{array}{lrl}
~ & x + (x + 42^o) &= 180^o \\
\iff & 2 x + 42^o &= 180^o \\
\iff & 2 x &= 180^o - 42^o \\
\iff & x &\displaystyle = \frac{180^o - 42^o}{2} \\
\iff & x &= 69^o \\
\end{array}$$
The larger angle is then \$x + 42^o = 69^o + 42^o = 111^o\$.

Verification: \$69^o + 111^o = 180^o\$, which is true.

Therefore, the answer is \$69^o\$ and \$111^o\$.

Alternate method, using a system of two equations in two variables:
$$\left\lbrace ~ \begin{aligned}
x + y &= 180^o \\
x + 42^o &= y \\
\end{aligned} \right.$$
Solve the lower one for for \$y\$, getting \$y = x + 42^o\$.  Substitute that to the upper, getting
$$x + (42^o + x) = 180^o$$
From this point onwards, the solution is the same as using the single-variable method, and the answer is \$x = 69^o\$ and \$y = 111^o\$.

Nominal Animal:
Sheet 1, B. 3)

Angle \$y\$ is five times angle \$x\$, and they form a linear pair.
$$\left\lbrace ~ \begin{aligned}
y &= 5 x \\
x + y &= 180^o \\
\end{aligned}\right . $$
Solve upper for \$y\$, getting \$y = 5 x\$.  Substitute into lower, getting
$$\begin{array}{lrl}
~ &  x + ( 5 x ) & = 180^o \\
\iff & 6 x & = 180^o \\
\iff & x & \displaystyle = \frac{180^o}{6} \\
\iff & x & = 30^o \\
\end{array}$$
Substituting \$x = 30^o\$ back to \$y = 5 x\$ yields \$y = 5 \times 30^o = 150^o\$.

Verification: \$30^o + 150^o = 180^o\$, which is true.

Therefore, the answer of \$30^o\$ and \$150^o\$ is correct.  The only reason for the mark can be that the teacher wanted "the one angle" (which is five times "the other angle") first, and "the other angle" second.

Side note:

In many languages, it is easy to confuse "y is five times x" and "y is five times more than x", corresponding to \$y = 5 x\$ and \$y = x + 5 x = 6 x\$; i.e. to confuse direct comparison and the comparison of the increase.  In this case, the opposite error -- treating the equation as if \$y = 4 x\$ -- yields an answer of \$36^o\$ and \$144^o\$.  Teachers can make mistakes, too, and when you need them to acknowledge an error, it often helps to understand how/why they made the error in the first place. (And you can then just note that "it is easy to accidentally do Z here, which would be an error, because ...", which is more palatable to fragile-egoed people than "you made an error ...".)

Nominal Animal:
Sheet 2, B. 1) As above, \$144^o\$ and \$36^o\$; correct.

Sheet 2, B. 2) Linear pair of angles, with one angle \$24^o\$ more than the other.

Similar to Sheet 1, B. 2), but the difference is now \$24^o\$ instead of \$42^o\$.  So, we have
$$\left\lbrace ~ \begin{aligned}
y &= x + 24^o \\
x + y &= 180^o \\
\end{aligned} \right.$$
Substituting \$y\$ into the lower gives us
$$\begin{array}{lrl}
~ & x + ( x + 24^o) & = 180^o \\
\iff & 2 x + 24^o & = 180^o \\
\iff & 2 x & = 180^o - 24^o \\
\iff & x & \displaystyle = \frac{180^o - 24^o}{2} \\
\iff & x & = 78^o \\
\end{array}$$
Substituting \$x = 78^o\$ back to \$y = x + 24^o\$ yields \$y = 78^o + 24^o = 102^o\$.

Verification: \$102^o + 78^o = 180^o\$, which is true.

Therefore, the angles are \$102^o\$ and \$78^o\$.

Nominal Animal:
I really don't like homework (or questions in general!) that only asks for the answer; the proper way is to see how the answer is worked out.  The answer itself is irrelevant, the procedure/algorithm is the interesting and important thing.

That's why I wrote the above answers: to show how one should work them out.  No insult/offense intended for the other members here.

The question I anticipate any learner to ask, is "how did you choose which one to solve first in the systems of two equations?".

The answer is, "whichever looks simpler".  Choosing either one (in a set of two equations with two unknowns) will work, and you will get the same answers.  The only thing that varies is the amount of work needed.  If one of the equations already isolates one of the variables (like \$x + 24^o = y\$ does, it gives us \$y = x + 24^o\$), it is always a good one to start.  Just remember to use the other equation for the rest, because if you reuse the same equation, you'll end up with something non-useful like \$0 = 0\$.

(I so wish StackOverflow hadn't gone full SJW retard; I miss doing this at math.stackexchange.com.)

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