Is there a reason why the answer isn't very related to 40?
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Module 2 Week 3 Day 10 Your Turn
Like why wasn't it 2x or 3x of 40? I don't really get why still.
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@TSS-Graviser No problem! I'm going to guess that you are asking about the Your Turn question. There were quite a few steps in this problem; a lot of puzzle pieces to fit together, but it's really neat how they each play a part in getting the answer.
We're going to collect a bunch of facts that all relate to the given angle, \(40^{\circ},\) and then use them to solve for the value of the \(\textcolor{orange}{\text{yellow}}\) angle marked with a "?." This is similar to how you might solve an algebra problem by collecting enough equations in order to solve for a number of variables.
Since the \(\textcolor{orange}{\text{yellow}}\) angle is inscribed in the \(\textcolor{orange}{\text{yellow}}\) arc, we just need to find the measure of the \(\textcolor{orange}{\text{yellow}}\) arc.
If we complete the circle by drawing a \(\textcolor{purple}{\text{purple}}\) arc like so, then we know that together these two arcs add up to a complete circle, or \(360^{\circ}.\)
Wait a second; how does this relate to the angle that we are given, \(40^{\circ}?\) Well, the trick is to look for angles that inscribe the \(\textcolor{orange}{\text{yellow}}\) arc and the \(\textcolor{purple}{\text{purple}}\) arc. Let's draw the line that would make this possible, and label the corresponding angles with the matching color of the arc that it inscribes. Now, pulling a fact that we learned from before, we know that the measure of an inscribed angle is half of the measure of the arc that it inscribes, so, labeling our \(\textcolor{orange}{\text{yellow}}\) arc as \(\textcolor{orange}{\text{a}},\) its inscribed angle is \(\frac{a}{2}.\) In the same way, labeling our \(\textcolor{purple}{\text{purple}}\) arc as \(\textcolor{purple}{\text{b}},\) its inscribed angle is \(\frac{b}{2}.\)
Now we get a nice triangle with two known angles and an exterior angle. What could we use here now? The Exterior Angle Theorem!
This is great, because we have another relation between \(a, b,\) and \(40^{\circ}.\) Visually, below, I've labeled the angle \(40^{\circ}\) as the difference of the other two angles. Since the third angle of the triangle is collinear with the \(\textcolor{orange}{\text{yellow}}\) angle, I've labeled the third angle of the triangle as \(180^{\circ} - \frac{a}{2}.\)
You can check for yourself that the angles of this triangle add up to \(180^{\circ}.\)
At this point, our question turns into an algebra problem, and we don't have to worry about the geometry part anymore. We simply have two equations, with two variables:
$$ \frac{a}{2} = \frac{b}{2} + 40^{\circ} $$
$$ a + b = 360^{\circ} $$
Recall that we wanted to find the value of the inscribed \(\textcolor{orange}{\text{yellow}}\) angle, which was \(\frac{a}{2}.\)
We can do this by eliminating \(b\) through substitution of \(b = 360^{\circ} - a,\) which gives us
$$\begin{aligned} \frac{a}{2} &= \frac{360^{\circ} - a}{2} + 40^{\circ}\\\\ \frac{a}{2} &= \frac{360^{\circ}}{2} - \frac{a}{2} + 40^{\circ}\\\\ \frac{a}{2} + \frac{a}{2} &= \frac{360^{\circ}}{2} - \frac{a}{2} + \frac{a}{2} + 40^{\circ} \\\\ a &= \frac{360^{\circ}}{2} + 40^{\circ} \\\\ a &= 180^{\circ} + 40^{\circ} \\\\ a &= 220^{\circ} \\\\ \end{aligned} $$
Don't forget that we wanted the inscribed angle of the arc \(\textcolor{orange}{\text{a}},\) so we really want half of the value of arc \(\textcolor{orange}{\text{a}}.\)
Thus
$$ \angle \text{?} = \frac{220^{\circ}}{2} = \boxed{110^{\circ}}. $$
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Wow how do you solve and explain how to solve all these questions
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@TSS-Graviser I didn't learn this when I was your age; I learned it as an adult. So, you have a head start! After spending enough time, you can learn anything!
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Well, it is kind of related. If we take the general example, calling the 40 degree angle x, then the answer is (180 + x)/2, or 110.
