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I have a book that covers more advanced trigonometry, unfortunately it does it in a rather terrible way. I have two problems that make me think that I am forgetting some sort of property/rule because I can't solve them quickly. The speed at which I can answer these is crucial.

I realise I have this formatted like it is homework, but it is just because I have two problems that I think use the same principal, hopefully this will help you narrow down what the principal is rather than brute force solving.

Problem 1:

What are all the values of side a in the figure below such that two triangles can be constructed. enter image description here For some reason the answer is a range. 4sqrt(3)

Problem 2: Given the following data which can form two triangles

I.   Angle C = 30 degrees, c = 8, b = 12
II.  Angle B = 45 degrees, a = 12sqrt(2), b = 15sqrt(2)
III. Angle C = 60 degrees, b = 12, c = 5sqrt(3)

The answer to that one is only I can make two triangles, II and III can only make one triangle.

From what I can tell both of these can be solved with some concept called an "altitude to base"

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  • $\begingroup$ Two triangles are congruent if two sides and the angle opposite to the LARGER site coincide. If it is the angle opposite the smaller site, this is not necessarily true. $\endgroup$
    – Peter
    Oct 5 '16 at 20:49
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If $a$ is too short, it won't reach the other side. It will just barely reach the other side if it forms a right angle with it. So $a$ has to be at least $8 \sin 60^\circ = 4\sqrt{3}.$ (In this case $a$ is an altitude of the triangle to the base which is the far side.) If $a$ is longer, then we can imagine the arc drawn by swinging it back and forth. It intersects the far side in two places. Well, if it gets too long then it won't intersect on the left side of your picture, but it will on the right. That is, if it's longer than 8, there is only one solution. So $4\sqrt{3} < a < 8$.

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The law of sines gives for $I:$

$\sin(B)=\sin(C)\cdot \frac{b}{c}=0.75$

The larger angle, $131,4°$ , is still small enough to allow a triangle. So, we can construct two triangles.

$II:$

$sin(A)=sin(B)\cdot \frac{a}{b}=0.56568\cdots$

Here, the larger angle leads to a sum of angles greater than $180°$. We only have one triangle.

$III:$

$sin(B)=sin(C)\frac{b}{c}=1.2$

So, we cannot construct a triangle with these pieces.

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So in a triangle there are 180° right and with one angle being 60° the other angle has to be less then 120° to form a triangle. With knowing the maximum angle, we can use the law of sines which gives the maximum length as 8. Now the shortest possible length would be when the angle is 90°, giving the solutions mentioned before.

Now to answer your second question, you would use sine rules to find the missing angles. If the sum of the angle given and the one you find less than 180° then it forms a triangle. So the first one the two possible angles found when the sine rule is used are 48 and 132. 48+30<180 and 132+30<180, so two possible triangles. When the same process is repeated, it can be said that the second one only produces one triangle and the last one zero.

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