In triangle $ABC$, we are given an angle $A = 42°$, and its opposite side length, $a = 38$.
i) For what values of adjacent side $b$ such that we
have one unique triangle?
ii) For what values $b$ do we have $2$ different triangles?
I was drawing a bunch of triangles with different cases where side $b$ varies, and came up with the following, but I'm not sure if this is correct and am wondering if someone more knowledgeable could verify, and/or correct my misunderstandings:
i) Here we can draw an altitude $h$ from $C$ meeting $AB$ at point $X$, making two triangles, $ACX$ and $BCX$. What I noticed is that if $a > b$, then we only get one unique triangle (since if we reflect $BCX$ across $h$, it lies beyond point $A$). Hence the constraint should be $\boldsymbol{b<38}$.
ii) Again if we draw the same altitude $h$ from $C$ meeting $AB$ at point $X$ to make two triangles $ACX$ and $BCX$, I notice that if $a < b$, then when reflecting $BCX$ across $h$, $BX$ will lie between $AX$, creating a second (smaller triangle) that fits the given criteria (however, if $a$ is too small compared to $b$, it will not make any triangle at all, i.e if $a < h$, then it won't reach $AB$). So a must be greater than $h$, but smaller than $b$, so, $h < a < b$. So now I express $h$ in terms of $b$ using right triangle $ACX$: $\sin(42°) = h/b \implies h = b\sin(42°)$. Since $h < a$, substituting we get $b\sin(42°) < 38 \implies b < 38/\sin(42°)$, giving $b < 56.79$ But since $a < b$, we have $b > 38$, so the constraint should be $\boldsymbol{38 < b < 56.79}$
Please let me know if any of this doesn't make sense, or I'm missing anything.
Best Answer
Your solution is basically correct. I think it'd be easier to draw a circle with center in the vertex $C$ and fixed radius = 38 and look at its intersections with the ray $AB$ while the vertex $C$ is moving from the vertex $A$ into infinity.
Also please think about boundary cases - for example, the triangle with maximal $b$ will be unique again.