The following image shows a problem out of my math book, Trigonometry, by Michael Corral (problem 13, section 1.3 for those who may be interested).


My analysis is that the $1.5$ (as far as I can tell) is an adjacent or opposite side, where as the hypotenuse is unknown. I can't see a use for $54^\circ$. However, judging from the dashed-split as shown, I'm thinking that's a hint suggesting the $54^\circ$ be divided in half. Still, even so I have yet to actually find and see where that potential $27^\circ$ should be used.

I've tried a number of things, from dividing the $2\frac18$ in half and using that to form a triangle with $1\frac12$, to using both whole fractions as their values and then setting the hypotenuse to be paralell with the radius, starting at the bottom tangent vertex. Still, no dice so far.

Can someone point me in the right direction in terms of how to solve this?


Let $x$ be the horizontal distance between the point at which the tangents meet to the right of the rounded tip, and the points at which the tangents meet the horizontal sides of the punch. Let $h=\frac12 \times 2 \frac18"$ be the half-height of the die punch, and $d = 1 \frac12 "$ be the length of the die punch. Then

$$x = h \tan{63^{\circ}}$$

Also let $y$ be the distance from the circular edge of the punch to the point at which the tangents meet, and $r$ be the radius of the punch. Then

$$y = x-d = h \tan{63^{\circ}}-d$$


$$\cos{63^{\circ}} = \frac{r}{y+r}$$

Solving for $r$, I get

$$r = \frac{h \sin{63^{\circ}} - d \cos{63^{\circ}}}{1-\cos{63^{\circ}}}$$

or $r \approx 0.487"$.

  • $\begingroup$ Would you mind explaining your process behind coming up with the solution? For instance, where does the $cos63^\circ$ come from? I assume the $63^\circ$ is the compliment of the $27^\circ$. Also, how did you come up with the first equation with the cosine and the radius? Regardless, thanks for the post. $\endgroup$ – about blank Jan 20 '14 at 17:42
  • $\begingroup$ Nevermind, I see what you were talking about. Thanks again! $\endgroup$ – about blank Jan 20 '14 at 17:56
  • $\begingroup$ @aboutblank: excellent - cheers! $\endgroup$ – Ron Gordon Jan 20 '14 at 18:29
  • $\begingroup$ sorry: I hate to go back on that, but I am still confused about the $cos63^\circ$. How does the r become the adjacent? Is its triangle a sub triangle within the original triangle denoted by sides $x$ and $h$? I ask this because $r$ seems too small to scale properly with $h$. Also, how can $y+r$ become a hypotenuse when $y$ comes from $x-d$ which are both opposite to the $63^\circ$? I seriously appreciate it! $\endgroup$ – about blank Jan 20 '14 at 18:35
  • 1
    $\begingroup$ Remember that the tangent is perpendicular to the radius at the point of tangency. $\endgroup$ – Ron Gordon Jan 20 '14 at 19:57

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