The ratio of an isosceles triangle to its altitude is 3:4. find the measures of the angles of triangle? So, I have the altitude formula where $h$= $\sqrt s^2-(\frac{s}{2})^2$ so I was thinking when I have the altitude the triangle will be into $2$ right triangles and I think I need to get the sides first then derive the formula for Area which is $A$=$\frac{1}{2}$ $absin$ but I'm confuse on how to get the sides if it only says base is $3:4$ to its altitude
[Math] Ratio of an isosceles triangle to its altitude is 3:4. find the measures of the angles
solid-geometry
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Alright, I've come up with a proof in what I think is the right flavor.
Take a sphere with radius $r$, and consider the upper hemisphere. For each $n$, we will construct a solid out of stacks of pyramidal frustums with regular $n$-gon bases. The stack will be formed by placing $n$ of the $n$-gons perpendicular to the vertical axis of symmetry of the sphere, centered on this axis, inscribed in the appropriate circular slice of the sphere, at the heights $\frac{0}{n}r, \frac{1}{n}r, \ldots,\frac{n-1}{n}r $ . Fixing some $n$, we denote by $r_\ell$ the radius of the circle which the regular $n$-gon is inscribed in at height $\frac{\ell}{n}r$ . Geometric considerations yield $r_\ell = \frac{r}{n}\sqrt{n^2-\ell^2}$ .
As noted in the question, the area of this polygonal base will be $\frac{n}{2}r_\ell^2 \sin\frac{2\pi}{n}$ for each $\ell$ . I am not sure why (formally speaking) it is reasonable to assume, but it appears visually (and appealing to the 2D case) that the sum of the volumes of these frustums should approach the volume of the hemisphere.
So, for each $\ell = 1,2,\ldots,n-1$, the term $V_\ell$ we seek is $\frac{1}{3}B_1 h_1 - \frac{1}{3}B_2 h_2 $, the volume of some pyramid minus its top. Using similarity of triangles and everything introduced above, we can deduce that $$ B_1 = \frac{n}{2}r_{\ell-1}^2 \sin\frac{2\pi}{n}~,~B_2 = \frac{n}{2}r_\ell^2 \sin\frac{2\pi}{n} ~,~h_1 = \frac{r}{n}\frac{r_{\ell-1}}{r_{\ell-1}-r_{\ell}}~,~h_2=\frac{r}{n}\frac{r_{\ell}}{r_{\ell-1}-r_{\ell}} ~~. $$ So, our expression for $V_\ell$ is $$ \frac{r}{6} \sin\frac{2\pi}{n} \left\{ \frac{r_{\ell-1}^3}{r_{\ell-1}-r_{\ell}} - \frac{r_{\ell}^3}{r_{\ell-1}-r_{\ell}} \right\} = \frac{\pi r}{3n} \frac{\sin\frac{2\pi}{n}}{2\pi/n} \left\{ r_{\ell-1}^2 + r_\ell^2 + r_{\ell-1}r_\ell \right\} $$ $$ = \frac{\pi r^3}{3n^3} \frac{\sin\frac{2\pi}{n}}{2\pi/n} \left\{ (n^2 - (\ell-1)^2) + (n^2-\ell^2) + \sqrt{(n^2-\ell^2)(n^2-(\ell-1)^2)} \right\} ~~. $$ So, we consider $ \lim\limits_{n\to\infty} \sum_{\ell=1}^{n-1} V_\ell$ . The second factor involving sine goes to 1, and we notice that each of the three terms in the sum is quadratic in $\ell$, and so the sum over them should intuitively have magnitude $n^3$. Hence, we pass the $\frac{1}{n^3}$ into the sum and evaluate each sum and limit individually, obtaining 2/3, 2/3, and 2/3 respectively (the first two are straightforward, while the third comes from the analysis in this answer).
Thus, we arrive at $\frac{\pi r^3}{3} (2/3+2/3+2/3) = \frac{2}{3}\pi r^3$ as the volume of a hemisphere, as desired.
So was this too excessive or perhaps worth it? I'll leave that to all of you. :)
What you need to look at is the second theorem of Pappus. When you rotate things in a circle, you have to account for the difference in the distance traveled by the point furthest from the axis (the bottom corner of our triangle) which goes the full $2\pi$ around. However, the vertical side doesn't actually move, so it doesn't get the full $2\pi$.
Pappus' theorem says you can use the radius of the centroid as your revolution radius, and it just so happens the the centroid of this triangle is a distance $r/3$ from the axis.
Another good read is this.
Best Answer
Notice you need only the angle (which is a ratio). We have that the ratio of the base to its altitude is $3:4$. Then the ration of half the base to its altitude is $3:8$. This ratio is the cotangent of the angle at the base. If we have the cotangent we have the angle.