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For reference:

The measures of the sides of a triangle are $3$, $5$ and $7$. Calculate the measure of the largest angle of a triangle whose sides are the inverses of the heights of the first triangle (Answer:$120^\circ$)

*I checked in geogebra and the answer is correct.

My progress: I tried but I don't know if that's the way to go.

$p = \frac{3+5+7}{2}=\frac{15}{2}\\p(p-a)p(p-b)(p-c)=\frac{15}{2}(\frac{9.}{2}\frac{5}{2}.\frac{1}{2}) = \frac{675}{16}$

Heron's Formula: $ h_b=\frac{2}{7}\sqrt{\frac{675}{16}}=\frac{2.15\sqrt3} {7.4}=\frac{15\sqrt3}{14}$

Similarly:

$h_a = \frac{2}{5}\cdot\frac{15\sqrt3}{4} = \frac{3\sqrt3}{2}\\h_c = \frac{2}{3}\cdot\frac{15\sqrt3}{4}=\frac{5\sqrt3}{2} \\ \frac{1}{h_a}=\frac{2}{3\sqrt3}\implies (\frac{1}{h_c})^2 = \frac{4}{27}\\ \frac{1}{h_b} = \frac{14}{15\sqrt3}\implies (\frac{1}{h_b})^2=\frac{196}{675}\\ \frac{1}{h_c} = \frac{2}{5\sqrt3} \implies (\frac{1}{h_c})^2=\frac{4}{75}$

By Law of Cosines:

$\frac{196}{675} = \frac{4}{27}+\frac{4}{75} - 2\cdot\frac{2}{3\sqrt3}\frac{2}{5\sqrt3}\cdot\cos B\implies\\ \cos B=-0.5\\ \therefore B = 120^\circ$

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    $\begingroup$ You have to square them when you apply cosine law $\endgroup$
    – Math Lover
    Nov 9, 2021 at 12:53
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    $\begingroup$ Hint: Notice that the triangle with sides $\frac{1}{h_a}, \frac{1}{h_b}, \frac{1}{h_c}$ is similar to the triangle with sides $a, b, c$. (Why?) $\endgroup$ Nov 9, 2021 at 12:54
  • $\begingroup$ $c^2 = a^2 + b^2 - 2 ab \cos C$, what you have written is $c = a + b - 2 ab \cos C$ $\endgroup$
    – Math Lover
    Nov 9, 2021 at 12:54
  • $\begingroup$ @MathLover See...$a =( \frac{1}{h_a})^2$ but I had already calculated this value before to make the equation easier $\endgroup$ Nov 9, 2021 at 13:45
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    $\begingroup$ OK in that case, $- 2ab\cos C$ is wrong. I think you have $ - 2 a^2 b^2 \cos C$ $\endgroup$
    – Math Lover
    Nov 9, 2021 at 14:01

1 Answer 1

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Elaborating @Stinking Bishop's hint, consider triangle $ABC$ with sides $a$, $b$ and $c$ (in standard notation). Let altitudes from each vertex to the opposite side be $h_a$, $h_b$ and $h_c$. Now, if the area of the triangle is $A$, we get, $$\frac1{h_a}=\frac{a}{2A}$$ $$\frac1{h_b}=\frac{b}{2A}$$ $$\frac1{h_c}=\frac{c}{2A}$$

It is clear that the triangle formed by the inverses of the heights is similar to the original triangle with similarity ratio $\frac1{2A}$.

Now we can find the required angle considering the above fact.

The angle opposite to the longest side is the largest angle. Now using law of cosines, $$\cos\theta=\frac{3^2+5^2-7^2}{2\cdot3\cdot5}=-\frac12$$ $$\implies\theta=120^\circ$$

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    $\begingroup$ Exactly...that's the easiest way...I went the more complicated way...thank you $\endgroup$ Nov 9, 2021 at 14:04
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    $\begingroup$ yes esp. Heron's formula is usually cumbersome $\endgroup$
    – Math Lover
    Nov 9, 2021 at 14:15

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