# Find the measure of the largest angle of a triangle formed by the reciprocals of the altitudes of a given triangle

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:

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

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

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 $$\frac1{2A}$$.
The angle opposite to the longest side is the largest angle. Now using law of cosines, $$\cos\theta=\frac{3^2+5^2-7^2}{2\cdot3\cdot5}=-\frac12$$ $$\implies\theta=120^\circ$$