Do altitudes produce similar triangles(Method for solving question)?

Question

Question: If triangle PQR has sides 40, 60, and 80, then the shortest altitude is K times the longest altitude. Find the value of K.

Original Method: I've tried to do the problem by just draw a altitude, and then use cos90 or tan90 to find the altitude's length, and then repeat for all three altitudes, and then find the ratio between the longest and shortest altitude. However, this did not work, as I realized cos90 = 0, and tan90 = undefined.

Problem: I can think of only one other way to solve the problem: using similar triangles, to produce some ratios which show that the longest altitude is k times larger than the shortest. So,

I was wondering if the altitude of any triangle always produces similar triangles, and if that fact could be used to solve this problem. If not, how should I go about working out a solution(please provide a method, or hints, not a full solution)?

Answer 1

Calculating area of PQR in two different ways, then we have 40H = 80h; where H is the longest altitude and h is the shortest altitude.

Answer 2

Given a triangle $ABC$ with sides $a$ opposite angle $A$ etc. Let the altitude from $B$ to side $b$ intersect at $D$. Then via right triangle $ABD$ we see the altitude, $BD$ is equal to $c\sin A$ and right triangle $BDC$ that $BD$ is equal $a\sin C$.

So if $A,B,C = 40, 60, 80$ respectively then the three altitudes will be $d = c \sin 40 = a\sin 80$ and $e = b \sin 40 = a\sin 60$ and $f = c \sin 60 = b \sin 80$.

We know $\sin 80 > \sin 60 > \sin 40$ so $c > a; b>a; c > b$ so $c >b>a$. So $e = b\sin 40 < d = c\sin 40 < f= c\sin 60$. So $K = \frac {shortest}{longest} = \frac ef = \frac {b\sin 40 = a\sin 60}{c\sin 60 = b\sin 80} = \frac {\sin 40}{\sin 80}$.