$[1]$ In a triangle $ABC$ , lengths of the two larger sides are $10$ and $9$. If the angles are in $A.P.$, what can be the length of the third side$?$

$[2]$ In a Triangle $ABC$ if $sin A, sin B, sin C$ are in $A.P$. Find the relationship between altitudes.


For [1], let the largest side, 10, be $AB$, and the side with length 9 be $AC$. The shortest side must be $BC$. Also, we know that the largest angle corresponds to the largest side and the smallest angle corresponds to the smallest side, so let angle $C$ be $\theta + a$, angle $B$ be $\theta$, and angle $A$ be $\theta - a$.

By the sine rule we have: $\frac{sin(\theta + a)}{10} = \frac{sin(\theta)}{9} = \frac{sin(\theta - a)}{BC}$, so if we can create an expression for $sin(\theta - a)$ we can find BC.

By equating the first two terms, we get $sin(\theta + a) = \frac{10}{9}sin(\theta)$, and thus $a = arcsin(\frac{10}{9}sin(\theta)) - \theta$. Thus:

$sin(\theta - a) = sin(2\theta -arcsin(\frac{10}{9}sin(\theta)))$

From this, we can see that once we find a value of $\theta$ it is trivial to find $BC$, as $BC = \frac{9sin(2\theta -arcsin(\frac{10}{9}sin(\theta)))}{sin(\theta)}$.

Also, we know that $(\theta + a) + (\theta) + (\theta - a) = \pi$, and thus $\theta = \frac{\pi}{3}$.

Now, substituting in $\theta$, we get $BC = 6\sqrt{3}sin(\frac{2\pi}{3}-arcsin(\frac{5\sqrt{3}}{9})=5+\sqrt{6}$, which is approximately $7.45$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.