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$[1]$ 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.

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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 $7.45$.

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