Question

If $\alpha ,\beta ,\gamma$ are the lengths of the altitudes of a triangle ABC with area $△$, then ${\displaystyle \frac{{△}^2}{R^2}}\left({\displaystyle \frac{1}{{\alpha }^2}}+{\displaystyle \frac{1}{{\beta }^2}}+{\displaystyle \frac{1}{{\gamma }^2}}\right)=$
A). ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$
B). ${\cos }^{2}A+{\cos }^{2}B+{\cos }^{2}C$
C). ${\tan }^{2}A+{\tan }^{2}B+{\tan }^{2}C$
D). ${\cot }^{2}A+{\cot }^{2}B+{\cot }^{2}C$

Answer 1

Hint: We know that the area of triangle is given by $△={\displaystyle \frac{1}{2}}\times base\times height$. Considering the bases of the triangles as a, b, c. We will be now able to determine the values of altitudes given that is $\alpha ,\beta ,\gamma$. On substituting these $\alpha ,\beta ,\gamma$values in the ${\displaystyle \frac{{△}^2}{R^2}}\left({\displaystyle \frac{1}{{\alpha }^2}}+{\displaystyle \frac{1}{{\beta }^2}}+{\displaystyle \frac{1}{{\gamma }^2}}\right)$, we will get the require solution. We might use some trigonometric rules, i.e., sine rule which relates the lengths of the sides of a triangle to the sines of its angles.

Complete step-by-step solution:
Given $\alpha ,\beta ,\gamma$ are the lengths of the altitudes of a triangle ABC with area$△$

${\displaystyle \frac{{△}^2}{R^2}}\left({\displaystyle \frac{1}{{\alpha }^2}}+{\displaystyle \frac{1}{{\beta }^2}}+{\displaystyle \frac{1}{{\gamma }^2}}\right)=$? … (1)
We know that the area of the triangle can be given by $△={\displaystyle \frac{1}{2}}\times base\times height$
$△={\displaystyle \frac{1}{2}}a\alpha ={\displaystyle \frac{1}{2}}b\beta ={\displaystyle \frac{1}{2}}c\gamma$ (Using$△={\displaystyle \frac{1}{2}}\times base\times height$)
$\alpha ={\displaystyle \frac{2△}{a}}$,$\beta ={\displaystyle \frac{2△}{b}}$,$\gamma ={\displaystyle \frac{2△}{c}}$
Substituting in equation (1), we get
$\Rightarrow {\displaystyle \frac{{△}^2}{R^2}}\left({\displaystyle \frac{1}{{\alpha }^2}}+{\displaystyle \frac{1}{{\beta }^2}}+{\displaystyle \frac{1}{{\gamma }^2}}\right)$
$\Rightarrow {\displaystyle \frac{{△}^2}{R^2}}\left({\left({\displaystyle \frac{a}{2△}}\right)}^2+{\left({\displaystyle \frac{b}{2△}}\right)}^2+{\left({\displaystyle \frac{c}{2△}}\right)}^2\right)$
$\Rightarrow {\displaystyle \frac{{△}^2}{R^2}}\left({\displaystyle \frac{a^2}{4{△}^2}}+{\displaystyle \frac{b^2}{4{△}^2}}+{\displaystyle \frac{c^2}{4{△}^2}}\right)$
Therefore, $4{△}^2$ is common and we will take it out from the bracket
$\Rightarrow {\displaystyle \frac{{△}^2}{4{△}^2{R}^2}}\left(a^2+b^2+c^2\right)$
Here ${△}^2$ is cancelled, which now gives us
$={\displaystyle \frac{1}{4R^2}}\left(a^2+b^2+c^2\right)$
$={\displaystyle \frac{a^2+b^2+c^2}{4R^2}}$
$={\displaystyle \frac{a^2}{4R^2}}+{\displaystyle \frac{b^2}{4R^2}}+{\displaystyle \frac{c^2}{4R^2}}$……... (2)
From the sine rule, we know that
${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{c}{\sin C}}$
The sine rule used when we are given either a) two angles and one side, or b) two sides and a non-included angle. To solve a triangle is to find the lengths of each of its sides and all its angles we use this sine rule.
From the extended law of sines we know that,
 ${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{c}{\sin C}}=2R$
The Extended Law of Sines is used to relate the radius of the circumcircle of a triangle to and angle/opposite side pair.
Therefore from the extended law of sines,
$$\sin A={\displaystyle \frac{a}{2R}}$$,$\sin B={\displaystyle \frac{b}{2R}}$,$\sin C={\displaystyle \frac{c}{2R}}$
Now equation (2) can be re-written as
$\Rightarrow {\left({\displaystyle \frac{a}{2R}}\right)}^2+{\left({\displaystyle \frac{b}{2R}}\right)}^2+{\left({\displaystyle \frac{c}{2R}}\right)}^2$
=${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$
Therefore ${\displaystyle \frac{{△}^2}{R^2}}\left({\displaystyle \frac{1}{{\alpha }^2}}+{\displaystyle \frac{1}{{\beta }^2}}+{\displaystyle \frac{1}{{\gamma }^2}}\right)={\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C$

Note: The same way we have cosine rule which is used when we are given either a) three sides or b) two sides and the included angle. The cosine rule gives $a^2=b^2+c^2-2bc\cos A$, $b^2=a^2+c^2-2ac\cos B$ and $c^2=a^2+b^2-2ab\cos C$.