Mashaal Masha

Question:

 

The third angle of an isosceles triangle is $16^{\circ}$ less than the sum of the two equal angles. Find three angles of the triangle.

Difficulty: Easy

Solution:

In an isosceles triangle, two angles are of equal measure. Let each equal angle be of $x$ degrees and the third angle be of $y$ degrees.

According to the given conditions:

The third angle $y$ is $16^{\circ}$ less than $2x$ (the sum of the two equal angles).

$2x-16=y$

$2x-y=16 \quad\quad\quad\quad\quad...~(1)$

 

The sum of all the three angles of a triangle is $180^{\circ}$

$x+x+y=180$

$2x+y=180 \quad\quad\quad\quad\quad...~(2)$

 

Method 1: Matrix Inversion Method

Write the linear equations $(1)$ and $(2)$ in matrix form.

$AX = B$

$\left[ \begin{matrix} 2 & -1 \\ 2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 16 \\ 180 \\ \end{matrix} \right]$

According to matrix inversion method, solution $X$ can be found by the following formula:

$X=A^{-1}B$

 

$\left| \text{A} \right|=\left| \begin{matrix} 2 & -1 \\ 2 & 1 \\ \end{matrix} \right|=2\times 1-\left( -1 \right)\times 2=2+2=4~\ne ~0$

The coefficient matrix $A$ is non-singular therefore $A^{-1}$ exists.

 

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = ${{\text{A}}^{-1}}\left[ \begin{matrix} 16 \\ 180 \\ \end{matrix} \right]$

= $\frac{1}{\left| \text{A} \right|}~Adj~\text{A}\left[ \begin{matrix} 16 \\ 180 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 1 & 1 \\ -2 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 16 \\ 180 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 1~\times 16+1\times 180 \\ \left( -2 \right)\times 16+2\times 180 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 16+180 \\ -32+360 \\ \end{matrix} \right]$

= $\frac{1}{4}\left[ \begin{matrix} 196 \\ 328 \\ \end{matrix} \right]$

$\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]=\left[ \begin{matrix} 49 \\ 82 \\ \end{matrix} \right]$

Therefore $x=49$ and $y=82$

 

Hence the angles of the isosceles triangle are $49^{\circ}$, $49^{\circ}$ and $82^{\circ}$.

 

Method 2: Cramer’s Rule

Write the linear equations $(1)$ and $(2)$ in matrix form.

$AX = B$

$\left[ \begin{matrix} 2 & -1 \\ 2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ y \\ \end{matrix} \right]$ = $\left[ \begin{matrix} 16 \\ 180 \\ \end{matrix} \right]$

According to Cramer's Rule:

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}$

 

$\left| \text{A} \right|=\left| \begin{matrix} 2 & -1 \\ 2 & 1 \\ \end{matrix} \right|=2\times 1-\left( -1 \right)\times 2=2+2=4~\ne ~0$

The coefficient matrix $A$ is non-singular therefore solution exists.

 

For ${{A}_{x}}$, the $x$ column is replaced with the constant column $B$

${{A}_{x}}=\left[ \begin{matrix} 16 & -1 \\ 180 & 1 \\ \end{matrix} \right]$

$\left| {{A}_{x}} \right|=\left| \begin{matrix} 16 & -1 \\ 180 & 1 \\ \end{matrix} \right|=16\times 1-\left( -1 \right)\times 180=16+180=196$

$x=\frac{\left| {{A}_{x}} \right|}{\left| A \right|}=\frac{196}{4}=49$

 

For ${{A}_{y}}$, the $y$ column is replaced with the constant column $B$

${{A}_{y}}=\left[ \begin{matrix} 2 & 16 \\ 2 & 180 \\ \end{matrix} \right]$

$\left| {{A}_{y}} \right|=\left| \begin{matrix} 2 & 16 \\ 2 & 180 \\ \end{matrix} \right|=2\times 180-16~\times 2=360-32=328$

$y=\frac{\left| {{A}_{y}} \right|}{\left| A \right|}=\frac{328}{4}=82$

 

Hence the angles of the isosceles triangle are $49^{\circ}$, $49^{\circ}$ and $82^{\circ}$.