Question

If a triangle has angles ${45}^{\circ },{45}^{\circ }$, and ${90}^{\circ }$, what is the ratio of the sides of the triangle opposite to these angles respectively? [2 marks]

Answer 1

Let us assume the length of side AB of the triangle to be x cms.
Applying trignometric ratios to the sides, we get :
$sin{45}^{\circ }=\left(\frac{x}{AC}\right)\Rightarrow \frac{1}{\sqrt{2}}=\left(\frac{x}{AC}\right)\Rightarrow AC=x\sqrt{2}...\left(i\right)\left[1Mark\right]Similarly,tan{45}^{\circ }=\left(\frac{x}{BC}\right)\Rightarrow 1=\left(\frac{x}{BC}\right)\Rightarrow BC=x...\left(ii\right)$
So, the ratios of the sides of the triangle with angles ${45}^{\circ },{45}^{\circ }\&{90}^{\circ }=x:BC:AC$
$=x:x:x\sqrt{2}\left(from\right(i\left)\&\right(ii\left)\right)$
$=1:1:\sqrt{2}$ [1 Mark]

An alternate and shortcut method of solving this question is:

For the given triangle, as two angles are equal; the two sides opposite to these angles will also be equal.
And as the third angle is ${90}^{\circ }$, the triangle is right - angled triangle.
Let us assume the length of the equal sides is equal to x cms.
So, length of the hypotenuse = $\sqrt{(}x+x)$ = $\sqrt{(}2x)$
So, Ratio of the sides of the triangle = x : x : $\sqrt{2}x$
$\to$ Ratio of the sides of the triangle = 1 : 1 : $\sqrt{2}$
[2 Marks]