Solution:
Consider the Triangle ABC in which , ∠A=60°, ∠B=90°,∠C=30°
Let , Perpendicular = P
We will use trigonometric ratio to prove that ratio of sides will always be same by using either of angles that is 30° or 60°.
Sin C= [tex]\frac{\text{Perpendicular}}{\text{Hypotenuse}}[/tex]
Sin 30°= [tex]\frac{1}{2}[/tex]
So, Ratio of Perpendicular and Hypotenuse is always constant, that is always [tex]\frac{1}{2}[/tex].
Therefore , Hypotenuse will be = 2 P
Now, using Pythagoras Theorem
⇒(Perpendicular)² + (Base)² = (Hypotenuse)²
⇒P²+ (Base)²=(2 P)²
⇒ (Base)²= 4 P²- P²
⇒(Base)²= 3 P²
⇒ Base= √3 P
So, ratio of Perpendicular to Base or Base to Perpendicular is [tex]\frac{1}{\sqrt{3}} or \frac{\sqrt{3}}{1}[/tex].
Sin 60°= [tex]\frac{BC}{AC}=\frac{\sqrt3P}{P}=\frac{\sqrt3}{1}[/tex]
⇒If you consider either angle of 30° or 60° to find the side length ratios, , then also ratio of sides is always that is BC: AB or BC: AC or AB: AC →→ AB : BC : AC = 1 : √3 : 2 .
now, consider Triangle P QR, in which ∠P=∠R= 45°,∠Q=90°
To find the side length ratio we will use trigonometric ratio.
→Tan 45°= [tex]\frac{PQ}{QR}[/tex]
→1= [tex]\frac{PQ}{QR}[/tex]
→P Q= QR
Let , P Q= QR= K
Using Pythagoras theorem
⇒(Perpendicular)² + (Base)² = (Hypotenuse)²
⇒ K² + K²= (Hypotenuse)²
⇒(Hypotenuse)²= 2 K²
Hypotenuse= √2 K
⇒If you consider either angle that is of 45° to find the side length ratios, , then also ratio of sides is always that is P Q: QR or P Q: PR or QR: PR →→ P Q : QR : PR = 1 : 1:√2.
