Answer:
The measure of side BC = 5 unit and The measure of side AC = 5 unit
Step-by-step explanation:
Given as :
In a triangle ΔABC
The measure of side = AB = 5[tex]\sqrt{2}[/tex] unit
The measure of angle A = ∠A = 45°
The measure of angle C = ∠C = 30°
Now, For a triangle the sum of three angles of triangle = 180°
∴, ∠A + ∠B + ∠C = 180°
Or, 45° + ∠B +30° = 180°
Or , ∠B = 180° - ( 45° + 30°)
Or, ∠B = 180° - 75°
i.e ∠B = 105°
Now, From figure
Sin 45° = [tex]\dfrac{\textrm BC}{\textrm AB}[/tex]
Or, [tex]\frac{1}{\sqrt{2} }[/tex] = [tex]\dfrac{\textrm BC}{5\sqrt{2}}[/tex]
Or, BC = [tex]\frac{5\sqrt{2} }{\sqrt{2} }[/tex]
∴ BC = 5
So, The measure of side BC = 5 unit
Again , from figure
Cos 45° = [tex]\dfrac{\textrm AC}{\textrm AB}[/tex]
Or, [tex]\frac{1}{\sqrt{2} }[/tex] =[tex]\dfrac{\textrm AC}{5\sqrt{2}}[/tex]
Or, AC = [tex]\frac{5\sqrt{2} }{\sqrt{2} }[/tex]
∴ AC = 5
So, The measure of side AC = 5 unit
Hence The measure of side BC = 5 unit and The measure of side AC = 5 unit Answer
