Given :
To Find :
Solution :
From point A,
[tex] \bf tan A = \dfrac{perpendicular}{base}[/tex]
[tex] \bf \implies tan A = \dfrac{BC}{AB}[/tex]
[tex] \bf \implies tan A = \dfrac{BC}{7}[/tex]
Now, we are given ∠A = 45°
[tex] \bf \implies tan A = tan 45^{\circ} [/tex]
[tex] \bf \implies tan 45^{\circ} = \dfrac{BC}{7}[/tex]
Now, we know that tan45° = 1
[tex] \bf \implies 1 = \dfrac{BC}{7}[/tex]
[tex] \bf \implies 1 \times 7 = BC [/tex]
[tex] \bf \implies BC = 7[/tex]
Now, by Pythagoras' theorem,
AC² = BC² + AB²
[tex] \bf \implies AC^{2} = (7)^{2} + (7)^{2}[/tex]
[tex] \bf \implies AC^{2} = 49 + 49[/tex]
[tex] \bf \implies AC^{2} = 98[/tex]
[tex] \bf \implies AC = \sqrt{98}[/tex]
[tex] \bf \implies AC = 7 \sqrt{2}[/tex]
[tex] \pink{\bf \therefore \: values \: of \: AC = 7 \sqrt{2} \: and \: BC = 7}[/tex]
