Question

5. In the given figure, Angle ACB = 90°, angle BDC =
90°, CD = 4 cm, BD = 3 cm, AC = 12 cm. cos A -

sin A is equal to :​

Answer 1

$Applying\: Pythagoras\:theorem \: :-$

${BC}^{2} = {CD}^{2} + {BD}^{2} = {4 cm}^{2} + {3cm}^{2}$

$BC = 5cm$

${AB}^{2} = {AC}^{2} + {B C}^{2} = {12cm}^{2} + {5cm}^{2}$

$AB\:=\:13cm$

$\therefore\:cosA\:-\:SinA$

$\frac{AC}{AB} - \frac{BC}{AB}$

$\frac{12}{1 3} - \frac{5}{13}$

$\orange{\:After\: solving\:we\:get\::-}$

$\frac{7}{13}$