Question

In the given figure, $\angle B={90}^{\circ }$ and $\angle ADB=x^{\circ }$, then find ${\cos }^2{C}^{\circ }+{\sin }^2{C}^{\circ }$.

• A. 1
• B. 2
• C. 0
• D. -1

Answer 1

Answer: (A)

1

In $△CAB$,

$CA=20$ and $AB=12$

Using Pythagoras theorem,
$C{A}^2=A{B}^2+B{C}^2$
${20}^2={12}^2+B{C}^2$
$BC=16$
${\cos }^{2}C+{\sin }^{2}C={\left(\frac{BC}{AC}\right)}^2+{\left(\frac{AB}{AC}\right)}^2$

= ${\left(\frac{16}{20}\right)}^2+{\left(\frac{12}{20}\right)}^2$

= $\frac{256+144}{400}$

= $\frac{400}{400}$

= $1$