Question

In an isosceles triangle $ABC$; $AB=AC=10$ cm and $BC=18$ cm. Find the value of :
${\sin }^{2}B+{\cos }^{2}C$

• A. $3$
• B. $0$
• C. $1$
• D. $7$

Answer 1

The correct option is C $1$

Given, $AB=AC=10$ and $BC=18$ cm

$\cos B=\frac{A{B}^2+B{C}^2-A{C}^2}{2AB\times BC}$

$\cos B=\frac{{10}^2+{18}^2-{10}^2}{2\times 10\times 18}$

$\cos B=\frac{9}{10}$

$\cos B=\frac{B}{H}=\frac{9}{10}$

Using Pythagoras Theorem,

$H^2=P^2+B^2$

${10}^2=P^2+9^2$

$P=\sqrt{19}$

$\sin B=\frac{P}{H}=\frac{\sqrt{19}}{10}$

Now, $\cos C=\frac{A{C}^2+B{C}^2-A{B}^2}{2AC\times BC}$

$\cos C=\frac{{10}^2+{18}^2-{10}^2}{2\times 10\times 18}$

$\cos C=\frac{9}{10}$

Thus, ${\sin }^{2}B+{\cos }^{2}C=(\frac{\sqrt{19}}{10}{)}^2+(\frac{9}{10}{)}^2$

${\sin }^{2}B+{\cos }^{2}C=\frac{19}{100}+\frac{81}{100}$

${\sin }^{2}B+{\cos }^{2}C=\frac{100}{100}=1$