Question

In the given figure, find the value of $sec\frac{A}{2}+\cos ec\frac{A}{2}$

• A. $\frac{12}{35}$
• B. $\frac{20}{12}$
• C. $\frac{37}{12}$
• D. $\frac{35}{12}$

Answer 1

The correct option is D $\frac{35}{12}$

Let’s draw perpendicular from vertex A.
We know that in an isosceles triangle, perpendicular bisects the unequal side, so BD = CD = 6 cm

$cosecA/2=\frac{AB}{BD}=\frac{10}{6}=\frac{5}{3}$
In $△ABD$,
Using Pythagoras theorem,
$A{B}^2=B{D}^2+A{D}^2$
${10}^2=6^2+A{D}^2$
$A{D}^2={10}^2-6^2$
$A{D}^2=100-36=64$
$AD=8cm$
$secA/2=\frac{AB}{AD}=\frac{10}{8}=\frac{5}{4}$
$cosecA/2+secA/2=\frac{5}{3}+\frac{5}{4}=\frac{35}{12}$