Question

Case study based
The Shivani wants to make a wall hanging in the shape of an isosceles triangle for her room as shown in figure. She has cut a piece of cardboard with the given measurements. By looking at the cardboard a few questions rose in her mind

Question

Find sinB + cosC
OPTIONS

A - 1/root2
B - 2/1
C - root2
D - 2

Answer 1

Correct Answer: (d) root2

Given: Triangle ABC is an isosceles triangle with

            AB = AC=5$\sqrt{2$

            BC=10

To Find: sinB + cosC

Solution: As AD bisects the line BC then, BD=DC=5

             Using pythagoras theorem

       $(AB)^{2}$=$(AD)^{2} + (BD)^{2}$

       $(5\sqrt{2})^{2} =(AD)^{2} +(5)^{2}$

       $50=(AD)^{2}+25\\\\50-25=(AD)^{2} \\\\25=(AD)^{2} \\\\5=AD$

     sinB = AD/AB          sinB = $\frac{5}{5\sqrt{2} }$                         (1)                                                  

     cosC = DC/AC        cosC=$\frac{5}{5\sqrt{2} }$                          (2)

                                                     

     From 1 and 2 -

sinB + cosC = $\frac{5}{5\sqrt{2} }$  +  

                      $\frac{10}{5\sqrt{2} } = \frac{2}{\sqrt{2} } \\\\= \sqrt{2}$

  sinB+sinC=$\sqrt{2}$ = root2