SOLUTION:
Step 1:
For Question No 15, we have that:
Step 2:
Now, let
[tex]\begin{gathered} m\text{ }\angle FBD\text{ = x} \\ m\angle DBE\text{ =y} \\ \text{such that x+ y = 90}^0\text{ ( complementary angles)} \end{gathered}[/tex]
Then,
[tex]\begin{gathered} m\angle FBD\text{ =2x} \\ m\angle DBE=\text{ x} \\ \text{such that } \\ 2x+x=90^0 \\ 3x=90^0 \\ \text{Divide both sides by 3, w}e\text{ have that:} \\ x\text{ = }\frac{90^0}{3} \\ x=30^0 \end{gathered}[/tex]
Then,
[tex]\begin{gathered} \sin ce^{} \\ x=30^0 \\ \text{Then,} \\ m\angle FBD=2x=2X30^0=60^{0\text{ }}(OPTIONC)^{} \end{gathered}[/tex]
