Question

Trignometry is helpful in triangle based mathematical analysis. One of various trigonometric relation is tan (A+B) $=\frac{tanA+tanB}{1-tanAtanB}$.

Now consider a diagram representing a triangular roof frame ABC with a window frame EFC. BC and EF are horizontal and AD and FC are vertical.

Calculate FC.

• A. 6.72
• B. 18.21
• C. 2.78
• D. 13.12

Answer 1

Answer: (A)

6.72

Given

$EF=1.8m$

$FC\perp DC$

$⟹\angle ECF+\angle ECD={90}^{\circ }$

$⟹\angle ECF={90}^{\circ }-{75}^{\circ }={15}^{\circ }$

In $\Delta EFC$

$\tan \angle ECF=\frac{EF}{FC}$

$\tan 15=\frac{1.8}{FC}$

$FC=\frac{1.8}{\tan 15}$

$FC=\frac{1.8}{\tan (45-30)}$

$FC=\frac{1.8}{\frac{\tan 45-\tan 30}{1+\tan 45\tan 30}}$

$FC=\frac{1.8}{\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}}$

$FC=1.8\times \frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}$

$FC=1.8\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$

$FC=6.72m$