## Trigonometry 7th Edition

(a) CH = $\sqrt 2$ (b) CF = $\sqrt 3$
(a) Each face of a cube is a square and a diagonal divides it into two 45°–45°–90° triangles. Therefore- In 45°–45°–90° triangle CDH Diagonal CH = $\sqrt 2 \times$ DH CH = $\sqrt 2 \times$ 1 = $\sqrt 2$ (b) CHF is a right triangle, right angled at H. Applying Pythagorean theorem- $CF^{2}$ = $CH^{2}$ + $HF^{2}$ $CF^{2}$ = $(\sqrt 2)^{2}$ + $1^{2}$ [HF is the side of cube] $CF^{2}$ = $2 + 1$ = 3 Therefore- CF = $\sqrt 3$