## Trigonometry 7th Edition

(a) GD = $5\sqrt 2$ (b) GB = $5\sqrt 3$
(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 GCD Diagonal GD = $\sqrt 2 \times$ CD GD = $\sqrt 2 \times$ 5 ( CD is the edge of cube given as 5) GD = $5\sqrt 2$ (b) GDB is a right triangle, right angled at D. Applying Pythagorean theorem- $GB^{2}$ = $GD^{2}$ + $DB^{2}$ $GB^{2}$ = $(5\sqrt 2)^{2}$ + $5^{2}$ [DB is the side of cube given equal to 5] $GB^{2}$ = $50 + 25$ = 75 Therefore- GB = $\sqrt 75$ = $\sqrt (25\times3)$ GB = $\sqrt 25 \times \sqrt 3$ GB = $5\sqrt 3$