## Trigonometry 7th Edition

BD = $\sqrt 89 \approx 9.43$
In given figure 23, we can see two right triangles BCD and BCA, both right angled at C. Given- BC = 5, AB = 13 , AD = 4 Applying Pythagoras theorem in right triangle BCA- $AB^{2}$ = $BC^{2}$ + $AC^{2}$ $AC^{2}$ = $AB^{2}$ - $BC^{2}$ $AC^{2}$ = $13^{2}$ - $5^{2}$ $AC^{2}$ = 169 - 25 = 144 $AC = \sqrt (144)$ = 12 Now we can calculate DC as- DC = AC - AD = 12 - 4 = 8 Now applying Pythagoras theorem in right triangle BCD- $BD^{2}$ = $BC^{2}$ + $DC^{2}$ $BD^{2}$ = $5^{2}$ + $8^{2}$ $BD^{2}$ = 25 + 64 = 89 Therefore BD = $\sqrt 89 \approx 9.43$