## Trigonometry 7th Edition

DB =$\sqrt 52$
Consider the triangle ABC is a right angled triangle According to the Pythagoras theorem $(AC)^{2}=(AB)^{2}+(BC)^{2}$ .......(1) Given AC = 5, BC = 3, DA = 6 Substitute these values in equation (1) we get $(5)^{2}=(AB)^{2}+(3)^{2}$ $(AB)^{2}$ = 25 -9 $AB = \sqrt 16$ $AB = 4 cm$ From the figure Consider DAB is a right angled triangle According to the Pythagoras theorem $(DB)^{2}=(DA)^{2}+(AB)^{2}$ .......(2) Substitute DA and AB values in equation (2) we get $(DB)^{2}=(6)^{2}+(4)^{2}$ $(DB)^{2}= 36+16$ $DB = \sqrt 52$