Question

In the diagram given below, triangle ABC is right angled at B and BD is prependicular AC, Find.(i) cos∠DBCcos∠DBC

Answer 1

In $\Delta ABC$ and $\Delta CBD$

$\angle ACB=\angle BCA$

$\angle ABC=\angle CBD$

$\therefore \Delta ABC\sim \Delta CBD$

$\therefore \frac{ABC!}{BCA!}=\frac{BDC!}{BCD!}$

$\therefore \frac{AC}{AB}=\frac{BC}{BD}$

$\frac{13}{12}=\frac{5}{BD}$

$BD=\frac{5\times 12}{13}$

$BD=\frac{60}{13}$

i, $\cos DBC!=\frac{adj}{hyp}=\frac{BD}{BC}=\frac{60}{13}\times \frac{1}{5}=\frac{12}{13}$

ii, $\cot DBA!=\frac{adj}{opp}=\frac{BD}{AD}=\frac{60}{13}\times \frac{1}{12}=\frac{5}{13}$