Question

In triangle $ABC$ is right-angled at $B$ and $BD$ is perpendicular to $AC$, then find:

$\cos \angle DBC$ is $\frac{12}{m}$. value of m is,

Answer 1

In $△ABC$

$A{B}^2+B{C}^2=A{C}^2$ (By Pythagoras theorem)

${12}^2+5^2=A{C}^2$

$A{C}^2=169$

$AC=13$
In $△ABC$

$\angle A+\angle B+\angle C=180$

$\angle A+\angle C=90$ (As $\angle B=90$).....(i)

In $△ABD$,

$\angle A+\angle BDA+\angle DBA=180$

$\angle A+\angle DBA=90$ (As $\angle BDA=90$).....(i)
From (i) and (ii)

$\angle C=\angle DBA$

Similarly, $\angle DBC=\angle A$
Now, $\cos \angle DBC=\cos \angle A=\frac{B}{H}=\frac{AB}{AC}=\frac{12}{13}$

$\therefore valueof{}^{\prime }{m}^{\prime }is13$