Question

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

• A. $6$
• B. $13$
• C. $9$
• D. $17$

Answer 1

Answer: (B)

$13$

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