Question

If in $a\Delta ABC,\Delta =a^2-(b-c{)}^2,$ then tan A is equal to

• A. $15/16$
• B. $8/15$
• C. $8/17$
• D. $1/2$

Answer 1

The correct option is B $8/15$

$△=a^2-{(b-c)}^2$

$a=2s-(b+c)$

$△={\left[2s-(b+c)\right]}^2-{(b-c)}^2$

$△=[4s^2+{(b+c)}^2-4s(b+c\left)\right]-{(b-c)}^2$

$△=4s^2-4s(b+c)+4bc$

$△=4s^2-4sb-4sc+4bc$

$△=4s(s-b)-4c(s-b)(4s-4c)(s-b)$

$\frac{1}{4}=\frac{(s-b)(s-c)}{△}$

$\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

$\sqrt{s(s-a)}\tan \frac{A}{2}=\sqrt{(s-b)(s-c)}$

Multiplying bot sides $\sqrt{(s-b)(s-c)}$

$(s-b)(s-c)=\tan \frac{A}{2}△(\because \sqrt{(s-a)(s-b)(s-c)}=△)$

$\frac{(s-b)(s-c)}{△}=\tan \frac{A}{2}$

$\tan \frac{A}{2}=\frac{1}{4}$

$\tan A=\frac{2\tan \frac{A}{2}}{1-{\tan }^2\frac{A}{2}}$

$\tan A=\frac{2\times \frac{1}{4}}{1-\frac{1}{16}}$

$\tan A=\frac{8}{15}$