Question

in a right angle triangle ABC ,right angle at B. If AB + AC =9 and BC =3 , then find cotC and secC​

Answer 1

I hope it will help u...................

Answer 2

Step-by-step explanation:

Right triangle at B

If AB+AC=9 and BC=3

then find cotc and secC

In /\ABC

AC

${ac}^{2} = {ab}^{2} + {bc}^{2}$

(9-AB)2=(AB)2+(3)2

81+AB2-18AB=AB2+9

81=AB2+9

AB2=81-9

AB2=72

AB=

$\sqrt{72}$

AB=8.4

then CotC=

$\frac{b}{p}$

$\frac{3}{8.4}$

SecC=3/0.6