Question

Pls help me with trig :)

Answer 1

Answer:

Step-by-step explanation:

let ∠GEF=x

then ∠GFE=90-x

$\frac{GH}{EH}=tan x\\GH=EH~ tan x\\or~ GH=a ~tan x=6 tan x\\\frac{GH}{FH}=tan (90-x)=cot x\\GH=FH*cot x\\GH=b*cot x=27 cot x\\6 tan x=27 cotx\\\frac{tan x}{cot x}=\frac{27}{6}\\tan ^2 x=\frac{27}{6}=\frac{9}{2}\\sec^2x-tan^2x=1\\sec^2 x=1+tan ^2x=1+\frac{9}{2}=\frac{11}{2}\\\frac{EG}{EH}=sec x\\ EG=EH sec x=a sec x= 6 sec x=6\sqrt{\frac{11}{2} } =3\sqrt{22} \approx 14.07$