Find a solution for the equation Assume that all angles in w

Find a solution for the equation Assume that all angles in which. an unknown appears are acute angles cot alpha = tan(alpha + 30 degree) alpha = degree (Simply your answer. Type an integer or a decimal.)

Solution

We have given cot(alpha)=tan(alpha+30)

1/tan(alpha) =(tan(alpha)+tan(30))/1-tan(alpha)tan(30)

By cross multiplication we get

1-tan(alpha)tan(30) =tan²(alpha)+tan(alpha)tan(30)

1-tan²(alpha) =2tan(alpha)tan(30)

1-tan²(alpha) =2tan(alpha)*(1/sqrt(3))

(2/sqrt(3))*tan(alpha)+tan²(alpha) -1=0 since we are using formula ax²+bx+c=0 then x=[-b(+ or -)sqrt(b²-4ac)]/(2*a)

tan(alpha)=[-(2/sqrt(3)) (+ or -) sqrt(4/3 -4*1*(-1))]/(2*1) =[-(2/sqrt(3)) (+ or -) sqrt(4/3 +4]/2

tan(alpha)=[-(2/sqrt(3))+sqrt(16/3)]/2 =[2/sqrt(3)]/2 and tan(alpha)=[-(2/sqrt(3))-sqrt(16/3)]/2 =[-6/sqrt(3)]/2

tan(alpha)=1/sqrt(3) and tan(alpha)=-3/sqrt(3)

alpha =30⁰ and alpha=arctan(-3/sqrt(3))=-60⁰

Therefore

alpha =30⁰ and -60⁰