In triangle ABC: B , C are two acute angles

Question

AD is drawn perpendicular to BC ,m(BAD)=alpha,and m(CAD)=beta

Prove that:AD=a*cos(alpha)*cos(beta)/sin(alpha+beta)

 

Answer 1

let BD = x

let DC = a-x

let AD = h

 

tan alpha = x/h

x = h(tan alpha)

 

tan beta = (a-x)/ h

a-x = h(tan beta)

 

a - h(tan alpha) = h(tan beta)

a = h(tan alpha) + h(tan beta)

a = h(tan alpha + tan beta)

a = h [(sin alpha / cos alpha) + (sin beta / cos beta)]

a = h[(sin alpha cos beta + sin beta cos alpha)/(cos alpha + cos beta)]

a = h[sin(alpha + beta)/(cos alpha + cos beta)

a(cosα + cosβ) / sin(α + β) = h