Solve : In a right angled ∆ the hypotenuse is 4 times as long as perpendicular drawn to it from opposite vertex .Its acute angles are

Question

My try :

c2 = h2 + (a-x)2 b2 = h2 + x2 on adding, we have  2h2 + x2 + (a-x)2 = a2 ..(1) also a = 4h  and cos B = b/a = x/b so x = b2/a on solving (1), we have  h2 + x2 -4hx = 0 substituting above gives b4 + a4/16 -a2.b2 =0 .. (2) as cos B = b/a on solving eqn. (2) we have ,  Cos2B = 0.933 or 0.067 B = 15 or 75 degrees as B is acute , B is 15 degrees I want one more solution different from it.

Answer 1

A hint:

Draw the base $AB$ and then a (Thales) semicircle with diameter $AB$. Find the locus of all points $C$ with height ${|AB|\over4}$ over the base. Stare at the resulting figure. In the end no computing is necessary.

Answer 2

\begin{align} |CH|=|C'H|&=\tfrac14|AB|=\tfrac12R ,\\ |CC'|&=|OC|=|OC'|=R ,\\ \text{hence, }\quad \angle C'OC&=60^\circ ,\\ \angle C'BC&=\tfrac12 \angle C'OC =30^\circ ,\\ \angle ABC&= \beta=\tfrac12 \angle C'BC =15^\circ ,\\ \angle CAB&= \alpha=90^\circ-\beta =75^\circ . \end{align}