In the diagram TX is perpendicular to UW UX 1cm... - WAEC Mathematics 2016 Question

In (\Delta) UXT, tan(\alpha) = (\frac{1}{\sqrt{3}})

(\alpha) = tan^-1((\frac{1}{\sqrt{3}}))

= 30^o

In (\Delta)WXT, tan(\beta) (\frac{\sqrt{3}}{\sqrt{3}}) = 1

(\beta) = tan^-1(1) = 45^o

Hence, < UTW = (\alpha) + (\beta)

= 30^o + 45^o = 75^o