JAMB - Mathematics (1993 - No. 50)
From the figure, calculate TH in centimeters
\(\frac{5}{\sqrt{3} + 1}\)
\(\frac{5}{\sqrt{3} - 1}\)
\(\frac{5}{\sqrt{3}}\)
\(\frac{\sqrt{3}}{5}\)
توضیح
TH = tan 45o, TH = QH
\(\frac{TH}{5 + QH}\) = tan 30o
TH = (b + QH) tan 30o
QH = 56 (5 + QH) \(\frac{1}{\sqrt{3}}\)
QH(1 - \(\frac{1}{\sqrt{3}}\)) = \(\frac{5}{\sqrt{3}}\)
QH = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)
= \(\frac{5}{\sqrt{3} - 1}\)
\(\frac{TH}{5 + QH}\) = tan 30o
TH = (b + QH) tan 30o
QH = 56 (5 + QH) \(\frac{1}{\sqrt{3}}\)
QH(1 - \(\frac{1}{\sqrt{3}}\)) = \(\frac{5}{\sqrt{3}}\)
QH = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)
= \(\frac{5}{\sqrt{3} - 1}\)