From the figure calculate TH in centimeters - JAMB Mathematics 1993 Question
A
\(\frac{5}{\sqrt{3} + 1}\)
B
\(\frac{5}{\sqrt{3} - 1}\)
C
\(\frac{5}{\sqrt{3}}\)
D
\(\frac{\sqrt{3}}{5}\)
correct option: b
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 = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)
= \(\frac{5}{\sqrt{3} - 1}\)
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