(a) Solve the inequality \(\frac{1}{3}x - \frac{1}{4} (x + 2) \geq 3x - 1\frac{1}{3}\)
(b)
In the diagram, ABC is right-angled triangle on a horizontal ground. |AD| is a vertical tower. < BAC = 90\(^o\), < ACB = 35\(^o\), < ABD = 52\(^o\) and |BC| = 66cm.
Find, correct to two decimal places:
(I) the height of the tower
(ii) the angle of elevation of the top of the tower from C
\(\frac{1}{3}x - \frac{1}{4}(x + 2) \geq 3x -1\frac{1}{3}\)
\(\frac{1}{3}x - \frac{(x + 2)}{12} \geq 3x - \frac{4}{3}\)
\(\frac{4x - 3x - 6}{12} \geq 3x - \frac{4}{3}\)
\(\frac{x - 6}{12} \geq \frac{3x}{1} - \frac{4}{3}\)
\(\frac{x - 6}{12} \geq \frac{9x - 4}{3}\)
3(x - 6) \(\geq\) 12(9x - 4)
3x - 18 \(\geq\) 108x - 48
3x - 108x \(\geq\) -48 + 18
\(\frac{-105x}{-105} \geq \frac{-30}{-105}\)
x \(\geq \frac{10}{35}\)
x \(\geq\) \(\frac{2}{7}\)
(b)(i)
C\(B\)A = 108\(^o\) - (90\(^o\) + 35\(^o\))
= 180\(^o\) - 125\(^o\) = 55\(^o\)
sin35\(^o\) = \(\frac{|AB|}{66}\)
|AB| = 66 x sin30\(^o\)
|AB| = 37.856m
cos 35\(^o\) = \(\frac{|AC|}{66}\)
|AC| = 66 x cos35\(^o\)
|AC| = 55.06m
Tan52\(^o\) = \(\frac{|AD|}{37.856}\)
|AD| = 37.856 x Tan52\(^o\)
= 48.45m the height of the tower to 2d.p
(ii) Tan A\(C\)D = \(\frac{48.453}{54.06}\)
=0.8962
A\(C\)D = tan\(^{-1}\)0.8962
= 41.867\(^o\)
Angle of elevation of the tower from
C = 35\(^o\) + 41.867\(^o\)
= 76.87\(^o\) to 2d.p
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