(a) Solve the inequality \( \frac{1}{3}x - \frac{1}{4}(x + 2) \ge 3x - 1\frac{1}{3} \) (b) In the diagram, ABC is right-angled triangle on a horizontal grou...

Answer Details

(a) Solve \(\;\dfrac{1}{3}x-\dfrac{1}{4}(x+2)\ge 3x-1\tfrac{1}{3}.\)

Write \(1\tfrac{1}{3}=\dfrac{4}{3}\) and multiply every term by the LCM of the denominators, \(12\):

\[12\cdot\tfrac{1}{3}x-12\cdot\tfrac{1}{4}(x+2)\ge 12\cdot 3x-12\cdot\tfrac{4}{3}.\]

\[4x-3(x+2)\ge 36x-16.\]

\[4x-3x-6\ge 36x-16.\]

\[x-6\ge 36x-16.\]

Collect like terms:

\[-6+16\ge 36x-x\;\Rightarrow\;10\ge 35x.\]

\[x\le\frac{10}{35}=\frac{2}{7}.\]

\(\therefore x\le\dfrac{2}{7}.\)

(b) In right-angled triangle \(ABC\) on horizontal ground, \(\angle BAC=90^{\circ}\), \(\angle ACB=35^{\circ}\) and \(|BC|=66\text{ m}\). \(AD\) is a vertical tower and \(\angle ABD=52^{\circ}\) is the angle of elevation of the top \(D\) from \(B\).

(i) Height of the tower \(|AD|\).

In triangle \(ABC\), \(BC\) is the hypotenuse and \(AB\) is opposite the \(35^{\circ}\) angle at \(C\):

\[\sin 35^{\circ}=\frac{|AB|}{|BC|}\;\Rightarrow\;|AB|=66\sin 35^{\circ}=66\times0.5736=37.86\text{ m}.\]

Also \(|AC|=66\cos 35^{\circ}=66\times0.8192=54.06\text{ m}.\)

The tower \(AD\) is vertical and \(AB\) horizontal, so triangle \(ABD\) is right-angled at \(A\):

\[\tan 52^{\circ}=\frac{|AD|}{|AB|}\;\Rightarrow\;|AD|=37.86\times\tan 52^{\circ}=37.86\times1.2799=48.45\text{ m}.\]

\(\therefore\) the height of the tower \(|AD|\approx 48.45\text{ m}\) (2 d.p.).

(ii) Angle of elevation of the tower from \(C\).

In right-angled triangle \(ACD\):

\[\tan(\angle ACD)=\frac{|AD|}{|AC|}=\frac{48.45}{54.06}=0.8962,\]

\[\angle ACD=\tan^{-1}(0.8962)=41.87^{\circ}.\]

Adding the horizontal angle \(\angle ACB=35^{\circ}\), the angle of elevation of the tower measured from \(C\) is

\[35^{\circ}+41.87^{\circ}=76.87^{\circ}\text{ (2 d.p.).}\]