(a) Solve the inequality 13x?14(x+2)?3x?113 (b) In the diagram, ABC is right-angled triangle on a horizontal ground. |AD| is a vertical tower. < BAC = 90o ?...

Answer Details

13x−14(x+2)≥3x−113 $\frac{\mathrm{<br>}}{}$

13x−(x+2)12≥3x−43 $\frac{\mathrm{<br>}}{}$

4x−3x−612≥3x−43 $\frac{4𝑥-3𝑥-6}{12}\ge 3𝑥-\frac{4}{3}$

x−612≥3x1−43 $\frac{𝑥-6}{12}\ge \frac{3𝑥}{1}-\frac{4}{3}$

x−612≥9x−43 $\frac{𝑥-6}{12}\ge \frac{9𝑥-4}{3}$ 

3(x - 6) ≥ $\ge$ 12(9x - 4)

3x - 18 ≥ $\ge$ 108x - 48

3x - 108x ≥ $\ge$ -48 + 18

−105x−105≥−30−105 $\frac{-105𝑥}{-105}\ge \frac{-30}{-105}$ 

x ≥1035 $<br>\frac{\mathrm{<br>}}{}$

x 
$\ge$ 27

 27

(b)

(i) 

CB $𝐵$A = 108o ${}^{𝑜}$ - (90o ${}^{𝑜}$ + 35o ${}^{𝑜}$)

= 180o ${}^{𝑜}$ - 125o ${}^{𝑜}$ = 55o

sin35o ${}^{𝑜}$ = |AB|66 $\frac{|𝐴𝐵|}{66}$

|AB| = 66 x sin30o ${}^{𝑜}$ 

|AB| = 37.856m

cos 35o ${}^{𝑜}$ = |AC|66 $\frac{|𝐴𝐶|}{66}$

|AC| = 66 x cos35o ${}^{𝑜}$ 

|AC| = 55.06m

Tan52o ${}^{𝑜}$ = |AD|37.856 $\frac{|𝐴𝐷|}{37.856}$

|AD| = 37.856 x Tan52o ${}^{𝑜}$

= 48.45m the height of the tower to 2d.p

(ii)

Tan AC $𝐶$D = 48.45354.06

=0.8962

AC $𝐶$D = tan−1 ${}^{-1}$0.8962

= 41.867o ${}^{𝑜}$

Angle of elevation of the tower from

C = 35o ${}^{𝑜}$ + 41.867o ${}^{𝑜}$ 

= 76.87o ${}^{𝑜}$ to 2d.p