Chapter 2.8, Problem 15PSC

To determine

To Find: The measure of $\angle 1$ from the given figure.

Answer to Problem 15PSC

Measure of angle $\text{ 1}={\left(132\frac{3}{4}\right)}^{°}\text{ or 140}°$

Explanation of Solution

Given information: In the given figure $\text{( }{x}^2\text{ - 6}x)°\text{ , }{\left(\frac{1}{2}x\text{ + 42}\right)}^{°}$ are vertically opposite angles. So they are equal .Also $\angle 1$ is making linear pair with both $\text{( }{x}^2\text{ - 6}x)°\text{ , }{\left(\frac{1}{2}x\text{ + 42}\right)}^{°}$ ,so it is supplement to both the angles

Concept used: Rule of vertically opposite angles, supplementary angles and solving quadratic equations by splitting method is used

Calculation: As per the given question $\text{( }{x}^2\text{ - 6}x)°\text{ , }{\left(\frac{1}{2}x\text{ + 42}\right)}^{°}$ are vertically opposite angle thus, $\text{( }{x}^2\text{ - 6}x)°\text{ = }{\left(\frac{1}{2}x\text{ + 42}\right)}^{°}\mathrm{..................................}(\text{i)}$

Also $\angle 1$ is making linear pair with both, so it is supplement to both the angles$m\angle \text{1 = 180}°-\text{ ( }{x}^2\text{ - 6}x)°$ or

  $m\angle \text{1 = 180}°-{\left(\frac{1}{2}x\text{ + 42}\right)}^{°}\mathrm{........................................}(\text{ii)}$

Considering (i) and solving for $x$ :

  $\begin{array}{l}\text{(}{x}^2\text{ - 6}x)\text{ - }\left(\frac{1}{2}x\text{ + 42}\right)\text{ = 0}\\ \\ \text{ (}{x}^2\text{ - 6}x\text{ - }\frac{1}{2}x\text{ + 42})\text{ = 0}\\ \\ \text{ 2}{x}^2\text{ - 12}x\text{ - }x\text{ - 84 = 0}\\ \\ \text{2}{x}^2\text{ - 13}x\text{ - 84 = 0}\\ \\ \text{2}{x}^2\text{ - 8}x\text{ - 21}x\text{ - 84 = 0}\\ \\ \text{2}x(x\text{ - 2) - 21 (}x\text{ + 4) = 0}\\ \\ \text{(2}x\text{ -21) (}x\text{ + 4) = 0}\\ \\ x\text{ = }\frac{21}{2}\text{ or - 4 }\mathrm{..........................................................}\text{(iii)}\end{array}$

  $$ Substituting both the values of $x$ in (ii) we get: Case $1:$ When $x\text{ = }\frac{21}{2}$

  $\begin{array}{c}m\angle \text{1 = 180}°\text{- }{\left(\frac{1}{2}x+42\right)}^{°}\\ m\angle 1\text{ = 180}°\text{ - 42}°\text{ - }{\left(\frac{1}{2}\times \frac{21}{2}\right)}^{°}\\ m\angle 1\text{ = 138}°\text{ - }{\left(\frac{21}{4}\right)}^{°}\\ m\angle 1\text{ = }{\left(\frac{552\text{ - 21}}{4}\right)}^{°}\\ m\angle 1\text{ = }{\left(132\frac{3}{4}\right)}^{°}\mathrm{................................................}(\text{iv)}\end{array}$

Case $2:$ When $x\text{ = - 4}$

  $\begin{array}{c}m\angle \text{1 = 180}°\text{- }{\left(\frac{1}{2}x+42\right)}^{°}\\ m\angle \text{1 = 180}°\text{- }{\left(\frac{1}{2}\times \text{ (-4) }+42\right)}^{°}\\ m\angle \text{1 = 180}°\text{- }{\left(\text{-2 }+42\right)}^{°}\\ m\angle \text{1 = 180}°{\text{- 40}}^{°}\\ m\angle \text{1 = 140}°\mathrm{...................................................................}\text{(v)}\end{array}$

Combining result of (iv) and (v) we can say:

  $m\angle \text{1 = }{\left(132\frac{3}{4}\right)}^{{}^{°}}\text{ or 140}°$

Conclusion:

Measure of angle $\text{ 1}={\left(132\frac{3}{4}\right)}^{°}\text{ or 140}°$