Ex.5.1 Q12 Lines and Angles - NCERT Maths Class 7

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Find the values of the angles \(x, y\) and \(z\) in each of the following:


 Video Solution
Lines & Angles
Ex 5.1 | Question 12

Text Solution

(i) Reasoning

There are two operations done in sequence. First, if one angle is \(55^\circ\) then the angle opposite to it will also be \(55^\circ \)as vertically opposite angles are equal.

 Also sum of \(\angle x + \angle y = 180^\circ\) and \(\angle z + 55^\circ = 180^\circ.\) Now, \(\angle x,\angle y\) and \(\angle z\) can be easily calculated.


Solve for \(\angle x,\angle y\) and \(\angle z:\)

(i) \(\angle x = 55^\circ\) (Vertically opposite angle)

\[\begin{align} \angle x + \angle y &= 180^\circ  \rm (Linear \,pair) \\55 ^\circ+ \angle y &= 180^\circ\\ \angle y &= 180^\circ- 55 ^\circ\\ \angle y &= 125 ^\circ\end{align}\]

Therefore\(\angle y= \angle z = 125^\circ\) (Vertically opposite angle)

Hence, \(\angle x = 55^\circ,\angle y= 125^\circ,\angle z = 125^\circ\)

(ii) Reasoning

By using angle sum property find the value of \(x \) and then find the value of \(y\) and \(z.\) Since the sum of \(y + z = 180^\circ.\) Now, it’s a matter of finding \(y\) and \(z.\)

By using angle sum property,

\[\begin{align}40^\circ \!+ \!x \!+ \!25^\circ &= 180^\circ \\ \text{(Angles on } &\text{straight line)}\\x + 65^\circ &= 180^\circ\\ x &= 180^\circ - 65^\circ = 115^\circ\end{align}\]


\[ \begin{align}40^\circ + y &= 180^\circ\text{(Linear pair)}\\ y &= 180^\circ - 40^\circ\\ y &= 140^\circ\\y + z &= 180^\circ \text{(Linear pair)}\\140^\circ + z &= 180^\circ (y = 140^\circ)\\ z &= 180^\circ- 140^\circ\\ z &= 40^\circ \end{align} \]

Thus, \( x = 115^\circ,y = 140^\circ {\text {and}} \,\,z = 40^\circ\)