Question

I have been finding this so difficult to do. Please help me

Answer 1

Given the diagram below

Introducing 'a', as shown above, it can be observed that

$a=61^0(\text{corresponding angles)}$

It can also be observed that

$\begin{gathered} 3y-41=a=61^0(\text{vertically opposites angles are equal)} \\ 3y-41=61 \\ 3y=61+41 \\ 3y=102 \\ y=\frac{102}{3}=34 \end{gathered}$

It can also be observed that

$(3y-41)^0+z=180^0(\text{angles on a straight line)}$

Substitute for y to get z

$\begin{gathered} 3(34)-41+z=180 \\ 102-41+z=180 \\ 61+z=180 \\ z=180-61 \\ z=119^0 \end{gathered}$

Hence, the value of y is 34°, while z is 119°.