Question

Find the value of the angles a,b, c, and d in the given figure.

Answer 1

Answer:

<a=26°

<b=48°

<c=26°,

<d=106°

Step-by-step explanation:

Given,

TS||PR,RS and QS is a transversal,

Now,

<PQS+<SQR=180° (linear pair).

=154°+<c=180°

=<c=180°-154

=<c=26°

Now,

<c=<a ( alternate interior angle)

so,

<a=26°

Again,

<TSU+<TSR=180° (linear pair)

=<TSU+<TSQ+<QSR=180° ( by figure)

=48°+26°+<d=180° (putting the value)

=74°+<d=180°

=<d=180°-74°

=<d=106°

Now,

In ∆QSR,

<c+<b+<d=180° (angle sum property)

=26°+<b+106°=180°

=<b+132°=180°

=<b=180°-132°

=<b=48°

Hope it is helpful.