Question

In the figure p II q and r II s. Find the values of x, y and z.pls help me .​

Answer 1

➩Given

pII q (pt II qm)

r II s (nr II ks)

➩To Find

The value of x,y,z.

➩Solution

(i) Finding the value of x

∠kCt=60°(given)

Also,

∠kCt=∠ACD(vertically opposite angles are always equal)

so, ∠ACD=60°

we know,pt is a transversal crossing the parallel lines ks and nr.

∠PAB=∠ACD(Corresponding angles are equal if a transversal crosses at least two parallel lines)

∠PAB=∠ACD

11x+5=60°

11x=60°-5

11x=55

x=5

(ii) Finding the value of z

∠PAB+∠BAC=180°(Linear pair)

11x+5+∠BAC=180°

(11×5)+5+∠BAC=180°

60°+∠BAC=180°

∠BAC=180°-60°

∠BAC=120°

Now if nr II ks

then, AB II CD(Parts of parallel lines)

similarly AC II BD

so this means ABCD is a parallelogram having opposite side parallel.

∠BAC=∠CDB(Opposite angles of a parallelogram are equal)

∠BAC=∠CDB

120°=2z

$z = \frac{120}{2}$

z=60

(iii) Finding the value of y

∠CDB=∠sDm ( vertically opposite angles are always equal)

2z=136-8y

2×60=136-8y

120°=136-8y

8y=136-120

8y=16

$y = \frac{16}{8}$

y=2