Question

In the triangle abc, be is perpendicular to ac , angle EBC =40 and angle DAC=30 find the value of x,y and z

Answer 1

Answer:

x=50°,y=80°,z=120°

Step-by-step explanation:

In ΔAEZ= 30°+90°+x=180°

a+120°=180°

a=180°-120°

a=60°

In line of EB

Z+60°=180°

z=120°

In line AD

120+c=180°

c=60°

InΔBDC

40+60+y=180

y=80

In polygon EDCB

x+90+100+120=360

x+310=360

x=50°°°°°°°

Answer 2

Given:

ΔABC

∠EBC $=40$°

∠DAC $=30$°

BE ⊥ AC

To Find:

Values of $x,y,z$

Solution:

Let AD and BE intersect at point P.

Since BE ⊥ AC, ∠E $=90$°.

Hence, ΔBEC and ΔAEP are right-angled triangles.

Sum of angles in a triangle is 180°.

In ΔBEC, ∠B and ∠E are known and ∠C can be determined using angle sum property. From the figure, ∠C is given by $x$.

∠B + ∠E + ∠C $=180$°

$40+90+x=180$

$130+x=180$

$x=180-130$

$x=50$°

In ΔAEP, ∠A $=30$°, ∠E $=90$° and ∠EPA is unknown. The sum of angles of ΔAEP is $180$°.

∠A + ∠E + ∠EPA$=180$

$30+90+<EPA=180$

$120+<EPA=180$

$<EPA=180-120$

$<EPA=60$°

∠EPA and ∠BCA are angles on a straight line BE, i.e., they are supplementary angles. If angles are supplementary, then their angle sum will be 180°. ∠BCA is given by $z$.

∠EPA + ∠BCA $=180$   (∵ supplementary angles)

$60+z=180$

$z=180-60$

$z=120$°

∠BPD and ∠BPA are angles lying on a straight line AD. Hence, their angle sum is $180$°.

∠BPD + ∠BPA $=180$

$<BPD+z=180$          (∵ supplementary angles)

$<BPD+120=180$

$<BPD=180-120$

$<BPD=60$°

∠BPD is an angle of ΔBPD. Sum of angles in this triangle is $180$°. ∠PBD and ∠BPD are known whereas ∠PDB is unknown and is given by $y$.

$<PBD+<BPD+<PDB=180$

$40+60+y=180$

$100+y=180$

$y=180-100$

$y=80$

Thus, $x=50,y=80,z=120$

In ΔABC, the values of $x,y,z$ are determined as 50°,80° and 120° respectively.