Exercise 4.4

 

1.In the given triangles, find out x,  y and  z

Sol. In fig(i)

$$x^{\circ}=50^{\circ}+60^{\circ}$$

($$\therefore$$  exterior angle is equal to sum of the opposite interior angles)

$$\therefore$$ x=$$110^{\circ}$$

In fig (ii)

$$z^{\circ}=60^{\circ}+70^{\circ}$$

($$\therefore$$  exterior angle is equal to sum of the opposite interior angles)

$$\therefore x=130^{\circ}$$

In fig(iii)

$$y^{\circ}=35^{\circ}+45^{\circ}$$=$$80^{\circ}$$

($$\therefore$$  exterior angle is equal to sum of the opposite interior angles)

2.In the given figure AS || BT; ∠4 = ∠5$$\overline{SB}$$  bisects ∠AST. Find the measure of ∠1.

Sol. Given AS || BT

$$\angle 4=\angle 5$$ and $$\overline{SB}$$ bisects $$\angle AST$$.

$$\therefore$$ BY PROBLEM

$$\angle 2=\angle 3$$.........(i)

for the lines $$AS\parallel BT$$

$$\angle 2=\angle 5$$($$\because$$ alternate interior angles)

$$\therefore$$ In $$\Delta BST$$

$$\angle 3=\angle 5=\angle 4$$

Hence $$\Delta BST$$ is equilateral triangle and each of its angle is equal to $$60^{\circ}$$

$$\therefore \angle 3=\angle 2=60^{\circ}$$

Now$$ \angle 1+\angle 2+\angle 3=180^{\circ}$$

$$\therefore \angle 1+60^{\circ}+60^{\circ}=180^{\circ}$$

$$\because$$ angle at a point on a line

$$\therefore \angle$$ 1= $$180^{\circ}-120^{\circ}=60^{\circ}$$

3.In the given figure$$ AB || CD; BC || DE $$ then find the values of x and y.

Sol.Given that $$ AB || CD; BC || DE $$.

$$\therefore$$ 3x=$$105^{\circ}$$ ($$\because$$ alternate interior angles for $$AB\parallel CD$$)

X= $$\frac{105^{\circ}}{3}$$ = $$35^{\circ}$$

Also $$BC\parallel DE$$

$$\therefore \angle D$$=$$105^{\circ}$$

$$\because$$ alternate interior angles

Now in $$\Delta CDE $$

$$24^{\circ}+105^{\circ}+Y$$=$$108^{\circ}$$

$$\because$$ angle sum property

$$\therefore y$$=$$180^{\circ}-129^{\circ}$$=$$51^{\circ}$$

4.In the adjacent figure $$BE ⊥ DA$$ and $$CD⊥DA$$ then prove that $$m∠1 ≅ m∠3$$.

Sol.Given that $$CD⊥DA$$ and $$BE ⊥ DA$$.

$$\Rightarrow$$ Two lines CD and BE are perpendicular to the same line DA.

$$\Rightarrow CD\parallel BE$$(or)

$$\angle D=\angle E\Rightarrow CD\parallel BE$$

$$\because$$ corresponding angles for CD and BE and DA are transversal

Now $$m\angle 1=m\angle 3$$

$$\because$$ altenate interior angles for the lines $$CD\parallel BE ; DB$$ are transversal

Hence proved.

5.Find the values of x, y for which the lines AD and BC become parallel.

Sol.For the loines AD and BC to be parallel $$x-y- 30^{\circ}$$(corresponding angles) ...........(1)

2x=8y ........(2)

($$\because$$  alternate interior angles)

solving(1)&(2)

(1)x(2)=2x-2y=60

2         =2x-8y=0

                 6y=60

y=$$\frac{60}{6} =10^{\circ}$$

substituting y=$$10^{\circ}$$ in eq.(1)

$$x-10^{\circ}=30^{\circ}$$

$$\Rightarrow x=40^{\circ}$$

$$\therefore x=40^{\circ}$$ and $$y=10^{\circ}$$

6. Find the values of x and y in the figure.

Sol.From the figure$$ y+ 140^{\circ}=180^{\circ}$$

$$\because$$ linear pair of angles

$$\therefore y=180^{\circ}-140^{\circ}=40^{\circ}$$

and $$X^{\circ}=30^{\circ}+40^{\circ}=70^{\circ}$$

7.In the given figure segments shown by arrow heads are parallel. Find the values of x and y.

Sol.From the figure

$$X^{\circ}=30^{\circ}$$ ($$\because$$ alternate interior angles)

$$y^{\circ}=45^{\circ}+X^{\circ}$$ ($$\because$$ exterior angles of a triangle =sum of opp.interior angles)

y=$$45^{\circ}+30^{\circ}=75^{\circ}$$

8. In the given figure sides QP and RQ of $$∠PQR$$ are produced to points S and T respectively. If  $$∠SPR = 135°$$ and  $$∠PQT = 110°$$, find $$∠PRQ$$.

Sol.Given that$$\angle SPR =135^{\circ} and \angle PQT =110^{\circ}$$

From the figure

$$\angle SPR+\angle RPQ=180^{\circ}$$

$$\angle PQT+\angle PQR=180^{\circ}$$

[$$\because$$ linear pair of angles ]

$$\Rightarrow \angle RPQ=180^{\circ}-\angle SPR$$

=$$180^{\circ}-135^{\circ}$$=$$45^{\circ}$$

$$\Rightarrow \angle PQR=180^{\circ}-\angle PQT$$

=$$180^{\circ}-110^{\circ}=70^{\circ}$$

Now in $$\Delta PQR$$

$$\angle RPQ+\angle PQR+\angle PRQ=180^{\circ}$$

($$\because$$ angle sum property)

$$\therefore 45^{\circ}+70^{\circ}+\angle PQR$$=$$180^{\circ}$$

$$\therefore \angle PQR$$=$$180^{\circ}-115^{\circ}$$=$$65^{\circ}$$

9.In the given figure,  $$∠X = 62°,  ∠XYZ = 54°$$. In $$ΔXYZ$$ If YO and ZO are the bisectors of  $$∠XYZ $$and $$∠XZY$$ respectively find  $$∠OZY and  ∠YOZ$$.

Sol.Given that $$∠X = 62°$$,  $$∠Y = 54°$$. YO and ZO are the bisectors of  $$∠Y $$ and $$∠Z$$ in  $$\Delta XYZ$$

$$\angle X+\angle XYZ+\angle XZY$$=$$180^{\circ}$$

$$62^{\circ}+54^{\circ}+\angle XZY$$=$$180^{\circ}$$

$$\Rightarrow \angle XYZ$$=$$180^{\circ}-116^{\circ}$$=$$64^{\circ}$$

Also in $$\Delta OYZ$$

$$\angle OYZ$$=$$\frac{1}{2}\angle XYZ$$=$$\frac{1}{2}X54^{\circ}$$=$$27^{\circ}$$

($$\because$$ YO is bisector of $$\angle XYZ$$)

$$\angle OZY$$=$$\frac{1}{2}\angle XZY$$=$$\frac{1}{2}X64^{\circ}$$=$$32^{\circ}$$

($$\because$$ OZ is bisector of $$\angle XYZ$$)

And $$\angle OYZ+\angle OZY+\angle YOZ$$=$$180^{\circ}$$

($$\because$$ angle sum property, $$\Delta OYZ$$

$$\Rightarrow 27+32^{\circ}+\angle YOZ$$=$$180^{\circ}$$

$$\Rightarrow \angle YOZ$$=$$180^{\circ}-59^{\circ}$$=$$121^{\circ}$$

10.In the given figure if $$AB || DE,  ∠BAC = 35° and ∠CDE = 53°, find  ∠DCE$$.

Sol.Given that $$AB || DE$$, $$∠CDE = 53°$$;$$∠BAC = 35°$$

Now $$\angle E$$=$$35^{\circ}$$

($$\because$$ alternate interior angles)

Now in $$\Delta CDE$$

$$\angle C+\angle D+\angle E$$=$$180^{\circ}$$

($$\because$$ ANGLE SUM PROPERTY,$$\Delta CDE$$)

$$\therefore \angle DCE+53^{\circ}+35^{\circ}$$=$$180^{\circ}$$

$$\Rightarrow \angle DCE$$=$$180^{\circ}-88^{\circ}$$=$$92^{\circ}$$

11.In the given figure  if line segments PQ and RS intersect at point T, such that $$∠PRT = 40°,  ∠RPT = 95° and ∠TSQ = 75°, find  ∠SQT$$.

Sol.Given that$$ ∠PRT = 40°,  ∠RPT = 95° and ∠TSQ = 75°$$

In $$\Delta PRT\angle P+\angle R+\angle PTR$$=$$180^{\circ}$$

($$\because$$ angle sum property)

$$95^{\circ}+40^{\circ}+\angle PTR$$=$$180^{\circ}$$

$$\Rightarrow \angle PTR$$=$$180^{\circ}-135^{\circ}$$=$$45^{\circ}$$

Now $$\angle PTR$$=$$\angle STQ$$

($$\because$$ vertically opposite angles)

In $$ \Delta STQ\angle S+\angle Q+\angle STQ$$=$$180^{\circ}$$

($$\because$$ angle sum property)

$$75^{\circ}+\angle SQT+45^{\circ}$$=$$180^{\circ}$$

$$\therefore \angle SQT$$=$$180^{\circ}-120^{\circ}$$=$$60^{\circ}$$

12.In the adjacent figure, ABC is a triangle in which $$∠B = 50° and ∠C = 70°$$.  Sides AB and AC are produced. If ‘z’ is the measure of the angle between the bisectors of the exterior angles so formed, then find ‘z’

Sol.Given that $$\angle B$$ = $$50^{\circ}$$ ; $$\angle C=70^{\circ}$$

Angle between bisectors of exterior angles B and C is Z.

From the figure

$$50^{\circ}+2x$$=$$180^{\circ}$$

70+2y=$$180^{\circ}$$

($$\because$$ linear pair of angles)

$$\therefore$$ 2x=$$180^{\circ}-50$$

2x=$$130^{\circ}$$

x=$$\frac{130}{2}$$

=$$65^{\circ}$$

2y=$$180^{\circ}-70^{\circ}$$

2y=$$110^{\circ}$$

y=$$\frac{110^{\circ}}{2}$$

=$$55^{\circ}$$

Now in $$\Delta BOC$$

x+y+z=$$180^{\circ}$$ ($$\because$$ angle sum property)

$$65^{\circ}+55^{\circ}+z$$=$$180^{\circ}$$

z=$$180{\circ}-120^{\circ}$$=$$60^{\circ}$$

 

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