Important Geometry theorems

###
__Theorem 3 __ : Angle sum property of a triangle.

__Theorem 3__: Angle sum property of a triangle.

The sum of the angles of a triangle is 180

^{0}.
In the picture above, PQR is a triangle with angles 1, 2 and
3

Then according to the theorem

Angle 1+Angle 2 +Angle 3 =180

^{0}###
__Theorem 4__

__Theorem 4__

**If a side of a triangle is produced then the exterior angle so**

**formed is equal to the sum of two interior opposite angles.**

In the picture above XYZ is a triangle whose side YZ is
extended

to R. 1, 2, and 3 are the interior angles of a triangle. Angle 1
and

Angle 2 are the interior angles opposite to the exterior angle 4

Then according to the theorem

Angle 4 = Angle 1+ Angle 2

**Let us do some questions based on these theorems.**

**Ques. - In the figure if QT is perpendicular PR , Angle TQR = 40**

^{0}**and**

**Angle SPR =30**

^{0}, find x and y .**Solution**: In triangle QTR

Angle TQR +Angle QRT +Angle QTR =180

^{0}
40

^{0}+ y + 90^{0}=180^{0}
y =180

^{0}-130^{0}
= 50

^{0}
Angle QSP = Angle SPR +Angle SRP

__Reason:Exterior angle__= sum of interior opposite angles

x = 30

^{0 }+y
x = 30

^{0}+50^{0}
x =80

^{0}^{}**Ques. - In the figure below XY II MN , Angle YXZ =350 and angle ZMN =530, find angle MZN.**

**Solution:**

We know that XY is parallel to MN.

Angle MNZ = Angle ZXY ( alternate interior angles)

= 35

^{0}
Now in triangle MZN

Angle ZMN +Angle MNZ +Angle MZN = 180

^{0}
53

^{0}+35^{0}+Angle MZN =180^{0}^{}

Angle MZN = 180

^{0}-88^{0}^{}

Angle =92

^{0}^{}

**Ques - In the figure given below If PQ and RS intersect at T, such that angle PRT =40**

^{0}, angle RPT=95^{0}and angle TSQ=75^{0}, find SQT.**Solution:**

In triangle PRT

40

^{0}+95^{0}+Angle RTP =180^{0}^{}

Angle RTP =180

^{0}-135^{0}^{}

Angle RTP =45

^{0}^{}

Angle STQ =Angle RTP ( vertically Opposite angle )

=45

^{0}^{}

Again in triangle TQS

Angle STQ + Angle SQT + Angle TSQ = 180

^{0}( Angle sum property)
45

^{0}+ Angle SQT +75^{0}=180^{0}^{}

Angle SQT =180

^{0}-120^{0}^{}

Angle SQT = 60

^{0}**Ques. In the figure if PQ is perpendicular to PS, PQ II SR Angle SQR =28**

^{0}and Angle QRT =65^{0}, then find the values of x and y .**Solution:**

Since PQ II SR

Angle QRT = Angle PQR (alternate interior angles )

65

^{0 }= x +28^{0}^{}

X= 65

^{0}-28^{0}^{}

= 37

^{0}^{}

In triangle PQS

Angle PSQ +Angle PQS + QPS = 180

^{0}( angle sum property)
Y +x +90

^{0}=180^{0}^{}

Y+37

^{0}+90^{0}=180^{0}^{}

Y =180

^{0}-127^{0}^{}

= 53

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