1. In Fig. 6.39, sides QP and RQ of ▵PQR are produced to points S and T respectively. If ∠SPR=135° and ∠PQT=110°, find ∠PRQ.
solution :
Given,
∠SPR=135° and ∠PQT=110°
Find,
∠PRQ=?
In the Fig,
Side QP and RQ of â–µPQR are produced to points S and T respectively.
∠SPR+∠QPR=180°[ Linear pair of angels ]
⇒135°+∠QPR=180°[∴∠SPR=135°]
⇒∠QPR=180°-135°
⇒∠QPR=45°
Similarly,
∠PQT+∠PQR=180°[ Linear pair of angels ]
⇒110°+∠PQR=180°[∴∠PQT=110°]
⇒∠PQR=180°-110°
⇒∠PQR=70°
Now,
In â–µPQR,
∠PRQ+∠PQR+∠QPR=180°[ Angle sum property of a traingle ]
⇒∠PRQ+70°+45°=180°[∴∠PQR=70° and ∠QPR=45°]
⇒∠PRQ+115°=180°
⇒∠PRQ=180°-115°
⇒∠PRQ=65°
So,
The value of ∠PRQ is 65°.
2. In Fig. 6.40, ∠X=62°, ∠XYZ=54°. If YO and ZO are the bisectirs of ∠XYZ and ∠XZY respectively of ▵XYZ, find ∠OZY and ∠YOZ.
solution :
Given,
∠X=62°, ∠XYZ=54° and
YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ▵XYZ.
Find,
∠OZY=? and ∠YOZ=?
In â–µXYZ,
∠X+∠XYZ+∠XZY=180°[ Angle sum property of a traingle ]
⇒62°+54°+∠XZY=180°[∴∠X=62° and ∠XYZ=54°]
⇒116°+∠XZY=180°
⇒∠XZY=180°-116°
⇒∠XZY=64°
Now,
∠OZY=12∠XZY[Given]
⇒∠OZY=12×64°[∴∠XZY=64°]
⇒∠OZY=32°
Then,
∠OYZ=12∠XYZ[Given]
⇒∠OYZ=12×54°[∴∠XYZ=54°]
⇒∠OZY=27°
Now,
In â–µOZY,
∠OZY+∠ZYO+∠YOZ=180°[ Angle sum property of a traingle ]
⇒32°+27°+∠YOZ=180°[∴∠OZY=32° and ∠ZYO=27°]
⇒59°+∠YOZ=180°
⇒∠YOZ=180°-59°
⇒YOZ=121°
So,
The value of ∠OZY=27° and ∠YOZ=121°.
3. In Fig. 6.41, if AB ∣∣ DE, ∠BAC=35° and ∠CDE=53°, find ∠DCE.
solution :
Given,
AB ∣∣ DE, ∠BAC=35° and CDE=53°.
Find,
∠DCE=?
In the Fig,
∠BAC=∠CED[Alternate interior angles ]
⇒∠CED=35°[∴∠BAC=35°]
Now,
In â–µCDE,
∠CDE+∠CED+∠DCE=180°[ Angle sum property of a traingle ]
⇒53°+35°+∠DCE=180°[∴∠CDE=53° and ∠CED=35°]
⇒88°+∠DCE=180°
⇒∠DCE=180°-88°
⇒∠DCE=92°
So,
The value of ∠DCE is 92°.
4. In Fig. 6.42, if lines PQ and RS intersect at point T, such that ∠PRT=40°,∠RPT=95° and ∠TSQ=75°, find ∠SQT.
solution :
Given,
∠PRT=40°, ∠RPT=95° and ∠TSQ=75°
Find,
∠SQT=?
In â–µPRT,
∠PTR+∠TRP+∠RPT=180°[ Angle sum property of a traingle ]
⇒∠PTR+40°+95°=180°[∴∠TRP=40° and ∠RPT=95°]
⇒PTR+135°=180°
⇒∠PTR=180°-135°
⇒PTR=45°
In the Fig.
∠PTR=∠QTS[ Vertically opposit angles ]
∠QTS=45°[∴∠PTR=45°]
Now,
In â–µQTS,
∠SQT+∠TSQ+∠QTS=180°[ Angle sum property of a traingle ]
⇒∠SQT+75°+45°=180°[∴∠TSQ=75° and ∠QTS=45°]
⇒∠SQT+120°=180°
⇒∠SQT=180°-120°
⇒∠SQT=60°
So,
The value of ∠SQT is 60°.
5. In Fig. 6.43, if PQ⊥PS, PQ ∣∣ SR, ∠SQR=28° and ∠QRT=65°, then find the value of x and y.
solution :
Given,
PQ⊥PS, PQ ∣∣ SR, ∠SQR=28° and ∠QRT=65°.
Find,
x=? and y=?
In the Fig.
∠QRS+∠QRT=180°[ Linear pair of angels ]
⇒∠QRS+65°=180°[∴∠QRT=65°]
⇒∠QRS=180°-65°
⇒∠QRS=115°
Now,
In â–µQRS,
∠QSR+∠SQR+∠QRS=180°[ Angle sum property of a traingle ]
⇒∠QSR+28°+115°=180°[∴∠SQR=28° and ∠QRS=115°]
⇒∠QSR+143°=180°
⇒∠QSR=180°-143°
⇒∠QSR=37°
Then,
PQ ∣∣ SR and PS is transversal.
∠QPS+∠PSR=180°[ Interior angles on the same side of the transversal ]
⇒∠PSR+90°=180°[∴∠QPS=90°]
⇒∠PSR=180°-90°
⇒∠PSR=90°
⇒∠PSR=∠PSQ+∠QSR[ From Fig. ]
⇒90°=y+37°[ From Fig. and ∠QSR=37° ]
⇒y=90°-37°
⇒y=53°
Now,
In â–µPQS,
∠QSP+∠SPQ+∠PQS=180°[ Angle sum property of a traingle ]
⇒53°+90°+∠PQS=180°[∴∠QSP=53° and ∠SPQ=90°]
⇒143°+∠PQS=180°
⇒∠PQS=180°-143°
⇒∠PQS=37°
⇒x=37°[∴∠PQS=x]
So,
The value of x is 37° and y is 53°.
6. In Fig. 6.44, the side QR of ▵PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR=12∠QPR.
solution :
Given,
∠PQR bisect QT and ∠PRS bisect RT.
Find,
∠QTR=12∠QPR
In â–µPQR,
∠PRS=∠QPR+∠PQR[ Angle Sum Property of a Triangle ]
⇒∠PRS-∠PQR=∠QPR
⇒∠QPR=∠PRS-∠PQR-----(I)
Similarly,
In â–µQTR,
∠TRS=∠QTR+∠TQR[ Angle Sum Property of a Triangle ]
⇒∠TRS-∠TQR=∠QTR
⇒∠QTR=∠TRS-∠TQR-----(II)
Now,
∠PRS=∠PRT+∠TRS[ From Fig. ]
⇒∠PRS=∠TRS+∠TRS[ From Fig. ∠PRT=∠TRS ]
⇒∠PRS=2∠TRS
⇒∠TRS=12∠PRS-----(III)
Similarly,
∠PQR=∠PQT+∠TQR[ From Fig. ]
⇒∠PQR=∠TQR+∠TQR[ From Fig. ∠PQT=∠TQR ]
⇒∠PQR=2∠TQR
⇒∠TQR=12∠PQR-----(IV)
Now,
From (II), (III) and (IV).
∠QTR=12∠PRS-12∠PQR
⇒∠QTR=12(∠PRS-∠PQR)-----(V)
From (I) and (V).
⇒∠QTR=12∠QPR
Proved.
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