Lines Angles

Exercise 6.3

Question 1: In the following figure, sides QP and RQ of ΔPQR are produced to points S and T respectively. If ∠SPR = 135°and ∠PQT = 110°, find the value of ∠PRQ.

lines And angles 2

Answer: On the line QS, ∠QPR + ∠SPR = 180°
Or, ∠QPR = 180° - 135° = 45°
Similarly, on the line TR, ∠TQP + ∠PQR = 180°
Or, ∠PQR = 180° - 110° = 70°
Now we have values of two angles of the given triangle so value of the third angle can be calculated as follows:
∠PRQ = 180° - (70° + 45°) = 65°


Question 2: In the following figure ∠X = 62° and ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ΔXYZ, find values of ∠OZY and ∠YOZ.

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Answer: In ΔXYZ,
∠XYZ + ∠YXZ + ∠XZY = 180°
Or, ∠XZY = 180° - (62° + 54°) = 64°

As per the question YO and ZO are bisectors of ∠XYZ and ∠XZY respectively
Hence, ∠OYZ = 1/2∠XYZ = 54/2 = 27°
And, ∠OZY = 1/2∠XZY = 64/2 = 32°
Now, for ΔOYZ,
∠YOZ = 180° - (∠OYZ + ∠OZY)
= 180° - (27° + 32°) = 121°
Required answers are 32° and 121°


Question 3: In the following figure AB|DE, ∠BAC = 35° and ∠CDE = 53°, find the value of ∠DCE.

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Answer: ∠BAC = ∠CED = 35° (alternate angles)
In ΔDCE, ∠DCE + ∠CDE + ∠CED = 180° (angle sum of triangle)
Hence, ∠DCE = 180° - (53° + 35°) = 92°

Question 4: In the following figure, lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°. Find the value of ∠SQT.

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Answer: In ΔPRT,
∠PRT + ∠RPT + ∠PTR = 180°
Or, ∠PTR = 180° - (95° + 40°) = 45°

As we know opposite angles are equal so ∠PTR = ∠STQ = 45°
Now, in ΔQST,
∠QST + ∠STQ + ∠SQT = 180°
Or, ∠SQT = 180° - (75° + 45°) = 60°

Question 5: In the following figure, PQ⊥PS, PQ||SR, ∠SQR = 28° and ∠QRT = 65°. Find the values of x and y.

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Answer: PQ || ST (given)

So, ∠PQR = ∠QRT (alternate angles)

Or, `28°+x=65°`

Or, `x=65°-28°=37°`

In ΔSPQ:

`∠SPQ+x+y=180°`

Or, `90°+37°+y=180°`

Or, `127°+y=180°`

Or, `y=180°-127°=53°`

So, `x=37°` and `y=53°`

Question 6: In the following figure, the side QR of ΔPQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = 1/2∠QPR.

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Answer: In ΔPQR
∠SRP = ∠QPR + ∠PQR
Or, 1/2∠SRP = 1/2∠QPR + 1/2∠PQR
Or, ½ ∠QPR = 1/2∠SRP – 1/2∠PQR ……………(1)

In ΔTQR
∠SRT = ∠QTR + ∠TQR
Or, 1/2∠SRP = ∠QTR + 1/2∠PQR
Or, ∠QTR = 1/2∠SRP – 1/2∠PQR ……………..(2)

As RHS of both equations are same, so following can be written:
∠QTR = 1/2∠QPR



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