Question

in figure 6.43,if PQ parallel PS, PQ perpendicular SR, SQR=28 and QRT=65, then find the values of x and y​

Answer 1

Answer:

x+y+3=0

then

y=−3−x

x-101y-2-3-4

3x−2y+4=0

then

x=32y−4

x-125y-202The point of intersection is (-1, -2)

Answer 2

Answer:

x=37 and y=53

Explanation:

Given

∠SQR = 28°

∠QRT = 65°

To Find

Value of x and y

Solution

∠SRQ +∠QRT = 180° [ Linear Pair ]

➨∠SRQ + 65° = 180°

➨∠SRQ = 180° - 65°

➨∠SRQ = 115°

∠SQR + ∠SRQ + ∠QSR = 180° [Angle Sum Property of triangle ]

➨ 28° + 115° + ∠QSR = 180°

➨ ∠QSR + 143° = 180°

➨ ∠QSR = 180° - 143°

➨ ∠QSR = 37°

∠PSR = ∠QSR + ∠y

➨90° = 37° + ∠y

➨∠y = 90° - 37°

➨∠y = 53°

∠x + ∠y + ∠SPQ = 180° [ Angle Sum Property of triangle ]

➨∠x + 53° + 90° = 180°

➨∠x + 143° = 180°

➨∠x = 180° - 143°

➨∠x = 37°

Therefore, x = 37° and y = 53°