Question

In figure, if PQ ⊥ PS, PQ || SR,
∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.​

Answer 1

$\bf \huge {\underline {\underline \red{AnSwEr}}}$

Given

• ∠SQR = 28°

• ∠QRT = 65°

To Find

• Value of x and y

Solution

∠SQR +∠QRT = 180° [ Linear Pair ]

➨∠SQR + 65° = 180°

➨∠SQR = 180° - 65°

➨∠SQR = 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°

Answer 2

$\bf \huge {\underline {\underline \red{AnSwEr}}}$

Given

∠SQR = 28°

∠QRT = 65°

To Find

Value of x and y

Solution

∠SQR +∠QRT = 180° [ Linear Pair ]

➨∠SQR + 65° = 180°

➨∠SQR = 180° - 65°

➨∠SQR = 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°