Question

in the given figure if triangle PTR is congruent to QTS, then show that angle TQS is equal to angle QRT + angle QSR​

Answer 1

Since,

∆PTR ≅ ∆QTS,

By CPCT

PT = QT

TR = TS

PR = QS

∠PTR = ∠QTS

∠TRP = ∠TSQ

∠RPT = ∠SQT

Let,

a = ∠TRQ

b = ∠QSR

x = ∠PTQ

Thus,

In ∆TPQ

∠TPQ = ∠TQP = 90 - $\frac{x}{2\\}$                     [Angle Sum and Isoceles Triangle]

Similarly,

In ∆TRS,

∠TRS = ∠TSR = 90 - $\frac{x}{2\\}$                     [Angle Sum and Isoceles Triangle]

Now,

90 - $\frac{x}{2\\}$ + 90 - [Angle sum in quadrilateral]

a = 90 - $\frac{x}{2\\}$ - y

Also,

90 - $\frac{x}{2\\}$ - y + b = 90 - [Angle S]

y = b

∠TQS = 90 - $\frac{x}{2\\}$

          = 90 - $\frac{x}{2\\}$ - y + y        [Adding and subtracting 'y']

          = ∠QRT + ∠QSR               [∠QRT = a, ∠QSR = b]

Therefore Proved

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Answer 2

Answer:

∆PTR ≅ ∆QTS,

By CPCT

PT = QT

TR = TS

PR = QS

∠PTR = ∠QTS

∠TRP = ∠TSQ

∠RPT = ∠SQT

Let,

a = ∠TRQ

b = ∠QSR

x = ∠PTQ

Thus,

In ∆TPQ

∠TPQ = ∠TQP = 90 - \begin{gathered}\frac{x}{2\\}\end{gathered}

2

x

[Angle Sum and Isoceles Triangle]

Similarly,

In ∆TRS,

∠TRS = ∠TSR = 90 - \begin{gathered}\frac{x}{2\\}\end{gathered}

2

x

[Angle Sum and Isoceles Triangle]

Now,

90 - \begin{gathered}\frac{x}{2\\}\end{gathered}

2

x

+ 90 - [Angle sum in quadrilateral]

a = 90 - \begin{gathered}\frac{x}{2\\}\end{gathered}

2

x

- y

Also,

90 - \begin{gathered}\frac{x}{2\\}\end{gathered}

2

x

- y + b = 90 - [Angle S]

y = b

∠TQS = 90 - \begin{gathered}\frac{x}{2\\}\end{gathered}

2

x

= 90 - \begin{gathered}\frac{x}{2\\}\end{gathered}

2

x

- y + y [Adding and subtracting 'y']

= ∠QRT + ∠QSR [∠QRT = a, ∠QSR = b]