Question

important ha jaldi batto ​

Answer 1

GIVEN :-

• ∠PQR = ∠PRQ.

TO PROVE :-

• ∠PQS = ∠PRT

SOLUTION :-

By using the exterior angle property of triangle we have,

➳ ∠PQS = ∠QPR + ∠PRQ. ....Eq(1)

Similarly,

➳ ∠PRT = ∠QPR + ∠PQR. ....Eq(2)

Now substitute the given values in Equation 1.

➳ ∠PQS = ∠QPR + ∠PQR. ....Eq(1)

Hence the R.H.S of both the equation are equal . So the L.H.S of both the equation are also equal,

➳ ∠PRT = ∠PQS

Hence Proved

Answer 2

Diagram:

$\setlength{\unitlength}{1cm} \begin{picture}(6, 6) \thicklines\put(3, 2){\vector(1,0){3}} \put(3, 2){\vector( - 1,0){3}} \put(1, 2){\line(1,1){2}} \put(5, 2){\line( - 1,1){2}} \multiput(0.5,1.8)(5,0){2}{\line(0,1){0.5}}\put(3,4.1){\sf P} \put(1,1.6){\sf Q} \put(0.2,1.6){\sf S} \put(4.8,1.6){\sf R}\put(5.6,1.6){\sf T} \put(2,5){\Large @BLACKSKULL} \end{picture}$

Given:

• $\angle$ PQR = $\angle$ PRQ

Prove That:

• $\angle$ PQS = $\angle$ PRT

Solution:

In the given fig. ST is a straight line and ray QP stands on it.

So,

$\displaystyle{\spadesuit\sf \angle PQR + \angle PQS = {180}^{ \circ} \qquad \big\lgroup{ \because Linear \: Pair } \big\rgroup }$

Now,

$\displaystyle{\dashrightarrow\sf \angle PQR + \angle PQS = {180}^{ \circ}}$

$\displaystyle{\dashrightarrow\sf \angle PQR = {180}^{ \circ} - \angle PQS }$

$\displaystyle{\to\sf \angle PQR = {180}^{ \circ} - \angle PQS }........1$

Again

$\displaystyle{\spadesuit\sf \angle PRQ + \angle PRT = {180}^{ \circ} \qquad \big\lgroup{ \because Linear \: Pair } \big\rgroup}$

Now,

$\displaystyle{\dashrightarrow\sf \angle PRQ + \angle PRT = {180}^{ \circ}}$

$\displaystyle{\dashrightarrow\sf \angle PRQ = {180}^{ \circ} - \angle PRT }$

$\displaystyle{\to\sf \angle PRQ = {180}^{ \circ} - \angle PRT }......2$

It is given that $\angle$ PQR = $\angle$ PRQ

Using eq(1) and (2)

• $\angle$ PQR = 180° - $\angle$ PQS
• $\angle$ PRQ = 180° - $\angle$ PRT

we, get

$\displaystyle{\leadsto\sf {180}^{ \circ} - \angle PQS = {180}^{ \circ} - \angle PRT }$

$\displaystyle{\leadsto\sf - \angle PQS = - \angle PRT }$

$\displaystyle{\leadsto\sf \angle PQS = \angle PRT }$

$\displaystyle{\therefore\sf \angle PQS = \angle PRT }$

Hence, Proved