Question

In the given figure, if∠P= 25°,andPS=ST=QT=QR then the measure of ∠R is

Answer 1

Answer:

130

Step-by-step explanation:

because P is 25 so Q is also 25

a triangle has 180 so 180-50 is 130

Answer 2

Given:

In the given figure, if ∠P= 25°,andPS=ST=QT=QR then the measure of ∠R is?

Solution:

Know that the angle opposite to the equal sides are equal.

Refer the question figure,

Take triangle PST,

$PS=ST$   (given)

$% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiiIaTaam% 4uaiaadcfacaWGubGaeyypa0JaeyiiIaTaamiuaiaadsfacaWGtbaa% aa!3F44!\[\angle SPT = \angle PTS\]$ (angle opposite to the equal side)

$% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiiIaTaam% 4uaiaadcfacaWGubGaeyypa0JaaGOmaiaaiwdadaahaaWcbeqaaiab% lIHiVbaaaaa!3E02!\[\angle SPT = {25^ \circ }\]$ (given)

$% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyinIWLaey% iiIaTaamiuaiaadsfacaWGtbGaeyypa0JaaGOmaiaaiwdacqGHWcaS% aaa!3FC5!\[\therefore \angle PTS = 25^\circ \]$

Evaluate $% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiiIaTaam% iuaiaadofacaWGubaaaa!3A1A!\[\angle PST\]$,

$\[\begin{array}{l} \Rightarrow \angle PST = 180^\circ - 25^\circ - 25^\circ \\ \Rightarrow \angle PST = 130^\circ \end{array}\]$

Take triangle STQ,

$% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taey% iiIaTaamyuaiaadofacaWGubGaeyypa0JaaGymaiaaiIdacaaIWaGa% eyiSaaRaeyOeI0IaaGymaiaaiodacaaIWaGaeyiSaalaaa!46AC!\[ \Rightarrow \angle QST = 180^\circ - 130^\circ \]$

$% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taey% iiIaTaamyuaiaadofacaWGubGaeyypa0JaaGynaiaaicdacqGHWcaS% aaa!40E3!\[ \Rightarrow \angle QST = 50^\circ \]$

$% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeSynIeLaam% 4uaiaadsfacqGH9aqpcaWGrbGaamivaaaa!3B95!\[ST = QT\]$ (given)

$\[\begin{array}{l}\therefore \angle QST = \angle SQT\\\therefore \angle SQT = 50^\circ \end{array}\]$

Evaluate $% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiiIaTaam% 4uaiaadsfacaWGrbaaaa!3A1B!\[\angle STQ\]$,

$\[\begin{array}{l} \Rightarrow \angle STQ = 180^\circ - 50^\circ - 50^\circ \\ \Rightarrow \angle STQ = 80^\circ \end{array}\]$

Take triangle QTR,

$\[\begin{array}{l} \Rightarrow \angle QTR = 180^\circ - 80^\circ - 25^\circ \\ \Rightarrow \angle QTR = 75^\circ \end{array}\]$

Since, $% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeSynIeLaam% yuaiaadsfacqGH9aqpcaWGrbGaamOuaaaa!3B91!\[QT = QR\]$ (given)

$\[\begin{array}{l}\therefore \angle QRT = \angle QTR\\\therefore \angle QRT = 75^\circ \end{array}\]$

Hence, the measure of $% MathType!MTEF!2!1!+-% feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn% hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr% 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9% vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiiIaTaam% Ouaiabg2da9iaaiEdacaaI1aGaeyiSaalaaa!3CDD!\[\angle R = 75^\circ \]$.