in ∆PQR, angle PQR =90° .QS perpendicular to PR and ST perpendicular to QR. if PQ =6 cm and PS =3 cm, find the area∆QST/area∆RST About the author Remi

Answer: Final Answer : 1/3 Steps and Understanding : 1) In PQS, cos (P) = PS/PQ => cos(P) = 3/6 = 1/2 => P = 60° 2) Now, finding other angles as shown in pic by using properties of triangle. 3) In PQS, tan(60°) = QS/PS => QS = 3* tan(60°) = 3√3cm In QST, sin(60°) = ST/QS => ST =( 3√3) * (√3/2 ) => ST = 9/2 cm. cos(60°) = QT/QS => 1/2 = QT/QS => QT = 3√3 * (1/2) => QT = 3√3/2 cm 4) In RST, tan(30°) = ST/TR => 1/√3 = (9/2)/ TR => TR = 9√3/2 cm. 5) Now, Finally area ( QST) / area (RST) = ( 1/2 * QT* ST )/( 1/2 * RT* ST) =QT/RT (3√3/2)/(3√3/2) =1/3 Hope this will help you Plz mark this answer as brainliest Reply

Answer:

Final Answer : 1/3

Steps and Understanding :

1) In PQS,

cos (P) = PS/PQ

=> cos(P) = 3/6 = 1/2

=> P = 60°

2) Now, finding other angles as shown in pic by using properties of triangle.

3) In PQS,

tan(60°) = QS/PS

=> QS = 3* tan(60°) = 3√3cm

In QST,

sin(60°) = ST/QS

=> ST =( 3√3) * (√3/2 )

=> ST = 9/2 cm.

cos(60°) = QT/QS

=> 1/2 = QT/QS

=> QT = 3√3 * (1/2)

=> QT = 3√3/2 cm

4) In RST,

tan(30°) = ST/TR

=> 1/√3 = (9/2)/ TR

=> TR = 9√3/2 cm.

5) Now, Finally

area ( QST) / area (RST) =

( 1/2 * QT* ST )/( 1/2 * RT* ST)

=QT/RT

(3√3/2)/(3√3/2)

=1/3

Hope this will help you

Plz mark this answer as brainliest

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