Question

Side QR of triangle PQR has been produced to S. If angle P: angleQ: angleR= 3:2:1 and RT is perpendicular on PR find angle TRS

Answer 1

answer :- 60°

Step-by-step explanation:

By angle sum property ,

3x + 2x + x = 180°

6x = 180°

x = 180/6

= 30°

Since angles PRQ , PRT , TRS are linear pair of angles ,

Angle TRS = 180° - ( 90° + 30° )

= 60°

Answer 2

Answer:

We have

$\angle$P : $\angle$Q : $\angle$R = 3:2:1.

Let $\angle$P = 3x, $\angle$Q = 2x and

$\angle$R = x.

Clearly $\angle$P, $\angle$Q and $\angle$R are the angles of a ∆PQR, then

$\angle$P + $angle$Q + $\angle$R = 180°.

$\hookrightarrow$$\bf{3x+2x+x=180°}$

$\hookrightarrow$$\bf{6x=180°}$

$\hookrightarrow$$\bf x=\frac{180}{6}$

$\hookrightarrow$$\bf\underline{x=30°}$

In ∆PQR, we have

Exterior $\angle$PRS = $\angle$P + $\angle$Q

= $\angle$PRT + $\angle$TRS = 3x + 2x

= 90° + $\angle$TRS = 3 * 30° + 2 * 30°

= 90° + $\angle$TRS = 90° + 60°

= $\angle$TRS = 60°.