In the figure- angle QPR-67-0 and angle SPR-72-0 and RP is a diameter of the circle- then angle QRS is equal to-18-023-041-067-0

Question

Toppr Admin · Accepted Answer

Angle in a semicircle is 900-Thus -x2220-PQR-x2220-PSR-900-x2220-QPS-x2220-QPR-x2220-RPS-670-720-1390Sum of all angles of quadrilateral must be 3600-Thus- 900-900-1390-x2220-QRS-3600-x2234- -x2220-QRS-410

In the figure, if $∠ Q P R = 67^{0}$ and $∠ S P R = 72^{0}$ and $R P$ is a diameter of the circle, then $∠ Q R S$ is equal to:

$18^{0}$
$23^{0}$
$41^{0}$
$67^{0}$

Angle in a semicircle is $90^{0} .$
Thus $∠ P Q R = ∠ P S R = 90^{0}$
$∠ Q P S = ∠ Q P R + ∠ R P S = 67^{0} + 72^{0} = 139^{0}$
Sum of all angles of quadrilateral must be $360^{0} .$
Thus, $90^{0} + 90^{0} + 139^{0} + ∠ Q R S = 360^{0}$
$∴$ $∠ Q R S = 41^{0}$

In the figure, if ∠QPR=670 and ∠SPR=720 and RP is a diameter of the circle, then ∠QRS is equal to:180230410670

Angle in a semicircle is 900.Thus ∠PQR=∠PSR=900∠QPS=∠QPR+∠RPS=670+720=1390Sum of all angles of quadrilateral must be 3600.Thus, 900+900+1390+∠QRS=3600∴ ∠QRS=410

In the figure, if $∠ Q P R = 67^{0}$ and $∠ S P R = 72^{0}$ and $R P$ is a diameter of the circle, then $∠ Q R S$ is equal to:

$18^{0}$
$23^{0}$
$41^{0}$
$67^{0}$

Angle in a semicircle is $90^{0} .$
Thus $∠ P Q R = ∠ P S R = 90^{0}$
$∠ Q P S = ∠ Q P R + ∠ R P S = 67^{0} + 72^{0} = 139^{0}$
Sum of all angles of quadrilateral must be $360^{0} .$
Thus, $90^{0} + 90^{0} + 139^{0} + ∠ Q R S = 360^{0}$
$∴$ $∠ Q R S = 41^{0}$