Given: \(∠\)XYZ = 64° and Ray YQ bisects \(∠\)PYZ.
To Find: \(∠\)XYQ and Reflex \(∠\)QYP
It can be observed that PX is a line. Rays YQ and YZ stand on it.
Hence, \(∠\)XYZ +\(∠\)ZYP = 180°
By substituting, \(∠\)XYZ = 64°, we get
64° +\(∠\)ZYP = 180°
∴ \(∠\)ZYP = 116°
As YQ bisects \(∠\)ZYP,
\(∠\)ZYQ = \(∠\)QYP
Or \(∠\)ZYP = 2\(∠\)ZYQ
∴ \(∠\)ZYQ = \(∠\)QYP = 58°
Then \(∠\)XYQ = \(∠\)XYZ + \(∠\)ZYQ
\(∠\)XYQ = a + 64°
\(⇒\) \(∠\)XYQ = 58° + 64° = 122°.
\(∠\)XYQ = 122°
Now, reflex \(∠\)QYP = 180°+\(∠\)XYQ
Since \(∠\)XYQ = 122°
we have
∠QYP = 180°+122°
∴ \(∠\)QYP = 302°
A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?