We are given a triangle PQR with the following conditions:
We know the sum of angles in any triangle is $180^\circ$. Therefore:
$ \angle P + \angle Q + \angle R = 180^\circ $
Substitute Equation 1 into the triangle angle sum property:
$ ( \angle P + \angle R ) + \angle Q = 180^\circ $ $ 150^\circ + \angle Q = 180^\circ $
Solve for $ \angle Q $:
$ \angle Q = 180^\circ - 150^\circ $
$ \angle Q = 30^\circ $
Now, use Equation 2 and the value of $ \angle Q $ we found:
$ \angle P + 3\angle Q = 170^\circ $
Substitute $ \angle Q = 30^\circ $:
$ \angle P + 3(30^\circ) = 170^\circ $
$ \angle P + 90^\circ = 170^\circ $
Solve for $ \angle P $:
$ \angle P = 170^\circ - 90^\circ $
$ \angle P = 80^\circ $
Thus, $ \angle P $ is $80^\circ$.
A, B, C, D are points on a circle with AB=5 cm, BC=12 cm, AC=13 cm and AD=7cm. Then, the closest approximation of CD is
Point $A$ on a wheel of radius $r$ touches the horizontal plane at point $P$. It rolls without slipping, till point $A$ is at the highest position in the first turn. What is the final distance $AP$?
