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Question

In a triangle PQR, if $\angle P + \angle R = 150^\circ$ and $\angle P + 3\angle Q = 170^\circ$, then $\angle P$ is equal to :

The correct answer is
$80^\circ$

Triangle Angle P Calculation

We are given a triangle PQR with the following conditions:

  • $ \angle P + \angle R = 150^\circ $ (Equation 1)
  • $ \angle P + 3\angle Q = 170^\circ $ (Equation 2)

We know the sum of angles in any triangle is $180^\circ$. Therefore:

$ \angle P + \angle Q + \angle R = 180^\circ $

Step 1: Find Angle Q

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 $

Step 2: Find Angle P

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 $80^\circ$.

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Important Questions from Geometry (Notes)

  1. PQR is a triangle. The bisectors of the internal angle $\angle Q$ and external angle $\angle R$ intersect at M. If $\angle QMR = 40^\circ$, then $\angle P$ is :
  2. A $2\text{ m}$ long ladder is to reach a wall of height $1.75\text{ m}$. The largest possible horizontal distance of the ladder from the wall could be
  3. 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

  4. An inclined plane rests against a horizontal cylinder of radius $R$. If the plane makes an angle of $30^\circ$ with the ground, the point of contact of the plane with the cylinder is at a height of
  5. 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 $AP$?

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