In quadrilateral MNPQ, angle N = angle M + 10 degrees, angle P = angle N + 10 degrees and angle Q = angle P + 10 degrees. Find all four angles.

Step-by-step solution

1. 1

   Express every angle in terms of one unknown. Let $\widehat{M}=x$. Chaining the three conditions,

   $\widehat{N}=x+10^{\circ},\qquad \widehat{P}=(x+10^{\circ})+10^{\circ}=x+20^{\circ},\qquad \widehat{Q}=x+30^{\circ}$

   Each condition refers to the previous angle, so the offsets accumulate rather than staying at $10^{\circ}$.

2. 2

   Use the angle sum of a quadrilateral. Any quadrilateral's interior angles total $360^{\circ}$ (split it into two triangles):

   $x+(x+10^{\circ})+(x+20^{\circ})+(x+30^{\circ})=360^{\circ}$

3. 3

   Solve the linear equation.

   $4x+60^{\circ}=360^{\circ}\quad\Longrightarrow\quad 4x=300^{\circ}\quad\Longrightarrow\quad x=75^{\circ}$

4. 4

   Write out all four angles.

   $\widehat{M}=75^{\circ},\quad \widehat{N}=85^{\circ},\quad \widehat{P}=95^{\circ},\quad \widehat{Q}=105^{\circ}$

   $\boxed{75^{\circ},\ 85^{\circ},\ 95^{\circ},\ 105^{\circ}}$

5. 5

   Check the sum and note the shortcut. $75+85+95+105=360$ ✓. Because the four angles form an arithmetic sequence, their mean is $360\div 4=90^{\circ}$, which must sit halfway between the middle two — and indeed $\tfrac{85+95}{2}=90^{\circ}$. That gives a one-line route: the two middle angles are $90\mp 5$ and the outer two are $90\mp 15$.

6. 6

   Confirm the shape is drawable. All four angles are strictly between $0^{\circ}$ and $180^{\circ}$, so the quadrilateral is convex and can be drawn without any reflex corner.