What Is a Vertex Angle?
A vertex angle is the angle formed where two sides of a figure meet at a corner, called a vertex. In its most common and most useful sense, it is the angle between the two equal sides (the legs) of an isosceles triangle — the angle that sits opposite the base. The two remaining angles, one at each end of the base, are the base angles, and in an isosceles triangle they are always equal to each other.
The vertex angle is also called the apex angle, because it sits at the apex — the corner opposite the base. So "vertex angle" and "apex angle" name the same angle in an isosceles triangle.
Because the three angles of any triangle add to $180^\circ$, the vertex angle and the two base angles are locked together. Knowing one fixes the other two.
Vertex Angle vs Base Angles — Which Is Which?
This is the distinction that decides every problem, so it is worth settling directly: how do you tell the vertex angle from the base angles in an isosceles triangle?
The rule is about position, not size:
The vertex angle is the angle between the two equal sides. It is the odd one out — it has no equal partner.
The base angles are the two angles at the ends of the base, opposite the two equal sides. They are always equal to each other.
A common confusion: the vertex angle is not always the largest, and it is not always at the top. If you rotate the triangle, the vertex angle rotates with it — it stays the angle between the equal sides no matter how the triangle sits on the page. In a tall, narrow isosceles triangle the vertex angle is small and the base angles are large; in a short, wide one the vertex angle is large. What identifies it is its location between the equal sides, never its measure.
The Vertex Angle Formula
Because the two base angles are equal, the relationship between the vertex angle and the base angles comes straight from the angle-sum rule. Let the vertex angle be $V$ and each base angle be $B$. The three angles add to $180^\circ$:
$$V + B + B = 180^\circ \quad\Rightarrow\quad V + 2B = 180^\circ.$$
Rearranging gives the two forms you actually use. To find the vertex angle from a base angle:
$$V = 180^\circ - 2B.$$
To find each base angle from the vertex angle:
$$B = \frac{180^\circ - V}{2}.$$
Here $V$ is the single vertex (apex) angle and $B$ is each of the two equal base angles. The formula is nothing more than "the leftover after the vertex angle, split evenly between the two equal corners" — which is exactly what the equal-sides condition forces.
The Vertex Angle of a Polygon
The word "vertex angle" stretches beyond the isosceles triangle. At any corner of a polygon, the angle formed by the two sides meeting there is a vertex angle (also called an interior angle at that vertex). For a regular polygon — one with all sides and all angles equal — every vertex angle is the same, and there is a clean formula for it. A polygon with $n$ sides has interior angles summing to $(n-2)\times 180^\circ$, so each of its $n$ equal vertex angles is:
$$\text{vertex angle} = \frac{(n-2)\times 180^\circ}{n}.$$
For a regular hexagon ($n = 6$), each vertex angle is $\frac{4\times 180^\circ}{6} = 120^\circ$. So "vertex angle" has two everyday meanings: the apex angle of an isosceles triangle (its dominant use), and the interior angle at any corner of a polygon. Both describe an angle at a vertex — the context tells you which.
Examples of Vertex Angle
With the definition, the apex-versus-base distinction, and the formula in place, here is the vertex angle in worked problems, moving from a direct calculation up to a polygon and a multi-step setup.
Example 1 - Each base angle of an isosceles triangle is $50^\circ$. Find the vertex angle.
Use $V = 180^\circ - 2B$ with $B = 50^\circ$:
$$V = 180^\circ - 2(50^\circ) = 180^\circ - 100^\circ = 80^\circ.$$
Final answer: the vertex angle is $80^\circ$.
Example 2 - The vertex angle of an isosceles triangle is $40^\circ$. A student is asked for the base angles and writes $180^\circ - 40^\circ = 140^\circ$ for each base angle
A first instinct is to subtract the vertex angle from $180^\circ$ and stop, giving $140^\circ$ for a base angle. Check it against the triangle: $140^\circ + 140^\circ = 280^\circ$ already, before the vertex angle is even added — far past the $180^\circ$ a triangle is allowed. So a single base angle cannot be $140^\circ$.
The leftover $140^\circ$ is the total for the two base angles together, and since they are equal it must be split between them:
$$B = \frac{180^\circ - V}{2} = \frac{180^\circ - 40^\circ}{2} = \frac{140^\circ}{2} = 70^\circ.$$
Final answer: each base angle is $70^\circ$.
Example 3 - An isosceles triangle has a vertex angle of $90^\circ$. Find each base angle and name the triangle
$$B = \frac{180^\circ - 90^\circ}{2} = \frac{90^\circ}{2} = 45^\circ.$$
Final answer: each base angle is $45^\circ$. With a $90^\circ$ vertex angle and two $45^\circ$ base angles, this is the isosceles right triangle — the familiar $45^\circ$–$45^\circ$–$90^\circ$.
Example 4 - The vertex angle of an isosceles triangle is twice each base angle. Find all three angles
Let each base angle be $x$; the vertex angle is $2x$. The three angles sum to $180^\circ$:
$$x + x + 2x = 180^\circ \quad\Rightarrow\quad 4x = 180^\circ \quad\Rightarrow\quad x = 45^\circ.$$
So the base angles are $45^\circ$ each and the vertex angle is $2(45^\circ) = 90^\circ$.
Final answer: $90^\circ$, $45^\circ$, $45^\circ$.
Example 5 - Find the vertex angle (interior angle at each corner) of a regular pentagon
A pentagon has $n = 5$ sides, so each vertex angle is:
$$\frac{(5-2)\times 180^\circ}{5} = \frac{3\times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ.$$
Final answer: each vertex angle of a regular pentagon is $108^\circ$.
Example 6 - In an isosceles triangle the vertex angle is $30^\circ$ more than a base angle. Find all three angles
Let each base angle be $x$; the vertex angle is $x + 30^\circ$. The angle sum gives:
$$x + x + (x + 30^\circ) = 180^\circ \quad\Rightarrow\quad 3x + 30^\circ = 180^\circ \quad\Rightarrow\quad 3x = 150^\circ \quad\Rightarrow\quad x = 50^\circ.$$
So the base angles are $50^\circ$ each and the vertex angle is $50^\circ + 30^\circ = 80^\circ$.
Final answer: $80^\circ$, $50^\circ$, $50^\circ$. (Quick check: $80 + 50 + 50 = 180^\circ$.)
Why the Vertex Angle Matters
Naming one angle the "vertex angle" is not bookkeeping — it is the hinge that makes isosceles geometry, and a lot of what comes after it, work.
It carries the symmetry - The line drawn from the vertex angle to the midpoint of the base is the triangle's line of symmetry: it bisects the vertex angle, hits the base at a right angle, and splits the whole triangle into two identical right triangles. Almost every isosceles-triangle proof and construction starts from this single line, which is why finding the vertex angle is step one.
It turns roof and truss design into one number - A symmetric gable roof or an A-frame is an isosceles triangle, and its pitch is set by the vertex angle at the ridge. Engineers size the rafters and the load on each support from that one angle — change the vertex angle and both base angles, and the whole load distribution, change with it.
It is where the regular polygon comes from - Tiling, gears, bolt heads, and stop signs are regular polygons, and each is defined by its repeated vertex angle. A stop sign is a regular octagon precisely because all eight of its vertex angles are equal ($135^\circ$ each) — the vertex-angle formula is the rule that makes the shape regular.
It anchors the unit circle later - Splitting an isosceles triangle at its vertex angle is the move that produces the special right triangles ($30^\circ$–$60^\circ$–$90^\circ$, $45^\circ$–$45^\circ$–$90^\circ$) students lean on all through trigonometry.
For a Class 7 student, the vertex angle is the moment "this triangle has two equal sides" stops being a label and becomes a tool — one named angle that unlocks the other two.
Where Students Trip Up on the Vertex Angle
Mistake 1: Treating a base angle as the vertex angle
Where it slips in: A problem gives one angle and the student assumes it is the vertex angle without checking whether it sits between the equal sides.
Don't do this: Grab whichever angle is given and feed it into $V = 180^\circ - 2B$ as if it were a base angle (or use it as the vertex) without reading the figure.
The correct way: Identify the angle by position first. The vertex angle is between the two equal sides; the base angles are at the ends of the base and are equal. Only after you know which is which do you pick the right form of the formula.
Mistake 2: Forgetting to divide the leftover between two base angles
Where it slips in: Finding a base angle from a known vertex angle.
Don't do this: Compute $180^\circ - V$ and call that a single base angle. That value is the combined total of both base angles.
The correct way: The two base angles share the leftover equally, so divide by two: $B = \dfrac{180^\circ - V}{2}$. The memorizer who recalls "$180$ minus the vertex" but not "then halve it" is the one this trap catches.
Mistake 3: Assuming the vertex angle is always the biggest or always on top
Where it slips in: Identifying the vertex angle in a triangle drawn sideways or upside down.
Don't do this: Pick the topmost or largest angle by reflex.
The correct way: The vertex angle is defined only by being between the two equal sides. Rotate the triangle and it rotates too. In a wide isosceles triangle the vertex angle is the largest; in a tall one it is the smallest — measure is never the test, position is.
Key Takeaways
The vertex angle (or apex angle) is the angle between the two equal sides of an isosceles triangle, opposite the base.
The two base angles, at the ends of the base, are always equal to each other.
From a base angle, $V = 180^\circ - 2B$; from the vertex angle, each base angle is $B = \dfrac{180^\circ - V}{2}$.
The vertex angle is identified by its position between the equal sides, never by being largest or on top.
"Vertex angle" also names the interior angle at any corner of a polygon, equal to $\dfrac{(n-2)\times 180^\circ}{n}$ in a regular polygon.
Practice These Problems to Solidify Your Understanding
Each base angle of an isosceles triangle is $65^\circ$. Find the vertex angle.
The vertex angle of an isosceles triangle is $100^\circ$. Find each base angle.
Find the vertex (interior) angle of a regular hexagon.
Answer to Question 1: $50^\circ$. Answer to Question 2: $40^\circ$ each. Answer to Question 3: $120^\circ$. If Question 2 gave you $80^\circ$, remember to divide the leftover by two before stopping (see Mistake 2).
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