Triangles

Suppose we have a triangle with angles $30°$ and $50°$. Work out the third angle.

Solution:

Let's denote the missing angle by $\alpha$. Since the three angles in a triangle add up to $180°$, we can say:

$30°+50°+\alpha =180°$

Therefore,

$80°+\alpha =180°$.

Subtracting $110°$ from both sides, we obtain:

$\alpha =180°-80°=100°$

Therefore, the missing angle is $100°$.

Suppose the base of a triangle is $10cm$ and the height is $12cm$. Work out the area of the triangle.

Solution:

Using the fact that:

$AreaofaTriangle=\frac{1}{2}\times base\times height$

We can say that:

$AreaofaTriangle=\frac{1}{2}\times 10cm\times 12cm=60c{m}^2$

Therefore, the area of this triangle is $60c{m}^2$.

If we have a triangle with side lengths $3cm$, $4cm$, and $5cm$, what would the perimeter be?

Solution:

Using the formula for the perimeter, we have that:

$P=3+4+5=12cm$

So, the perimeter of this triangle would be $12cm$.

Suppose the three angles of a triangle are $90°$, $30°$, and $60°$. In this case, since one of the angles is a right angle, it is a right-angled triangle. However, since all three of the angles are different, it is also a scalene triangle.

Now, suppose we have another right-angled triangle with angles of $90°$, $45°$, and $45°$. In this case, it is a right-angled triangle and also an isosceles triangle because two of the angles are the same.

It's not possible for a triangle to be both equilateral and right-angled, however. To fit the definition of an equilateral triangle, all of the angles would need to be the same, and to fit the definition of a right-angled triangle, one of the angles would need to be $90°$. This means that the triangle would need to have three angles of$90°$, like so:

$90°+90°+90°=270°\ne 180$

However, the angles of a triangle have to add up to $180°$! Thus, right-angled triangles can also be classified as either isosceles or scalene.

Suppose we have the below triangle. Work out the size of the size labelled $x$:

Right-angled triangle with missing side - StudySmarter Originals

Solution:

For this right-angled triangle, we can see that $x$ is the hypotenuse, so we label it as $c$ to fit our formula. So, let's now label the other sides as $a=3$ and $b=4$.

Applying the Pythagorean theorem, we can say that:

$a^2+b^2=c^2$

Now, substituting in our values of $a$,$b$, and $c$, we get:

$3^2+4^2=x^2$

$9+16=x^2$

$25=x^2$

Taking the square root of both sides,

$x=\sqrt{25}=5$

Therefore, the length of the triangle's hypotenuse is $x=5cm$.

A triangle has two angles $52°$ and $38°$. Show that this triangle is right-angled.

Solution:

Let's first define the missing angle to be $x°$. Since angles in a triangle sum to $180°$, we have:

$52°+38°+x°=180°$

Therefore,

$90°+x°=180°$

Subtracting $90°$ from both sides, we obtain:

$x=180°-90°=90°$.

Thus, the missing angle is $90°$, which is a right angle. From this, we know that it is a right-angled triangle.

In the below isosceles triangle $MNO$, we know that $MN=OM$ and $\angle MNO=42°$. Work out the size of the other two angles.

Triangle example finding missing angle - StudySmarter Originals

Solution:

Since $MN=OM$, we know that$\angle MON=42°$. Now, since angles in a triangle sum to $180°$, we can say:

$42°+42°+\angle NMO=180°$.

Therefore,

$84°+\angle NMO=180°$

Subtracting $84°$ from both sides, we obtain:

$\angle NMO=180°-84°=96°$

So, $\angle MON=42°$ and $\angle NMO=96°$

In the below triangle, $△ADC$ is equilateral and $\angle CAB=32°$. Work out the size of $\angle ACB$ and $\angle ABC$.

Triangle example finding missing angles - StudySmarter Originals

Solution:

Firstly, since $△ADC$ is equilateral, we can say that each of the angles within it are $60°$. So, $\angle DCA=60°$.

Since angles on a straight line sum to $180°$, we have:

id="2869227" role="math" $$\begin{gathered} \angle ACB=180°-\angle DCA=120° \\ \angle ACB=180°-60°=120° \end{gathered}$$

With this information, we can work out $\angle ABC$:

id="2869236" role="math" $$\begin{gathered} \angle ACB+\angle CAB+\angle ABC=180° \\ 120°+32°+\angle ABC=180° \end{gathered}$$

$152°+\angle ABC=180°$

Subtracting $152°$ from both sides, we get:

$\angle ABC=180°-152°=28°$.

So $\angle ACB=120°$ and $\angle ABC=28°$.

A given isosceles triangle has an angle of $30°$. Work out two possibilities for the sizes of its other two angles.

Solution:

Firstly, since it is isosceles, two of the angles must be the same. If one of the angles is $30°$, then one of the other angles could be $30°$as well to meet this property. In this case, that would make the third and last angle $120°$ by the following calculation:

$180°-30°-30°=120°$

So, our isosceles triangle could have angles: $30°,30°,120°$.

Another possible scenario is that only one of the angles is $30°$. In this case, the other two angles would need to be the same. Since angles in a triangle sum to $180°$, the other two angles would need to sum to:

$180°-30°=150°$.

Since the two remaining angles are both the same, they would each be:

$150°÷2=75°$.

Therefore, our isosceles triangle could also have angles: $30°,75°,75°$.

So, the two possibilities are: $30°,30°,120°$ or $30°,75°,75°$.