Angles in a Triangle: Practice

Add the three angle expressions because they belong to one triangle.

$(2x+7)+(3x+1)+(x+4)=180.$

Combining terms gives $6x+12=180$. Subtract $12$ and divide by $6$.

$x=28.$

The measures $63^\circ,85^\circ,32^\circ$ are positive and total $180^\circ$.

The acute measures total $90^\circ$. Write them as $2k$ and $7k$ degrees.

$2k+7k=90.$

Thus $k=10$, and the smaller measure is $2(10)^\circ=20^\circ.$ The other measure is $70^\circ$, which checks the complementary sum.

The whole angle at $C$ comes from the triangle sum.

$m\angle ACB=180^\circ-49^\circ-62^\circ=69^\circ.$

Because $D$ is inside side $AB$, ray $CD$ divides that angle into the stated adjacent parts.

$m\angle DCB=69^\circ-26^\circ=43^\circ.$

First find the angle at $O$ in triangle $AOB$.

$m\angle AOB=180^\circ-46^\circ-58^\circ=76^\circ.$

Angles $AOB$ and $COD$ are vertical angles, so the lower triangle also has $76^\circ$ at $O$. Its angle sum gives

$m\angle OCD=180^\circ-76^\circ-67^\circ=37^\circ.$

Use the triangle sum to determine the third measure.

$180^\circ-57^\circ-81^\circ=42^\circ.$

Equate its expression to $42$, then solve.

$2x+6=42.$

Subtract $6$ and divide by $2$ to obtain $x=18$.

The two unknown measures total $180-103=77$ degrees. Let the smaller measure be $u$ degrees; the larger is $u+19$ degrees.

$u+(u+19)=77.$

Then $2u=58$, so $u=29$. The larger measure is $48^\circ$. Their difference is $19^\circ$ and their sum is $77^\circ$.

A triangle has three noncollinear vertices. Therefore C is outside the line through A and B.

The accepted foundation supplies a parallel through an external point in the same plane. The construction does not depend on equal sides or an acute angle.

Write the measures as $3k,4k,5k$ degrees. Their sum is $180^\circ$.

$12k=180.$

So $k=15$. The largest measure is $5(15)^\circ=75^\circ.$ The other measures are $45^\circ$ and $60^\circ$.

In $ABC$, the third measure is $180^\circ-31^\circ-88^\circ=61^\circ.$

Thus $\angle B$ matches $\angle P$ and $\angle C$ matches $\angle Q$. The third-angle theorem pairs $\angle A$ with $\angle R$; both measure $31^\circ$.

The equal lengths give two congruent sides, so the triangle is isosceles.

The definitions and angle-sum theorem do not turn these given side lengths into numerical angle measures. That would require an additional relationship beyond this lesson.

The two acute angles are complementary.

$(5z-7)+(3z+1)=90.$

So $8z-6=90$, giving $z=12$. Substitution gives $5(12)-7=53$ and $3(12)+1=37$.

The larger acute angle is $53^\circ$, and $53+37=90$.

The triangle sum gives $a+2a+30=180.$

Thus $3a=150$ and $a=50$. The measures are $50^\circ,100^\circ,30^\circ$.

Since one angle is greater than $90^\circ$, the triangle is obtuse.