Question

reasoning in geometry (2)
unit 7: problem solving and geometry
1 in a right - angled triangle, if one angle is 55°, find the other acute angle.
2 in a right - angled triangle the two shorter sides are equal. what is the size of each acute angle?
3 in a right - angled triangle, one acute angle is twice the size of the other. what is the size of each angle?
4 the angles of a triangle are x°, 2x° and 3x°. find the size of each angle.
5 abcd is a rectangle. if ∠bdc = 35°, find ∠dbc.
6 three angles of a quadrilateral are 120°, 70° and 110°. find the fourth angle.
7 if each of the two equal angles of an isosceles triangle is 68°, find the size of the remaining angle.
8 the angle between the two equal sides of an isosceles triangle is 86°. find the size of each of the remaining angles.
9 in △abc, ab = bc and bc = ac. what is the size of ∠a?

Explanation:

Step1: Recall triangle - angle sum property

The sum of angles in a triangle is 180°. In a right - angled triangle, one angle is 90°.

Step2: Solve for the other acute angle in question 1

Let the given acute angle be 55°. The other acute angle $\theta$ can be found using the formula $\theta=180^{\circ}-90^{\circ}-55^{\circ}=35^{\circ}$.

Step3: Analyze the isosceles right - angled triangle in question 2

In an isosceles right - angled triangle, let the acute angles be $x$ each. Then $90^{\circ}+x + x=180^{\circ}$. Combining like terms gives $90^{\circ}+2x = 180^{\circ}$. Subtracting 90° from both sides: $2x=180^{\circ}-90^{\circ}=90^{\circ}$. Dividing both sides by 2, we get $x = 45^{\circ}$.

Step4: Set up an equation for question 3

Let one acute angle be $x$ and the other be $2x$. Then $90^{\circ}+x + 2x=180^{\circ}$. Combining like terms: $90^{\circ}+3x = 180^{\circ}$. Subtracting 90° from both sides gives $3x=180^{\circ}-90^{\circ}=90^{\circ}$. Dividing by 3, $x = 30^{\circ}$, and $2x = 60^{\circ}$.

Step5: Solve for angles in question 4

Since the sum of angles in a triangle is 180°, we have $x + 2x+3x=180^{\circ}$. Combining like terms: $6x = 180^{\circ}$. Dividing both sides by 6, $x = 30^{\circ}$, $2x=60^{\circ}$, and $3x = 90^{\circ}$.

Step6: Analyze the rectangle in question 5

In rectangle ABCD, $\angle BCD = 90^{\circ}$. In $\triangle BCD$, we know $\angle BDC=35^{\circ}$. Using the angle - sum property of a triangle ($\angle BCD+\angle BDC+\angle DBC = 180^{\circ}$), and since $\angle BCD = 90^{\circ}$, then $\angle DBC=180^{\circ}-90^{\circ}-35^{\circ}=55^{\circ}$.

Step7: Find the fourth angle of the quadrilateral in question 6

The sum of angles in a quadrilateral is 360°. Let the fourth angle be $x$. Then $120^{\circ}+70^{\circ}+110^{\circ}+x = 360^{\circ}$. Combining like terms: $300^{\circ}+x = 360^{\circ}$. Subtracting 300° from both sides, $x = 60^{\circ}$.

Step8: Find the third angle of the isosceles triangle in question 7

Let the equal angles be 68° each. Using the angle - sum property of a triangle ($\angle A+\angle B+\angle C=180^{\circ}$), if $\angle A=\angle B = 68^{\circ}$, then the third angle $\angle C=180^{\circ}-68^{\circ}-68^{\circ}=44^{\circ}$.

Step9: Solve for the equal angles in question 8

Let the equal angles be $x$ each. The angle between the equal sides is 86°. Then $86^{\circ}+x + x=180^{\circ}$. Combining like terms: $86^{\circ}+2x = 180^{\circ}$. Subtracting 86° from both sides: $2x=180^{\circ}-86^{\circ}=94^{\circ}$. Dividing by 2, $x = 47^{\circ}$.

Step10: Analyze the equilateral triangle in question 9

Since $AB = BC$ and $BC = AC$, $\triangle ABC$ is an equilateral triangle. In an equilateral triangle, all angles are equal. Let $\angle A=\angle B=\angle C=x$. Then $x + x+x=180^{\circ}$, $3x = 180^{\circ}$, and $x = 60^{\circ}$.

Answer:

1. $35^{\circ}$
2. $45^{\circ}$
3. $30^{\circ},60^{\circ},90^{\circ}$
4. $30^{\circ},60^{\circ},90^{\circ}$
5. $55^{\circ}$
6. $60^{\circ}$
7. $44^{\circ}$
8. $47^{\circ}$
9. $60^{\circ}$