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In the shown figure (not to scale), \(\mathrm{AB}=\mathrm{CD}\) and \(\overline{\mathrm{AB}}\) and \(\overline{\mathrm{CD}}\) are produced to meet at the point \(\mathrm{P}\). If \(\angle \mathrm{BAC}=70^{\circ}\), then find \(\angle \mathrm{P}\). (1) \(30^{\circ}\) (2) \(40^{\circ}\) (3) \(45^{\circ}\) (4) \(50^{\circ}\)

Short Answer

Expert verified
Answer: (2) 40°

Step by step solution

01

Determine the angle in triangle ACP

Since triangle ACP forms a straight line, we know that the sum of angles in the triangle is 180 degrees. Thus, we can write the equation \(\angle BAC + \angle PAC + \angle APC = 180^{\circ}\)
02

Use given information to find angle APC

We know that ∠BAC = 70°. Therefore, our equation becomes \(70^{\circ} + \angle PAC + \angle APC = 180^{\circ}\) Since \(\mathrm{AB}=\mathrm{CD}\), the two triangles \(\triangle ABC\) and \(\triangle ADC\) are isosceles (with equal base angles). Therefore, \(\angle PAC = \angle BAC = 70^{\circ}\). Now our equation becomes \(70^{\circ} + 70^{\circ} + \angle APC = 180^{\circ}\) Subtract 140° from both sides to find the angle APC: \(\angle APC = 40^{\circ}\)
03

Find angle P (final answer)

Finally, we need to find the angle P. Since \(\angle P = \angle APC\), we can conclude that: Angle P = \(40^{\circ}\) Therefore, the correct answer is (2) \(40^{\circ}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isosceles Triangle
In geometry, an "isosceles triangle" is a polygon with at least two sides of equal length. This property comes with some neat features that make these triangles especially interesting and fun to solve problems with. In any isosceles triangle, the two angles opposite the equal sides are also equal. This is known as the "base angles theorem."
It means that if you know the measure of one of those angles, you automatically know the other.
This property is particularly useful when solving problems like the original exercise, where identifying the equal angles quickly leads to finding other unknown values.
  • An isosceles triangle must have at least two equal sides \( AB = AC \) if \( \angle B = \angle C \).
  • The sides can be referred to as the base and legs, with the base being the unequal side, if there is one.
  • The vertex angle is the angle opposite the base and is a crucial part of problems where angles are calculated.
Understanding these properties lets you break down complex triangle problems with ease.
Angle Calculation
Calculating angles in a triangle is an essential part of solving many geometric problems. A triangle's internal angles always add up to 180 degrees, which is a key fact that drives many solution strategies. In the original exercise, the known angle was \( \angle BAC = 70^{\circ} \), allowing us to set up a straightforward calculation.
By using some simple algebra, if two angles are known, finding the third one becomes a simple matter of subtraction.
Here's how you usually calculate angles within a triangle:
  • Write out the equation: \( \angle A + \angle B + \angle C = 180^{\circ} \).
  • Substitute known values into the equation to solve for the unknown angle.
  • For example, if \( \angle A = 70^{\circ} \) and you know another equal angle because of the isosceles property, you subtract their sum from 180.
In any triangle problem, precise angle calculation is usually just some quick and easy subtraction, making it both a reliable and efficient method.
Triangle Properties
"Triangle properties" encompass various attributes that can simplify geometric problems, making them a valuable asset in your math toolkit. For the problem at hand, it was key to understand that both \( \triangle ABC \) and \( \triangle ACP \) could be analyzed with properties like the base angles in isosceles triangles and the sum of angles in triangles.
Triangles, by definition, are three-sided polygons.
Each side and angle can tell you different things about the triangle itself.Here are some crucial triangle properties to remember:
  • The sum of all interior angles is always 180 degrees.
  • An isosceles triangle has two equal sides and two equal angles.
  • Triangles can be classified by side length (equilateral, isosceles, scalene) or by angles (acute, obtuse, or right).
  • The external angle of the triangle is equal to the sum of the two non-adjacent internal angles.
These properties serve as helpful guides when you tackle a range of problems that rely on structure and measurements within triangles.

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Most popular questions from this chapter

In the shown figure, \(\mathrm{O}\) is the centre of the circle and \(\mathrm{AD}\) is a tangent to the circle at \(\mathrm{A}\). If \(\angle \mathrm{CAD}=55^{\circ}\) and \(\angle \mathrm{ADC}=25^{\circ}\), then \(\angle \mathrm{ABO}=\) (1) \(10^{\circ}\) (2) \(15^{\circ}\) (3) \(20^{\circ}\) (4) \(25^{\circ}\)

In the diagram, \(\mathrm{O}\) is the centre of the circle and \(\angle \mathrm{OPA}=\) \(30^{\circ} .\) Find \(\angle \mathrm{ACB}\) and \(\angle \mathrm{ADB}\) respectively. (1) \(120^{\circ}, 60^{\circ}\) (2) \(60^{\circ}, 120^{\circ}\) (3) \(75^{\circ}, 105^{\circ}\) (4) \(35^{\circ}, 145^{\circ}\)

In two triangles \(\mathrm{ABC}\) and \(\mathrm{DEF}, \angle \mathrm{A}=\angle \mathrm{D}\). The sum of the angles \(\mathrm{A}\) and \(\mathrm{B}\) is equal to the sum of the angles \(\mathrm{D}\) and \(\mathrm{E}\). If \(\mathrm{BC}=6 \mathrm{~cm}\) and \(\mathrm{EF}=8 \mathrm{~cm}\), find the ratio of the areas of the triangles, \(\mathrm{ABC}\) and \(\mathrm{DEF}\). (1) \(3: 4\) (2) \(4: 3\) (3) \(9: 16\) (4) \(16: 9\)

\(\mathrm{PQ}\) is the direct common tangent of two circles \((\mathrm{S}, 9 \mathrm{~cm})\) and \((\mathrm{R}, 4 \mathrm{~cm})\) which touch each other externally. Find the area of the quadrilateral PQRS. (in \(\mathrm{cm}^{2}\) ) (1) 72 (2) 65 (3) 78 (4) 69

Two circles intersect at two points \(\mathrm{P}\) and \(\mathrm{S} . \mathrm{QR}\) is a tangent to the two circles at \(\mathrm{Q}\) and \(\mathrm{R}\). If \(\angle \mathrm{QSR}=72^{\circ}\), then \(\angle \mathrm{QPR}\) \(=\) (1) \(84^{\circ}\) (2) \(96^{\circ}\) (3) \(102^{\circ}\) (4) \(108^{\circ}\)

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