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Name three other angles, with reasons, equal to 32° and determine the values of M, T, and N.

Mathematics

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Answer

64∘\text{}64^\circ\text{}

32∘\text{}32^\circ\text{}

116∘\text{}116^\circ\text{}

Step-by-step solution

7 steps

Here's the solution to question 6.1:

6.1.1 Name three other angles, with reasons, equal to 32∘32^\circ.

Given ∠B=∠KBA=32∘\angle B = \angle KBA = 32^\circ.

  1. Angle: ∠KTA\angle KTA Reason: Angles in the same segment (subtended by arc KA) are equal.

    ∠KTA=∠KBA=32∘\angle KTA = \angle KBA = 32^\circ
  2. Angle: ∠KCA\angle KCA Reason: Angles in the same segment (subtended by arc KA) are equal.

    ∠KCA=∠KBA=32∘\angle KCA = \angle KBA = 32^\circ
  3. Angle: ∠NAC\angle NAC Reason: Given CN=NACN = NA, △CNA\triangle CNA is an isosceles triangle. The base angles opposite the equal sides are equal. Since ∠KCA=32∘\angle KCA = 32^\circ, then ∠NCA=32∘\angle NCA = 32^\circ. Therefore, ∠NAC=∠NCA\angle NAC = \angle NCA.

    ∠NAC=32∘\angle NAC = 32^\circ

6.1.2 Determine the values of the following:

6.1.2a) M
Assuming M refers to ∠KMA\angle KMA, the angle at the center.

  1. Identify the relationship between the angle at the center and the angle at the circumference.

    The angle at the center subtended by an arc is twice the angle at the circumference subtended by the same arc. Arc KA subtends ∠KMA\angle KMA at the center and ∠KBA\angle KBA at the circumference.

  2. Calculate the angle.

    ∠KMA=2×∠KBA\angle KMA = 2 \times \angle KBA ∠KMA=2×32∘\angle KMA = 2 \times 32^\circ ∠KMA=64∘\angle KMA = 64^\circ

    The value of M is 64∘\boxed{\text{}64^\circ\text{}}

    6.1.2b) T1T_1
    T1T_1 refers to ∠KTA\angle KTA.

  3. Identify the relationship between ∠KTA\angle KTA and ∠KBA\angle KBA.

    Both angles are subtended by the same arc KA in the circle.

  4. State the value.

    ∠KTA=∠KBA\angle KTA = \angle KBA ∠KTA=32∘\angle KTA = 32^\circ

    The value of T1T_1 is 32∘\boxed{\text{}32^\circ\text{}}

    6.1.2c) N
    N refers to ∠CNA\angle CNA.

  5. Use the given information about △CNA\triangle CNA.

    We are given CN=NACN = NA, which means △CNA\triangle CNA is an isosceles triangle.

  6. Determine the base angles of △CNA\triangle CNA.

    From 6.1.1, we found ∠NCA=32∘\angle NCA = 32^\circ and ∠NAC=32∘\angle NAC = 32^\circ.

  7. Calculate ∠CNA\angle CNA using the sum of angles in a triangle.

    The sum of angles in △CNA\triangle CNA is 180∘180^\circ.

    ∠CNA+∠NAC+∠NCA=180∘\angle CNA + \angle NAC + \angle NCA = 180^\circ ∠CNA+32∘+32∘=180∘\angle CNA + 32^\circ + 32^\circ = 180^\circ ∠CNA+64∘=180∘\angle CNA + 64^\circ = 180^\circ ∠CNA=180∘−64∘\angle CNA = 180^\circ - 64^\circ ∠CNA=116∘\angle CNA = 116^\circ

    The value of N is 116∘\boxed{\text{}116^\circ\text{}}

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