In the diagram below O is the center, A, C and B are points on the circumference of the circle. PCD and TBD are tangents to the circle at points C and B respectively. C1 = 36° and B1 = 62°.

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Asked on June 7, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-06-07T18:46:49.183Z

Here are the solutions to the geometry problems: 8.1.1 Give, with reasons, TWO other angles, each equal to 36^. Step 1: Identify the first angle using the Tangent-Chord Theorem. The angle between the tangent PC and the chord AC is C_1 = 36^. By the Tangent-Chord Theorem, this angle is equal to the angle in the alternate segment, which is ABC. ABC = C_1 = 36^ Reason: Angle between tangent and chord is equal to the angle in the alternate segment. Step 2: Identify the second angle by calculating angles in ADC. First, find BAC and ABC. The angle between the tangent TB and the chord BC is B_1 = 62^. By the Tangent-Chord Theorem, this angle is equal to the angle in the alternate segment, which is BAC. BAC = B_1 = 62^ We already found ABC = 36^. Next, find ACB using the sum of angles in ABC. ACB = 180^ - ( BAC + ABC) ACB = 180^ - (62^ + 36^) ACB = 180^ - 98^ = 82^ Now, consider ADC. We have DAC = BAC = 62^. We have ACD = ACB = 82^. Using the sum of angles in ADC: ADC = 180^ - ( DAC + ACD) ADC = 180^ - (62^ + 82^) ADC = 180^ - 144^ = 36^ Reason: Sum of angles in a triangle. The two other angles equal to 36^ are: • ABC = 36^ (Reason: Angle between tangent and chord is equal to angle in alternate segment) • ADC = 36^ (Reason: Sum of angles in ADC) 8.1.2 Determine with reasons the sizes of the following angles. a. O_1 Step 1: Relate O_1 to an angle at the circumference. O_1 is AOC. The angle subtended by arc AC at the center is AOC, and the angle subtended by the same arc at the circumference is ABC. From 8.1.1, we know ABC = 36^. O_1 = AOC = 2 × ABC O_1 = 2 × 36^ = 72^ Reason: Angle at the center is twice the angle at the circumference. O_1 = 72^ b. C_3 Step 1: Relate C_3 to angles in OBC. C_3 is OCB. In OBC, OB = OC (radii), so it is an isosceles triangle. First, find BAC. From 8.1.1, BAC = B_1 = 62^ (Angle in alternate segment). Next, find BOC. The angle subtended by arc BC at the center is BOC, and at the circumference is BAC. BOC = 2 × BAC BOC = 2 × 62^ = 124^ Reason: Angle at the center is twice the angle at the circumference. Now, in isosceles OBC: OCB = OBC = (180^ - BOC)/(2) C_3 = (180^ - 124^)/(2) = (56^)/(2) = 28^ Reason: Base angles of an isosceles triangle and sum of angles in a triangle. C_3 = 28^ c. D Step 1: Use the result from 8.1.1. D is ADC. We already calculated this in 8.1.1. ADC = 36^ Reason: Sum of angles in ADC. D = 36^ d. ACB Step 1: Use the sum of angles in ABC. From 8.1.1, we have: BAC = 62^ (Angle in alternate segment) ABC = 36^ (Angle in alternate segment) ACB = 180^ - ( BAC + ABC) ACB = 180^ - (62^ + 36^) ACB = 180^ - 98^ = 82^ Reason: Sum of angles in a triangle. ACB = 82^ 8.1.3 Give a reason why CD = BD Step 1: State the theorem for tangents from an external point. CD and BD are both tangents drawn from the external point D to the circle. Reason: Tangents drawn from an external point to a circle are equal in length. Tangents from an external point to a circle are equal in length. Drop the next question 📸

Accepted Answer · 2026-06-07T18:46:49.183Z

Here are the solutions to the geometry problems: 8.1.1 Give, with reasons, TWO other angles, each equal to 36^. Step 1: Identify the first angle using the Tangent-Chord Theorem. The angle between the tangent PC and the chord AC is C_1 = 36^. By the Tangent-Chord Theorem, this angle is equal to the angle in the alternate segment, which is ABC. ABC = C_1 = 36^ Reason: Angle between tangent and chord is equal to the angle in the alternate segment. Step 2: Identify the second angle by calculating angles in ADC. First, find BAC and ABC. The angle between the tangent TB and the chord BC is B_1 = 62^. By the Tangent-Chord Theorem, this angle is equal to the angle in the alternate segment, which is BAC. BAC = B_1 = 62^ We already found ABC = 36^. Next, find ACB using the sum of angles in ABC. ACB = 180^ - ( BAC + ABC) ACB = 180^ - (62^ + 36^) ACB = 180^ - 98^ = 82^ Now, consider ADC. We have DAC = BAC = 62^. We have ACD = ACB = 82^. Using the sum of angles in ADC: ADC = 180^ - ( DAC + ACD) ADC = 180^ - (62^ + 82^) ADC = 180^ - 144^ = 36^ Reason: Sum of angles in a triangle. The two other angles equal to 36^ are: • ABC = 36^ (Reason: Angle between tangent and chord is equal to angle in alternate segment) • ADC = 36^ (Reason: Sum of angles in ADC) 8.1.2 Determine with reasons the sizes of the following angles. a. O_1 Step 1: Relate O_1 to an angle at the circumference. O_1 is AOC. The angle subtended by arc AC at the center is AOC, and the angle subtended by the same arc at the circumference is ABC. From 8.1.1, we know ABC = 36^. O_1 = AOC = 2 × ABC O_1 = 2 × 36^ = 72^ Reason: Angle at the center is twice the angle at the circumference. O_1 = 72^ b. C_3 Step 1: Relate C_3 to angles in OBC. C_3 is OCB. In OBC, OB = OC (radii), so it is an isosceles triangle. First, find BAC. From 8.1.1, BAC = B_1 = 62^ (Angle in alternate segment). Next, find BOC. The angle subtended by arc BC at the center is BOC, and at the circumference is BAC. BOC = 2 × BAC BOC = 2 × 62^ = 124^ Reason: Angle at the center is twice the angle at the circumference. Now, in isosceles OBC: OCB = OBC = (180^ - BOC)/(2) C_3 = (180^ - 124^)/(2) = (56^)/(2) = 28^ Reason: Base angles of an isosceles triangle and sum of angles in a triangle. C_3 = 28^ c. D Step 1: Use the result from 8.1.1. D is ADC. We already calculated this in 8.1.1. ADC = 36^ Reason: Sum of angles in ADC. D = 36^ d. ACB Step 1: Use the sum of angles in ABC. From 8.1.1, we have: BAC = 62^ (Angle in alternate segment) ABC = 36^ (Angle in alternate segment) ACB = 180^ - ( BAC + ABC) ACB = 180^ - (62^ + 36^) ACB = 180^ - 98^ = 82^ Reason: Sum of angles in a triangle. ACB = 82^ 8.1.3 Give a reason why CD = BD Step 1: State the theorem for tangents from an external point. CD and BD are both tangents drawn from the external point D to the circle. Reason: Tangents drawn from an external point to a circle are equal in length. Tangents from an external point to a circle are equal in length. Drop the next question 📸

In the diagram below O is the center, A, C and B are points on the circumference of the circle. PCD and TBD are tangents to the circle at points C and B respectively. C1 = 36° and B1 = 62°.

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In the diagram below O is the center, A, C and B are points on the circumference of the circle. PCD and TBD are tangents to the circle at points C and B respectively. C1 = 36° and B1 = 62°.

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