Question

2. In a parallelogram ABCD, AD = 28 feet, AB = 20 feet, and diagonal AC=26 feet. What is the measure of angle BCD?

Answer 1

The measure of angle BCD is
`117^(0)`.

Explanation:

A parallelogram is a four sided shape with equal and parallel sides.

From the given dimension, we have two triangles ABC and ADC.

From ΔABC, applying the cosine rule;

`c^(2)` =
`a^(2)` +
`b^(2)` - 2ab Cos C

Let a = 28, b = 26 and c = 20. So that;

`20^(2)` =
`28^(2)` +
`26^(2)` - 2 × 28 × 26 Cos C

400 = 784 + 676 - 1450 Cos C

400 = 1460 - 1450 Cos C

1450 Cos C = 1460 - 400

1450 Cos C = 1060

Cos C = 0.7310

C =
`43.03^(0)`

⇒ <ACB =
`43.03^(0)`

Also from ΔADC, applying the Cosine rule;

`c^(2)` =
`a^(2)` +
`b^(2)` - 2ab Cos C

Let a = 20, b = 26 and c = 28

`28^(2)` =
`20^(2)` +
`26^(2)` - 2 × 20 × 26 Cos C

784 = 400 + 676 - 1040 Cos C

784 = 1076 - 1040 Cos C

1040 Cos C = 1076 - 784

1040 Cos C = 292

Cos C = 0.2808

C =
`73.69^(0)`

⇒ <ACD =
`73.69^(0)`

<BCD = <ACB + <ACD

=
`43.03^(0)` +
`73.69^(0)`

=
`116.72^(0)`

The measure of angle BCD is
`117^(0)`.

Answer 2

The measure of angle BCD is 117°.

Explanation:

A parallelogram is a quadrilateral with two pairs of side with same length and two pairs of angles. It is given that sum of all internal angles is equal to 360 degrees. According to the statement of problem, it is also given that AD = 28 feet, AB = 20 feet and AC = 26 feet (See attachment). As first step it is needed to find the value of the angle ABC by the Law of Cosine:

`AC^(2) = AB^(2) + AD^(2) - 2\cdot AB \cdot AD \cdot \cos B`

`\cos B = -(AC^(2)-AB^(2)-AD^(2))/(2\cdot AB \cdot AD)`

`\cos B = -(26^(2)-20^(2)-28^(2))/(2\cdot (20)\cdot (28))`

`\cos B = 0.454`

`B \approx 63^(\circ)`

The measure of the angle C can be obtained by using the following identity:

`2\cdot B + 2\cdot C = 360^(\circ)`

`2\cdot C = 360^(\circ) - 2\cdot B`

`C = 180^(\circ) - B`

`C = 180^(\circ) - 63^(\circ)`

`C = 117^(\circ)`

The measure of angle BCD is 117°.