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8 The sides of a parallelogram are 8 cm and
10 cm and the angle between them is 130°.
Use cosine rule to calculate
(a) the diagonals of the parallelogram.
(b) angle DBC.

Sagot :

GinnyAnswer
Let's solve the problem step-by-step:

### Given:
- Side [tex]\( a \)[/tex] = 8 cm
- Side [tex]\( b \)[/tex] = 10 cm
- Angle between the sides, [tex]\(\theta\)[/tex] = 130°

### To calculate:
- (a) The lengths of the diagonals of the parallelogram.
- (b) Angle [tex]\( \angle DBC \)[/tex].

### (a) Calculating the diagonals

In a parallelogram, the diagonals can be calculated using the Cosine Rule.

#### Diagonal [tex]\( d_1 \)[/tex]:

Using the Cosine Rule for the first diagonal [tex]\( d_1 \)[/tex]:

[tex]\[ d_1^2 = a^2 + b^2 - 2ab \cdot \cos(\theta) \][/tex]

Plug in the values:

[tex]\[ d_1^2 = 8^2 + 10^2 - 2 \cdot 8 \cdot 10 \cdot \cos(130°) \][/tex]

Next, we calculate [tex]\( \cos(130°) \)[/tex] which is a constant.

[tex]\[ \cos(130°) \approx -0.6428 \][/tex]

Now substitute this:

[tex]\[ d_1^2 = 64 + 100 - 2 \cdot 8 \cdot 10 \cdot (-0.6428) \][/tex]

[tex]\[ d_1^2 = 164 + 102.848 \][/tex]

[tex]\[ d_1^2 = 266.848 \][/tex]

Taking the square root to find [tex]\( d_1 \)[/tex]:

[tex]\[ d_1 \approx \sqrt{266.848} \][/tex]

[tex]\[ d_1 \approx 16.34 \, \text{cm} \][/tex]

#### Diagonal [tex]\( d_2 \)[/tex]:

Using the Cosine Rule for the second diagonal [tex]\( d_2 \)[/tex], but with a [tex]\(\cos\)[/tex] function that will change signs:

[tex]\[ d_2^2 = a^2 + b^2 + 2ab \cdot \cos(\theta) \][/tex]

Plug in the values:

[tex]\[ d_2^2 = 8^2 + 10^2 + 2 \cdot 8 \cdot 10 \cdot \cos(130°) \][/tex]

[tex]\[ d_2^2 = 64 + 100 + 2 \cdot 8 \cdot 10 \cdot (-0.6428) \][/tex]

[tex]\[ d_2^2 = 164 - 102.848 \][/tex]

[tex]\[ d_2^2 = 61.152 \][/tex]

Taking the square root to find [tex]\( d_2 \)[/tex]:

[tex]\[ d_2 \approx \sqrt{61.152} \][/tex]

[tex]\[ d_2 \approx 7.82 \, \text{cm} \][/tex]

### (b) Calculating angle [tex]\( \angle DBC \)[/tex]:

In a parallelogram, consecutive angles are supplementary. This means that [tex]\( \angle DBC \)[/tex] is the supplementary angle to the given angle [tex]\( 130° \)[/tex].

[tex]\[ \angle DBC = 180° - 130° \][/tex]

[tex]\[ \angle DBC = 50° \][/tex]

### Summary:

(a) The lengths of the diagonals of the parallelogram are:

[tex]\[ d_1 \approx 16.34 \, \text{cm} \][/tex]
[tex]\[ d_2 \approx 7.82 \, \text{cm} \][/tex]

(b) The angle [tex]\( \angle DBC \)[/tex] is:

[tex]\[ \angle DBC = 50° \][/tex]
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