Answer :
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]
### 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]
