Question

the adjacent sides of parallelogram are 4a, 3a. the angle b/w them is 60, then one of the diagonal of the parallelogram? ​

Answer 1

Required Answer:-

Two adjacent sides and the respective diagonal makes a triangle and here, we have the length of two sides and the angle between them$.$

By using cosine law,

(Refer to the attachment)

➙ d² = (4a)² + (3a)² - 2×4a×3a×cos 60°

➙ d² = 16a² + 9a² - 24a² × 1/2

➙ d² = 25a² - 12a²

➙ d² = 13a²

➙ d = √(13a²)

➙ d = a√13

➙ d ≈ 3.61 a (Ans)

One of the diagonal will have length 3.61a. Similarly you can find the length of other diagonal with the included angle as 120°.

Answer 2

Information provided with us:

• Adjacent sides of parallelogram are 4a and 3a
• Angle between them is 60⁰

What we have to calculate:

• We have to calculate the diagonals of that parallelogram

Options given:

1. √13a
2. 2√3a
3. 5√3a
4. 3√3a

Using Formula,

The Law of Cosines:-

• $\large{ \dag} \underline{\boxed{ \sf{\large{c {}^{2} = a {}^{2} + b {}^{2} - 2ab(cosC) }}}}$

Where,

• a , b , and c are the sides of parallelogram

Solution:

• As here we have been given with two sides those are 4a and 3a. In the above given formula c is third side of parallelogram which we are going to find out by substituting the values of a and b. Remember that b is 3a and a is 4a

We know that,

• $\red{ \boxed{ \tt{cos \: 60 {}^{0} \: = \: \dfrac{1}{2} }}}$

Substituting the values,

$: \longmapsto \: \sf{c {}^{2} \: = \: (4a) {}^{2} + (3a) {}^{2} - 2(4a)(3a) \times \dfrac{1}{2} }$

$: \longmapsto \: \sf{c {}^{2} \: = \: (4a) {}^{2} + (3a) {}^{2} - 2 \times (4a) \times (3a) \times \dfrac{1}{2} }$

$: \longmapsto \: \sf{c {}^{2} \: = \: (4a) {}^{2} + (3a) {}^{2} - 2 \times 4a \times 3a \times \dfrac{1}{2} }$

$: \longmapsto \: \sf{c {}^{2} \: = \: (4a \times 4a) + (3a \times 3a) - 2 \times 4a \times 3a \times \dfrac{1}{2} }$

$: \longmapsto \: \sf{c {}^{2} \: = \: (16a {}^{2} ) + (9a {}^{2} ) - 2 \times 4a \times 3a \times \dfrac{1}{2} }$

$: \longmapsto \: \sf{c {}^{2} \: = \: 16a {}^{2} + 9a {}^{2} - 2 \times 4a \times 3a \times \dfrac{1}{2} }$

$: \longmapsto \: \sf{c {}^{2} \: = \: 16a {}^{2} + 9a {}^{2} - 8a \times 3a \times \dfrac{1}{2} }$

$:\longmapsto \: \sf{c {}^{2} \: = \: 16a {}^{2} + 9a {}^{2} - 24a {}^{2} \times \dfrac{1}{2} }$

$: \longmapsto \: \sf{c {}^{2} \: = \: 16a {}^{2} + 9a {}^{2} - \cancel{24a} {}^{2} \times \dfrac{1}{ \cancel{2}} }$

$: \longmapsto \: \sf{c {}^{2} \: = \: 16a {}^{2} + 9a {}^{2} - 12a {}^{2} \times 1 }$

$: \longmapsto \: \sf{c {}^{2} \: = \: 16a {}^{2} + 9a {}^{2} - 12a {}^{2} }$

$: \longmapsto \: \sf{c {}^{2} \: = \: 16a {}^{2} \: - \: 3a {}^{2} }$

$:\longmapsto \: \sf{c {}^{2} \: = \: 13a {}^{2} }$

$:\longmapsto \: \sf{c \: = \: \sqrt{ 13a {}^{2} }}$

$:\longmapsto \: \sf{c \: = \: \sqrt{ 13a }}$

$\underline{\bf{Henceforth, \: one \: of \: the \: diagonals \: is \: \sqrt{13a}}}$