Question

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Answer 1

Question:

A(a cos 35° , 0) , B(0 , a cos 55°) then find $\overline{AB}$ ?

Solution:

We have to find the distance between points A and B.

$Distance \ between \ two \ points, \ \overline{AB} = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

x₁ = a cos 35°

x₂ = 0

y₁ = 0

y₂ = a cos 55°

$\overline{AB} = \sqrt{(0 - a cos35^{\circ} )^{2} + (a cos 55^{\circ} - 0)^{2}}$

$\overline{AB} = \sqrt{a^{2} cos^{2} 35^{\circ} + a^{2} cos^{2} 55^{\circ}}$

$\overline{AB} = \sqrt{a^{2} [cos^{2} 35^{\circ} + cos^{2} 55^{\circ}]}$

$\overline{AB} = \sqrt{a^{2} [cos^{2} 35^{\circ} + cos^{2} (90-55)^{\circ}]} \ \ \ \ [ We \ know \ that \ sin \theta = cos(90^{\circ} - \theta)]$

$\overline{AB} = \sqrt{a^{2} [cos^{2} 35^{\circ} + sin^{2} 35^{\circ}]} \ \ \ \ [ We \ know \ that \ sin \theta = cos(90^{\circ} - \theta)]$

$\overline{AB} = \sqrt{a^{2} [cos^{2} 35^{\circ} + sin^{2} 35^{\circ}]} \ \ \ \ [ We \ know \ that \ sin^{2} \theta + cos^{2} \theta = 1]$

$\overline{AB} = \sqrt{a^{2} * 1}$

$\overline{AB} = \sqrt{a^{2}}$

$\overline{AB} = a$

The distance between two points A and B ($\overline{AB}$) is a

Answer 2

Step-by-step explanation:

$\bf \huge \pink{dur} \red{ga} \: \green{cha}uhan$

Question:

A(a cos 35° , 0) , B(0 , a cos 55°) then find $\overline{AB}$ ?

Solution:

We have to find the distance between points A and B.

$Distance \ between \ two \ points, \ \overline{AB} = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$

x₁ = a cos 35°

x₂ = 0

y₁ = 0

y₂ = a cos 55°

$\overline{AB} = \sqrt{(0 - a cos35^{\circ} )^{2} + (a cos 55^{\circ} - 0)^{2}}$

$\overline{AB} = \sqrt{a^{2} cos^{2} 35^{\circ} + a^{2} cos^{2} 55^{\circ}}$

$\overline{AB} = \sqrt{a^{2} [cos^{2} 35^{\circ} + cos^{2} 55^{\circ}]}$

$\overline{AB} = \sqrt{a^{2} [cos^{2} 35^{\circ} + cos^{2} (90-55)^{\circ}]} \ \ \ \ [ We \ know \ that \ sin \theta = cos(90^{\circ} - \theta)]$

$\overline{AB} = \sqrt{a^{2} [cos^{2} 35^{\circ} + sin^{2} 35^{\circ}]} \ \ \ \ [ We \ know \ that \ sin \theta = cos(90^{\circ} - \theta)]$

$\overline{AB} = \sqrt{a^{2} [cos^{2} 35^{\circ} + sin^{2} 35^{\circ}]} \ \ \ \ [ We \ know \ that \ sin^{2} \theta + cos^{2} \theta = 1]$

$\overline{AB} = \sqrt{a^{2} * 1}$

$\overline{AB} = \sqrt{a^{2}}$

$\overline{AB} = a$

The distance between two points A and B ($\overline{AB}$) is a