Question

The distance between the point A ( p sin 25,0) and B ( 0 , p sin 65) is

Answer 1

Answer:

p units

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Answer 2

The distance between the point A (psin 25°,0) and B (0 , psin 65°) is p unit

Correct question : The distance between the point A (psin 25°,0) and B (0 , psin 65°) is

Given :

The points A (psin 25°,0) and B (0 , psin 65°)

To find :

The distance between the point A (psin 25°,0) and B (0 , psin 65°)

Formula :

For the given two points ( x₁ , y₁) & (x₂ , y₂) the distance between the points

$= \sf{ \sqrt{ {(x_2 -x_1 )}^{2} + {(y_2 -y_1 )}^{2} } }$

Solution :

Step 1 of 2 :

Write down the given points

Here the given points are A (psin 25°,0) and B (0 , psin 65°)

Step 2 of 2 :

Calculate the distance between the points

The distance between the point A (psin 25°,0) and B (0 , psin 65°)

$\displaystyle \sf{ = \sqrt{ {(0 - p \: sin {25}^{ \circ} )}^{2} +{(p \: sin {65}^{ \circ} - 0 )}^{2} } } \: \: unit$

$\displaystyle \sf{ = \sqrt{ { p}^{2} \: {sin }^{2} {25}^{ \circ}+{ p}^{2} \: {sin }^{2} {65}^{ \circ}} } \: \: unit$

$\displaystyle \sf{ = \sqrt{ { p}^{2}( \: {sin }^{2} {25}^{ \circ}+ \: {sin }^{2} {65}^{ \circ}}) } \: \: unit$

$\displaystyle \sf{ = \sqrt{ { p}^{2} [ \: {sin }^{2} {25}^{ \circ}+ \: {sin }^{2} ({90}^{ \circ} - {25}^{ \circ})}]} \: \: unit$

$\displaystyle \sf{ = \sqrt{ { p}^{2} [ \: {sin }^{2} {25}^{ \circ}+ \: {cos }^{2} {25}^{ \circ}}]} \: \: unit$

$\displaystyle \sf{ = \sqrt{ { p}^{2} \times 1}} \: \: unit\: \: \: \bigg[ \: \because \: {sin }^{2} \theta+ \: {cos }^{2} \theta = 1 \bigg]$

$\displaystyle \sf{ = \sqrt{ {p}^{2} } } \: \: unit$

$\displaystyle \sf{ = p } \: \: unit$

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