The distance between the points A a cos 20- b sin 20- 0 and B 0- a sin 20- b cos 20- is-A- -a2-b2B- -a2-b2C- a-bD- a-b

The correct option is B √a2+b2 AB^2 = (a cos 20^∘ + b sin 20^∘)^2 + (a sin 20^∘ b cos 20^∘)^2 AB^2 = (a^2 cos^2 20^∘ + b^2 sin^2 20^∘ +2ab cos20^∘ sin20^∘ ...

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Byju's Answer

Standard IX

Mathematics

If Two Sides of a Triangle Are Unequal, Then the Angle Opposite to the Greater Side Is Greater Than Angle Opposite to the Smaller Side

The distance ...

Question

The distance between the points A(a cos $20^{\circ}$ + b sin $20^{\circ}$ , 0) and B(0, a sin $20^{\circ}$ - b cos $20^{\circ}$ ) is:

a+b

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a-b

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√a²-b²

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√a²+b²

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Solution

The correct option is D
√a²+b²

$A B^{2}$ = ${(a c o s 20^{\circ} + b s i n 20^{\circ})}^{2}$ + ${(a s i n 20^{\circ} - b c o s 20^{\circ})}^{2}$
$A B^{2}$ = $(a^{2} c o s^{2} 20^{\circ} + b^{2} s i n^{2} 20^{\circ} + 2 a b c o s 20^{\circ} s i n 20^{\circ} + a^{2} s i n^{2} 20^{\circ} + b^{2} c o s^{2} 20^{\circ} - 2 a b c o s 20^{\circ} s i n 20^{\circ}$ )
$A B^{2}$ = $a^{2} (c o s^{2} 20^{\circ} + s i n^{2} 20^{\circ}) + b^{2} (c o s^{2} 20^{\circ} + s i n^{2} 20^{\circ})$

$A B^{2}$ = $a^{2}$ + $b^{2}$
AB = $\sqrt{a^{2} + b^{2}}$

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The distance between the points A(a cos 20∘ + b sin20∘, 0) and B(0, a sin20∘ - b cos20∘) is:

The correct option is D √a2+b2 AB2 = (acos20∘+bsin20∘)2 + (asin20∘−bcos20∘)2 AB2 = (a2cos220∘+b2sin220∘+2abcos20∘sin20∘+a2sin220∘+b2cos220∘−2abcos20∘sin20∘) AB2 = a2(cos220∘+sin220∘)+b2(cos220∘+sin220∘) AB2 = a2 + b2 AB = √a2+b2

The distance between the points A(a cos $20^{\circ}$ + b sin $20^{\circ}$ , 0) and B(0, a sin $20^{\circ}$ - b cos $20^{\circ}$ ) is: