If x-sin 130- cos 80- y-sin 80- cos 130- z-1-xy-which one of the following is true

X=sin 130^∘ cos 80^∘, y=sin 80^∘ cos 130^∘ ⇒ x=cos 40^∘ cos 80^∘, y= sin 80^∘ sin 40^∘ So, x gt; 0 and y lt; 0 and xy lt; 0 Now, z = 1 + xy ⇒ 0 lt; ...

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Standard XIII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

If x=sin 130∘...

Question

If $x = s i n 130^{\circ} c o s 80^{\circ}, y = s i n 80^{\circ} c o s 130^{\circ}, z = 1 + x y,$
which one of the following is true

x > 0, y > 0, z > 0

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x > 0, y < 0, 0 < z < 1

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x > 0, y < 0, z > 1

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x < 0, y < 0, 0 < z < 1

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Solution

The correct option is B x > 0, y < 0, 0 < z < 1
$x = s i n 130^{\circ} c o s 80^{\circ}, y = s i n 80^{\circ} c o s 130^{\circ}$
$\Rightarrow x = c o s 40^{\circ} c o s 80^{\circ}, y = - s i n 80^{\circ} s i n 40^{\circ}$
So, x > 0 and y < 0 and xy < 0
Now, z = 1 + xy $\Rightarrow$ 0 < z < 1.

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Similar questions

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x = s i n 130^{\circ} c o s 80^{\circ}, y = s i n 80^{\circ} c o s 130^{\circ}, z = 1 + x y,

which one of the following is true

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If x=sin 130∘ cos 80∘, y=sin 80∘ cos 130∘, z=1+xy, which one of the following is true

The correct option is B x > 0, y < 0, 0 < z < 1x=sin 130∘ cos 80∘, y=sin 80∘ cos 130∘ ⇒x=cos 40∘ cos 80∘, y=−sin 80∘ sin 40∘ So, x > 0 and y < 0 and xy < 0 Now, z = 1 + xy ⇒ 0 < z < 1.

If $x = s i n 130^{\circ} c o s 80^{\circ}, y = s i n 80^{\circ} c o s 130^{\circ}, z = 1 + x y,$
which one of the following is true

The correct option is B x > 0, y < 0, 0 < z < 1
$x = s i n 130^{\circ} c o s 80^{\circ}, y = s i n 80^{\circ} c o s 130^{\circ}$
$\Rightarrow x = c o s 40^{\circ} c o s 80^{\circ}, y = - s i n 80^{\circ} s i n 40^{\circ}$
So, x > 0 and y < 0 and xy < 0
Now, z = 1 + xy $\Rightarrow$ 0 < z < 1.