Question

How to find the value of cod theta if the value of theta is given?


Correct answerer will get thanks on all his/her answers. Please I am very needy.​

Answer 1

Explanation:

The value of cos theta depends on the angle theta. As theta changes the value of cos theta changes. Nevertheless, the value of cos theta will always be between -1 and + 1 both included.

Answer 2

Explanation:

$</p><p></p><p></p><p></p><p>\boxed{\bf{\frac{1}{f} =\frac{1}{v} -\frac{1}{u} }}}}</p><p></p><p>\begin{gathered}\longrightarrow\sf{\dfrac{1}{-10} =\dfrac{1}{v} -\dfrac{1}{-30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-10} +\dfrac{1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{-3+1}{30} }\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\cancel{\dfrac{-2}{30} }}\\\\\\\longrightarrow\sf{\dfrac{1}{v} =\dfrac{1}{-15} }\\\\\\\longrightarrow\sf{v=-15\:cm}\end{gathered} </p><p>⟶ </p><p>−10</p><p>1</p><p> </p><p> = </p><p>v</p><p>1</p><p> </p><p> − </p><p>−30</p><p>1</p><p> </p><p> </p><p>⟶ </p><p>v</p><p>1</p><p> </p><p> = </p><p>−10</p><p>1</p><p> </p><p> + </p><p>30</p><p>1</p><p> </p><p> </p><p>⟶ </p><p>v</p><p>1</p><p> </p><p> = </p><p>30</p><p>−3+1</p><p> </p><p> </p><p>⟶ </p><p>v</p><p>1</p><p> </p><p> = </p><p>30</p><p>−2</p><p> </p><p> </p><p> </p><p> </p><p>⟶ </p><p>v</p><p>1</p><p> </p><p> = </p><p>−15</p><p>1</p><p> </p><p> </p><p>⟶v=−15cm</p><p> </p><p> </p><p></p><p></p><p>\boxed{\bf{M \:A \:G \:N\: I \:F \:I \:C\: A\: T \:I \:O\: N :}} </p><p>MAGNIFICATION:</p><p> </p><p> </p><p></p><p>\begin{gathered}\mapsto\sf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:image}{Distance\:of\:object} =\dfrac{v}{u} }\\\\\\\mapsto\sf{m=\cancel{\dfrac{-30}{-15}} }\\\\\\\mapsto\bf{m=2\:cm}\end{gathered} </p><p>↦m= </p><p>Heightofobject(O)</p><p>Heightofimage(I)</p><p> </p><p> = </p><p>Distanceofobject</p><p>Distanceofimage</p><p> </p><p> = </p><p>u</p><p>v</p><p> </p><p> </p><p>↦m= </p><p>−15</p><p>−30</p><p> </p><p> </p><p> </p><p> </p><p>↦m=2cm</p><p> </p><p> </p><p></p><p>Thus;</p><p></p><p>The magnification will be 2 cm .</p><p></p><p>$