Question

Find the distance between the points $\left( a\cos {{25}^{\circ }},0 \right)$ and $\left( 0,a\cos {{65}^{\circ }} \right)$.

Answer 1

Hint: We use trigonometrical value to find out the coordinates. Then we use the distance formula between two points to find the solution. The points need to be represented in the formula to find the answer. Then we use trigonometric identities to find out the final answer.

Complete step by step answer:
The given points are $\left( a\cos {{25}^{\circ }},0 \right)$ and $\left( 0,a\cos {{65}^{\circ }} \right)$.
Let’s term them as $p\equiv \left( a\cos {{25}^{\circ }},0 \right)$ and $q\equiv \left( 0,a\cos {{65}^{\circ }} \right)$.
We know if two points are $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ then the distance between them will be \[\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}\] unit.
To understand the concept, we are taking an example. Let two points be $\left( 1,1 \right)$ and $\left( 2,2 \right)$.
The distance between these two points be \[\sqrt{{{\left( 2-1 \right)}^{2}}+{{\left( 2-1 \right)}^{2}}}=\sqrt{1+1}=\sqrt{2}\] unit.
The distance also can be geometrically defined as the length of the straight line joining those two points.
So, the distance between $p\equiv \left( a\cos {{25}^{\circ }},0 \right)$ and $q\equiv \left( 0,a\cos {{65}^{\circ }} \right)$ will be \[\sqrt{{{\left( a\cos {{25}^{\circ }}-0 \right)}^{2}}+{{\left( 0-a\cos {{65}^{\circ }} \right)}^{2}}}\] unit.
Now we find the solution.
So, \[\sqrt{{{\left( a\cos {{25}^{\circ }}-0 \right)}^{2}}+{{\left( 0-a\cos {{65}^{\circ }} \right)}^{2}}}=\sqrt{{{a}^{2}}{{\cos }^{2}}{{25}^{\circ }}+{{a}^{2}}{{\cos }^{2}}{{65}^{\circ }}}\]
To simplify the equation, we use trigonometric identities to change cos into sin form.
We know $\cos \left( \dfrac{\pi }{2}-\alpha \right)=\sin \alpha $. We place $\alpha ={{65}^{\circ }}$.
So, $\cos \left( {{90}^{\circ }}-{{65}^{\circ }} \right)=\sin {{65}^{\circ }}\Rightarrow \cos {{25}^{\circ }}=\sin {{65}^{\circ }}$
Now the equation becomes \[\sqrt{{{a}^{2}}{{\cos }^{2}}{{25}^{\circ }}+{{a}^{2}}{{\cos }^{2}}{{65}^{\circ }}}=\sqrt{{{a}^{2}}{{\sin }^{2}}{{65}^{\circ }}+{{a}^{2}}{{\cos }^{2}}{{65}^{\circ }}}=\sqrt{{{a}^{2}}\left( {{\sin }^{2}}{{65}^{\circ }}+{{\cos }^{2}}{{65}^{\circ }} \right)}\]
We know that \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow {{\sin }^{2}}{{65}^{\circ }}+{{\cos }^{2}}{{65}^{\circ }}=1\].
The distance becomes \[\sqrt{{{a}^{2}}\left( {{\sin }^{2}}{{65}^{\circ }}+{{\cos }^{2}}{{65}^{\circ }} \right)}=\sqrt{{{a}^{2}}}=a\] unit. Distance can’t be negative.
So, the distance between the points $\left( a\cos {{25}^{\circ }},0 \right)$ and $\left( 0,a\cos {{65}^{\circ }} \right)$ is a unit.

Note: We need to remember we can also change the points from its polar form to coordinate form in the start. Then we can use the distance formula to find the solution. But in this case the given angle is not well known to have a value. So, we can’t use that as the problem would become more tough to solve then. That’s why we first find out the distance then use the trigonometric identity to find the answer independent of any trigonometric form present.