Question

Find the distance between the points$\left( a\cos {{35}^{\circ }},0 \right),\left( 0,a\cos {{55}^{\circ }} \right)$. $$$$

Answer 1

We use the distance formula between two points $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ $AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ to find the distance and then use reduction formula of sine and cosine $\cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta$and Pythagorean trigonometric identity involving sine and cosine ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to find simplify.$$$$

Complete step by step answer:
We know that distance between two points is the measure of shortest path we need to move from one point to another and it is given by length of line segment joining those two points. The distance between two points $A$ and $B$ on a plane with their coordinates $A\left( {{x}_{1}},{{y}_{1}} \right)$ and $B\left( {{x}_{2}},{{y}_{2}} \right)$ is given by the formula
$AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
The distance is always a positive quantity and hence we take only positive square root. We know from Pythagorean trigonometric identity for sine and cosine with any acute angle $\theta$ we have;
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
We know the reduction formula about angle ${{90}^{\circ }}$ for two complimentary angles $\theta$ and ${{90}^{\circ }}-\theta$ as

& \sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta \\\ & \cos \left( {{90}^{\circ }}-\theta \right)=\sin \theta \\\ \end{aligned}$$ We are asked in the question to find the distance between$\left( a\cos {{35}^{\circ }},0 \right),\left( 0,a\cos {{55}^{\circ }} \right)$. Let us denote the points as $A\left( a\cos {{35}^{\circ }},0 \right),B\left( 0,a\cos {{55}^{\circ }} \right)$ and find the distance between them as $AB$ using the distance formula for${{x}_{1}}=a\cos {{35}^{\circ }},{{y}_{1}}=0,{{x}_{2}}=a\cos {{55}^{\circ }},{{y}_{2}}=0$. We have; $$\begin{aligned} & AB=\sqrt{{{\left( a\cos {{55}^{\circ }}-0 \right)}^{2}}+{{\left( 0-a\cos {{35}^{\circ }} \right)}^{2}}} \\\ & \Rightarrow AB=\sqrt{{{\left( a\cos {{55}^{\circ }} \right)}^{2}}+{{\left( -a\cos {{35}^{\circ }} \right)}^{2}}} \\\ \end{aligned}$$ We use the reduction formula for complimentary angles ${{35}^{\circ }}$ and ${{90}^{\circ }}-{{35}^{\circ }}={{55}^{\circ }}$and convert $\cos {{55}^{\circ }}$ to $\sin {{35}^{\circ }}$as $$\begin{aligned} & \Rightarrow AB=\sqrt{{{\left( a\cos \left( {{90}^{\circ }}-{{35}^{\circ }} \right) \right)}^{2}}+{{\left( a\cos {{35}^{\circ }} \right)}^{2}}} \\\ & \Rightarrow AB=\sqrt{{{\left( a\sin {{35}^{\circ }} \right)}^{2}}+{{\left( a\cos {{35}^{\circ }} \right)}^{2}}} \\\ & \Rightarrow AB=\sqrt{{{a}^{2}}{{\sin }^{2}}{{35}^{\circ }}+{{a}^{2}}{{\cos }^{2}}{{35}^{\circ }}} \\\ & \Rightarrow AB=\sqrt{{{a}^{2}}\left( {{\sin }^{2}}{{35}^{\circ }}+{{\cos }^{2}}{{35}^{\circ }} \right)} \\\ \end{aligned}$$ We use the Pythagorean trigonometric identity for sine and cosine with any angle $\theta ={{35}^{\circ }}$ to have; $$\begin{aligned} & \Rightarrow AB=\sqrt{{{a}^{2}}\cdot 1} \\\ & \Rightarrow AB=a \\\ \end{aligned}$$  **So the distance between the points $A\left( a\cos {{35}^{\circ }},0 \right),B\left( 0,a\cos {{55}^{\circ }} \right)$ is $a$ unit. $$$$** **Note:** We note that the given points $A\left( a\cos {{35}^{\circ }},0 \right),B\left( 0,a\cos {{55}^{\circ }} \right)$ lie on coordinate axes which we have shown in the figure for $a=5$ and the line joining them wills always in the first quadrant. We must be careful of confusion between the complementary angle for formula and shift by right angle formula which is given as $\cos \left( \theta \pm {{90}^{\circ }} \right)=\mp \sin \theta $