
A rectangle has dimensions 5.5cm by 2.8cm. How do you determine the measure of the angles at the point where the diagonals intersect?
Answer
529.5k+ views
Hint: In this problem, we have to find the measure of the angles at the point where the diagonals intersect in a rectangle has dimensions 5.5cm by 2.8cm. We can first draw the diagram with angles in it. We will get a triangle with which we can find the angles using the Pythagoras theorem.
Complete step by step solution:
We know that the given dimensions are,
Length = 5.5cm and breadth = 2.8cm.
We can draw the diagram.
Now we take the triangle 1 and we can draw it separately, we get
Now we can form right triangle from the above diagram to use the Pythagoras theorem, we get
We can see from the above diagram,
\[\begin{align}
& h=\dfrac{5.5}{2}=2.75 \\
& a=\dfrac{2.8}{2}=1.4 \\
\end{align}\]
We can now use the Pythagoras theorem for the above right triangle, we get
\[\begin{align}
& \Rightarrow {{c}^{2}}={{a}^{2}}+{{h}^{2}} \\
& \Rightarrow {{c}^{2}}={{\left( 1.4 \right)}^{2}}+{{\left( 2.75 \right)}^{2}}=9.52 \\
& \Rightarrow c=3.1cm \\
\end{align}\]
We can now use the trigonometric formula, we get
\[a=c\sin \left( \dfrac{\alpha }{2} \right)\]
We can substitute the values, we get
\[\begin{align}
& \Rightarrow \sin \left( \dfrac{\alpha }{2} \right)=\dfrac{1.4}{3.1}=0.45 \\
& \Rightarrow \dfrac{\alpha }{2}=\arcsin \left( 0.45 \right) \\
\end{align}\]
We can use calculator to find the above value, we get
\[\begin{align}
& \Rightarrow \dfrac{\alpha }{2}=26.98 \\
& \Rightarrow \alpha ={{53.96}^{\circ }} \\
\end{align}\]
We also know that,
\[\alpha +\alpha +\beta +\beta ={{360}^{\circ }}\]
We can write it as,
\[\begin{align}
& \Rightarrow 2\beta ={{360}^{\circ }}-2\alpha \\
& \Rightarrow \beta =\dfrac{{{360}^{\circ }}-2\left( 53.96 \right)}{2}={{126.04}^{\circ }} \\
\end{align}\]
Therefore, the angles are, \[\alpha ={{53.96}^{\circ }}\] and \[\beta ={{126.04}^{\circ }}\].
Note: Students make mistakes while using the Pythagoras theorem. We should first take the part of the right triangle from the given rectangular part to use the Pythagoras theorem. We can use calculators to find the sine values for some angles. We should concentrate on the diagram part while marking the points.
Complete step by step solution:
We know that the given dimensions are,
Length = 5.5cm and breadth = 2.8cm.
We can draw the diagram.
Now we take the triangle 1 and we can draw it separately, we get
Now we can form right triangle from the above diagram to use the Pythagoras theorem, we get
We can see from the above diagram,
\[\begin{align}
& h=\dfrac{5.5}{2}=2.75 \\
& a=\dfrac{2.8}{2}=1.4 \\
\end{align}\]
We can now use the Pythagoras theorem for the above right triangle, we get
\[\begin{align}
& \Rightarrow {{c}^{2}}={{a}^{2}}+{{h}^{2}} \\
& \Rightarrow {{c}^{2}}={{\left( 1.4 \right)}^{2}}+{{\left( 2.75 \right)}^{2}}=9.52 \\
& \Rightarrow c=3.1cm \\
\end{align}\]
We can now use the trigonometric formula, we get
\[a=c\sin \left( \dfrac{\alpha }{2} \right)\]
We can substitute the values, we get
\[\begin{align}
& \Rightarrow \sin \left( \dfrac{\alpha }{2} \right)=\dfrac{1.4}{3.1}=0.45 \\
& \Rightarrow \dfrac{\alpha }{2}=\arcsin \left( 0.45 \right) \\
\end{align}\]
We can use calculator to find the above value, we get
\[\begin{align}
& \Rightarrow \dfrac{\alpha }{2}=26.98 \\
& \Rightarrow \alpha ={{53.96}^{\circ }} \\
\end{align}\]
We also know that,
\[\alpha +\alpha +\beta +\beta ={{360}^{\circ }}\]
We can write it as,
\[\begin{align}
& \Rightarrow 2\beta ={{360}^{\circ }}-2\alpha \\
& \Rightarrow \beta =\dfrac{{{360}^{\circ }}-2\left( 53.96 \right)}{2}={{126.04}^{\circ }} \\
\end{align}\]
Therefore, the angles are, \[\alpha ={{53.96}^{\circ }}\] and \[\beta ={{126.04}^{\circ }}\].
Note: Students make mistakes while using the Pythagoras theorem. We should first take the part of the right triangle from the given rectangular part to use the Pythagoras theorem. We can use calculators to find the sine values for some angles. We should concentrate on the diagram part while marking the points.
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