Given \[\Delta ABC\] right angled at C in which AB = 29 units, BC = 21 units and \[\angle ABC=\theta \]. Find \[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta \].
Answer
681.9k+ views
Hint:Given two sides of the right angled triangle. Find the \[{{3}^{rd}}\] side, by using basic geometry. Find the value of \[\sin \theta \] and \[\cos \theta \] from the figure. Substitute these values in \[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta \] and get the value.
Complete step-by-step answer:
Consider the figure drawn,
From, \[\vartriangle ABC\], AB = 29 units and BC = 21 units, \[\angle ABC=\theta \].
Using Pythagoras theorem,
\[\begin{align}
& {{\left( Hypotenuse \right)}^{2}}={{\left( Height \right)}^{2}}+{{\left( Base \right)}^{2}} \\
& \Rightarrow A{{B}^{2}}=A{{C}^{2}}+A{{B}^{2}} \\
& A{{C}^{2}}=A{{B}^{2}}-B{{C}^{2}} \\
& A{{C}^{2}}={{\left( 29 \right)}^{2}}-{{\left( 21 \right)}^{2}} \\
\end{align}\]
Using, \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]
\[\begin{align}
& A{{C}^{2}}=\left( 29-21 \right)\left( 29+21 \right) \\
& A{{C}^{2}}=8\times 50=400 \\
& \therefore AC=\sqrt{400}=20 \\
\end{align}\]
Now, \[\sin \theta \]= \[\dfrac{Opposite side}{Hypotenuse side}\].
\[\sin \theta =\dfrac{AC}{AB}=\dfrac{20}{29}\]
\[\cos \theta \]= \[\dfrac{Adjacent side}{Hypotenuse side}\].
\[\cos \theta =\dfrac{BC}{AB}=\dfrac{21}{29}\]
We need to find the value of \[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta \].
Putting values of \[\cos \theta =\dfrac{21}{29}\] and \[\sin \theta =\dfrac{20}{29}\].
\[\begin{align}
& {{\cos }^{2}}\theta -{{\sin }^{2}}\theta ={{\left( \dfrac{21}{29} \right)}^{2}}-{{\left( \dfrac{20}{29} \right)}^{2}}=\dfrac{{{21}^{2}}-{{20}^{2}}}{{{29}^{2}}} \\
& \because {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \\
& \Rightarrow \dfrac{\left( 21-20 \right)\left( 21+20 \right)}{{{29}^{2}}}=\dfrac{1\times 41}{{{29}^{2}}}=\dfrac{41}{841} \\
& \therefore {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\dfrac{41}{841} \\
\end{align}\]
Note: Find altitude of the \[\Delta ABC\], which will give the values of \[\cos \theta \] and \[\sin \theta \]. Put the values in the entity to find the desired answer.Students should remember pythagoras theorem , trigonometric identities and trigonometric ratios for solving these types of problems.
Complete step-by-step answer:
Consider the figure drawn,
From, \[\vartriangle ABC\], AB = 29 units and BC = 21 units, \[\angle ABC=\theta \].
Using Pythagoras theorem,
\[\begin{align}
& {{\left( Hypotenuse \right)}^{2}}={{\left( Height \right)}^{2}}+{{\left( Base \right)}^{2}} \\
& \Rightarrow A{{B}^{2}}=A{{C}^{2}}+A{{B}^{2}} \\
& A{{C}^{2}}=A{{B}^{2}}-B{{C}^{2}} \\
& A{{C}^{2}}={{\left( 29 \right)}^{2}}-{{\left( 21 \right)}^{2}} \\
\end{align}\]
Using, \[{{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)\]
\[\begin{align}
& A{{C}^{2}}=\left( 29-21 \right)\left( 29+21 \right) \\
& A{{C}^{2}}=8\times 50=400 \\
& \therefore AC=\sqrt{400}=20 \\
\end{align}\]
Now, \[\sin \theta \]= \[\dfrac{Opposite side}{Hypotenuse side}\].
\[\sin \theta =\dfrac{AC}{AB}=\dfrac{20}{29}\]
\[\cos \theta \]= \[\dfrac{Adjacent side}{Hypotenuse side}\].
\[\cos \theta =\dfrac{BC}{AB}=\dfrac{21}{29}\]
We need to find the value of \[{{\cos }^{2}}\theta -{{\sin }^{2}}\theta \].
Putting values of \[\cos \theta =\dfrac{21}{29}\] and \[\sin \theta =\dfrac{20}{29}\].
\[\begin{align}
& {{\cos }^{2}}\theta -{{\sin }^{2}}\theta ={{\left( \dfrac{21}{29} \right)}^{2}}-{{\left( \dfrac{20}{29} \right)}^{2}}=\dfrac{{{21}^{2}}-{{20}^{2}}}{{{29}^{2}}} \\
& \because {{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right) \\
& \Rightarrow \dfrac{\left( 21-20 \right)\left( 21+20 \right)}{{{29}^{2}}}=\dfrac{1\times 41}{{{29}^{2}}}=\dfrac{41}{841} \\
& \therefore {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =\dfrac{41}{841} \\
\end{align}\]
Note: Find altitude of the \[\Delta ABC\], which will give the values of \[\cos \theta \] and \[\sin \theta \]. Put the values in the entity to find the desired answer.Students should remember pythagoras theorem , trigonometric identities and trigonometric ratios for solving these types of problems.
Recently Updated Pages
What is BLO What is the full form of BLO class 8 social science CBSE

Explain the Treaty of Vienna of 1815 class 10 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

A Paragraph on Pollution in about 100-150 Words

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

What is the Full Form of PVC, PET, HDPE, LDPE, PP and PS ?

