# Consider $\vartriangle {\text{ABC}}$, right-angled at C, in which AB = 29 units, BC = 21 units and $\angle {\text{ABC}} = \theta $. Determine the value of ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2}$.

Last updated date: 19th Mar 2023

•

Total views: 304.5k

•

Views today: 7.86k

Answer

Verified

304.5k+ views

Hint- Here, we will be proceeding by determining the length of the side AC which is not given in the problem using the Pythagoras theorem and then applying simple trigonometric formulas $\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$ and $\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$ to determine the value of the expression.

Complete step-by-step answer:

Given that we have a right triangle $\vartriangle {\text{ABC}}$ with right-angled at C which means $\angle {\text{ACB}} = {90^0}$

AB = 29 units and BC = 21 units

As we know in any right triangle, the side opposite to the right angle is termed as the hypotenuse, the side opposite to the considered angle ($\theta $ in this case) is termed as the perpendicular and the remaining side is termed as the base.

In $\vartriangle {\text{ABC}}$, side AB (opposite to right angle at C) is the hypotenuse of the triangle, side AC (opposite to angle $\theta $) is the perpendicular of the triangle and the remaining side BC is the base of the triangle.

According to Pythagoras theorem,

In any right angled triangle, ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$

\[

\Rightarrow {\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} + {\left( {{\text{BC}}} \right)^2} \\

\Rightarrow {\left( {{\text{29}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} + {\left( {{\text{21}}} \right)^2} \\

\Rightarrow {\left( {{\text{AC}}} \right)^2} = {\left( {{\text{29}}} \right)^2} - {\left( {{\text{21}}} \right)^2} = 841 - 441 = 400 \\

\Rightarrow {\text{AC}} = \sqrt {400} = 20{\text{ units}} \\

\\

\]

Also, $\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$ and $\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$

Using the above trigonometric formulas, we can write

$\sin \theta = \dfrac{{{\text{AC}}}}{{{\text{AB}}}} = \dfrac{{20}}{{29}}$ and $\cos \theta = \dfrac{{{\text{BC}}}}{{{\text{AB}}}} = \dfrac{{21}}{{29}}$

Now, let us substitute the values obtained above in the expression whose value is required.

So, ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = {\left( {\dfrac{{20}}{{29}}} \right)^2} + {\left( {\dfrac{{21}}{{29}}} \right)^2} = \dfrac{{400}}{{841}} + \dfrac{{441}}{{841}} = \dfrac{{400 + 441}}{{841}} = \dfrac{{841}}{{841}} = 1$

Hence, the value of the expression ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2}$ is 1.

This problem can also be solved in one line because ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1$ is an identity is always true.

Note- In this particular problem, only positive value of AC is considered (negative value of AC is neglected) because AC is the length of the side of the given triangle and length of side of any triangle is always positive. Also, the choice of perpendicular and base of any triangle depends upon the choice of the angle considered.

Complete step-by-step answer:

Given that we have a right triangle $\vartriangle {\text{ABC}}$ with right-angled at C which means $\angle {\text{ACB}} = {90^0}$

AB = 29 units and BC = 21 units

As we know in any right triangle, the side opposite to the right angle is termed as the hypotenuse, the side opposite to the considered angle ($\theta $ in this case) is termed as the perpendicular and the remaining side is termed as the base.

In $\vartriangle {\text{ABC}}$, side AB (opposite to right angle at C) is the hypotenuse of the triangle, side AC (opposite to angle $\theta $) is the perpendicular of the triangle and the remaining side BC is the base of the triangle.

According to Pythagoras theorem,

In any right angled triangle, ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$

\[

\Rightarrow {\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} + {\left( {{\text{BC}}} \right)^2} \\

\Rightarrow {\left( {{\text{29}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} + {\left( {{\text{21}}} \right)^2} \\

\Rightarrow {\left( {{\text{AC}}} \right)^2} = {\left( {{\text{29}}} \right)^2} - {\left( {{\text{21}}} \right)^2} = 841 - 441 = 400 \\

\Rightarrow {\text{AC}} = \sqrt {400} = 20{\text{ units}} \\

\\

\]

Also, $\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$ and $\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$

Using the above trigonometric formulas, we can write

$\sin \theta = \dfrac{{{\text{AC}}}}{{{\text{AB}}}} = \dfrac{{20}}{{29}}$ and $\cos \theta = \dfrac{{{\text{BC}}}}{{{\text{AB}}}} = \dfrac{{21}}{{29}}$

Now, let us substitute the values obtained above in the expression whose value is required.

So, ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = {\left( {\dfrac{{20}}{{29}}} \right)^2} + {\left( {\dfrac{{21}}{{29}}} \right)^2} = \dfrac{{400}}{{841}} + \dfrac{{441}}{{841}} = \dfrac{{400 + 441}}{{841}} = \dfrac{{841}}{{841}} = 1$

Hence, the value of the expression ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2}$ is 1.

This problem can also be solved in one line because ${\left( {\sin \theta } \right)^2} + {\left( {\cos \theta } \right)^2} = 1$ is an identity is always true.

Note- In this particular problem, only positive value of AC is considered (negative value of AC is neglected) because AC is the length of the side of the given triangle and length of side of any triangle is always positive. Also, the choice of perpendicular and base of any triangle depends upon the choice of the angle considered.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE