Question

The value of $\sin \left( {40^\circ 35'} \right)\cos \left( {19^\circ 25'} \right) + \cos \left( {40^\circ 35'} \right)\sin \left( {19^\circ 25'} \right) =$
(a) 1
(b) $\dfrac{{\sqrt 3 }}{2}$
(c) 0
(d) $- 1$

Answer 1

Here, we need to find the value of the given expression. First, we will rewrite the given angles in degrees. Then, using the formula for sine of the sum of two angles and simplifying, we will find the value of the given expression.

Formula Used:
We will use the formula of the sine of the sum of two angles, ${\rm{sin}}\left( {A + B} \right) = \sin A\cos B + \cos A\sin B$.

Complete step-by-step answer:
First, we will convert the angle $40^\circ 35'$ to degrees.
We can write the angle $40^\circ 35'$ as the sum of $40^\circ$ and $35'$.
Therefore, we get
$\Rightarrow 40^\circ 35' = 40^\circ + 35'$
We will use unitary method to convert 35 minutes to degrees.
We know that 1 degree is equal to 60 minutes.
Therefore, we get
60 minutes $=$ 1 degree
Dividing both sides by 60, we get
$\Rightarrow$1 minute $= \dfrac{1}{{60}}$ degree
Multiplying both sides by 35, we get
$\Rightarrow$35 minutes $= \dfrac{{35}}{{60}}$ degree
Simplifying the expression, we get
$\Rightarrow 35' = \dfrac{7}{{12}}$degree
Substituting $35' = \dfrac{7}{{12}}$ degree in the equation $40^\circ 35' = 40^\circ + 35'$, we get
$\Rightarrow 40^\circ 35' = 40^\circ + \dfrac{7}{{12}}$ degree
$\Rightarrow 40^\circ 35' = \left( {40 + \dfrac{7}{{12}}} \right)$ degree
Taking the L.C.M., we get
$\Rightarrow 40^\circ 35' = \left( {\dfrac{{480 + 7}}{{12}}} \right)$ degree
Adding the terms, we get
$\Rightarrow 40^\circ 35' = \left( {\dfrac{{487}}{{12}}} \right)$ degree
Now, we will convert the angle $19^\circ 25'$ to degrees.
We can write the angle $19^\circ 25'$ as the sum of $19^\circ$ and $25'$.
Therefore, we get
$\Rightarrow 19^\circ 25' = 19^\circ + 25'$
We will convert 25 minutes to degrees.
Multiplying both sides of the equation 1 minute $= \dfrac{1}{{60}}$ degree by 25, we get
$\Rightarrow$25 minutes $= \dfrac{{25}}{{60}}$ degree
Simplifying the expression, we get

Substituting $$25' = \dfrac{5}{{12}}$$ degree in the equation $$19^\circ 25' = 19^\circ + 25'$$, we get $$ \Rightarrow 19^\circ 25' = 19^\circ + \dfrac{5}{{12}}$$ degree $$ \Rightarrow 19^\circ 25' = \left( {19 + \dfrac{5}{{12}}} \right)$$ degree Taking the L.C.M., we get $$ \Rightarrow 19^\circ 25' = \left( {\dfrac{{228 + 5}}{{12}}} \right)$$ degree Adding the terms, we get $$ \Rightarrow 19^\circ 25' = \left( {\dfrac{{233}}{{12}}} \right)$$ degree Now, we will simplify the given expression. We can rewrite the equation $$\sin \left( {40^\circ 35'} \right)\cos \left( {19^\circ 25'} \right) + \cos \left( {40^\circ 35'} \right)\sin \left( {19^\circ 25'} \right)$$ as $$\sin {\left( {\dfrac{{487}}{{12}}} \right)^{\rm{o}}}\cos {\left( {\dfrac{{233}}{{12}}} \right)^{\rm{o}}} + \cos {\left( {\dfrac{{487}}{{12}}} \right)^{\rm{o}}}\sin {\left( {\dfrac{{233}}{{12}}} \right)^{\rm{o}}}$$. We know that the sine of the sum of two angles is given by the formula $${\rm{sin}}\left( {A + B} \right) = \sin A\cos B + \cos A\sin B$$. Substituting $$A = {\left( {\dfrac{{487}}{{12}}} \right)^{\rm{o}}}$$ and $$B = {\left( {\dfrac{{233}}{{12}}} \right)^{\rm{o}}}$$ in the formula, we get $${\rm{sin}}\left[ {{{\left( {\dfrac{{487}}{{12}}} \right)}^{\rm{o}}} + {{\left( {\dfrac{{233}}{{12}}} \right)}^{\rm{o}}}} \right] = \sin {\left( {\dfrac{{487}}{{12}}} \right)^{\rm{o}}}\cos {\left( {\dfrac{{233}}{{12}}} \right)^{\rm{o}}} + \cos {\left( {\dfrac{{487}}{{12}}} \right)^{\rm{o}}}\sin {\left( {\dfrac{{233}}{{12}}} \right)^{\rm{o}}}$$ Thus, we get $$ \Rightarrow {\rm{sin}}{\left( {\dfrac{{487 + 233}}{{12}}} \right)^{\rm{o}}} = \sin \left( {40^\circ 35'} \right)\cos \left( {19^\circ 25'} \right) + \cos \left( {40^\circ 35'} \right)\sin \left( {19^\circ 25'} \right)$$ Adding the terms, we get $$ \Rightarrow {\rm{sin}}{\left( {\dfrac{{720}}{{12}}} \right)^{\rm{o}}} = \sin \left( {40^\circ 35'} \right)\cos \left( {19^\circ 25'} \right) + \cos \left( {40^\circ 35'} \right)\sin \left( {19^\circ 25'} \right)$$ Simplifying the expression and rewriting the equation, we get $$ \Rightarrow \sin \left( {40^\circ 35'} \right)\cos \left( {19^\circ 25'} \right) + \cos \left( {40^\circ 35'} \right)\sin \left( {19^\circ 25'} \right) = {\rm{sin}}{60^{\rm{o}}}$$ The sine of angle measuring $${60^{\rm{o}}}$$ is equal to $$\dfrac{{\sqrt 3 }}{2}$$. Substituting $${\rm{sin}}{60^{\rm{o}}} = \dfrac{{\sqrt 3 }}{2}$$ in the equation, we get $$ \Rightarrow \sin \left( {40^\circ 35'} \right)\cos \left( {19^\circ 25'} \right) + \cos \left( {40^\circ 35'} \right)\sin \left( {19^\circ 25'} \right) = \dfrac{{\sqrt 3 }}{2}$$ $$\therefore $$ The value of the expression $$\sin \left( {40^\circ 35'} \right)\cos \left( {19^\circ 25'} \right) + \cos \left( {40^\circ 35'} \right)\sin \left( {19^\circ 25'} \right)$$ is $$\dfrac{{\sqrt 3 }}{2}$$. **The correct option is option (b).** **Note:** We used a unitary method to convert 35 minutes and 25 minutes to degrees. Unitary method is a method where first, the per unit quantity is calculated, and then the number of units are multiplied. Here, we first calculated the value of 1 minute in degrees, and then multiplied it by 35 and 25 to get the value of 35 minutes and 25 minutes in degrees respectively.