
Simplify the given trigonometric expression \[\sin 30{}^\circ 230'+\cos 5{}^\circ 33'\]
Answer
458.4k+ views
Hint: According to the Sexagesimal system, 1degree is considered as 60 minutes which in turn is considered as (60 \[\times \] 60) seconds. It means that \[{{1}^{\circ }}=60'=3600''\]. For suppose, if we have an angle \[x{}^\circ y'\] then we can convert the angle completely into degrees which is equal to \[\left( x+\dfrac{y}{60} \right){}^\circ \]. It can be converted completely into minutes which is equal to\[(60x+y)'\].
Complete step-by-step solution:
From the question, let the simplified value of \[\sin 30{}^\circ 230'+\cos 5{}^\circ 33'\]be (say) y=\[\sin 30{}^\circ 230'+\cos 5{}^\circ 33'\].
Let the angles in y be named as A=\[30{}^\circ 230'\] and angle B=\[5{}^\circ 33'\]
Now let us do the conversion of the given angles completely into degrees then we get A and B angles as
A = \[{{\left( 30+\dfrac{230}{60} \right)}^{\circ }}\] = \[{{\left( 30+3.83 \right)}^{\circ }}\] = \[(33.83){}^\circ \]
and the angle B will be
B = \[{{\left( 5+\dfrac{33}{60} \right)}^{\circ }}\] = \[{{\left( 5+0.55 \right)}^{\circ }}\] = \[5.55{}^\circ \]
On substituting these converted angles into the question given we get,
y = \[\sin 33.83{}^\circ +\cos 5.55{}^\circ \]
we can get the sine value of 33.83 degrees and the cosine value of 5.55 degrees from the sine table and cosine table respectively.
From the sine table we get the value of \[\sin 30{}^\circ \] is equal to 0.5 and then from the cosine table we get the value of \[\cos 5{}^\circ 30'\] equal to 0.9954. Also, the mean difference for 3 minutes that is,\[3'\] = 0.0001.
Then the value cos B = \[\cos 5{}^\circ 33'\] = \[\cos 5{}^\circ 30'\] – cos \[3'\] which is equal to
= 0.9954 – 0.0001
= 0.9953
That is \[\cos 5{}^\circ 33'\] = 0.9953 -------(1)
From sine table, we get \[\sin 33.83{}^\circ \] = 0.5567
That is \[\sin 30{}^\circ 230'\] = 0.5567 --------(2)
Hence, y = \[\sin 30{}^\circ 230'+\cos 5{}^\circ 33'\] = 0.5567 + 0.9953 = 1.5520 is the required answer.
Note: We need to be very careful while doing calculations. We can solve the above question by converting the angles completely into minutes. Using the mean difference value for \[3'\] = 0.0001, we can find the sine and cosine values by addition and subtraction respectively.
Complete step-by-step solution:
From the question, let the simplified value of \[\sin 30{}^\circ 230'+\cos 5{}^\circ 33'\]be (say) y=\[\sin 30{}^\circ 230'+\cos 5{}^\circ 33'\].
Let the angles in y be named as A=\[30{}^\circ 230'\] and angle B=\[5{}^\circ 33'\]
Now let us do the conversion of the given angles completely into degrees then we get A and B angles as
A = \[{{\left( 30+\dfrac{230}{60} \right)}^{\circ }}\] = \[{{\left( 30+3.83 \right)}^{\circ }}\] = \[(33.83){}^\circ \]
and the angle B will be
B = \[{{\left( 5+\dfrac{33}{60} \right)}^{\circ }}\] = \[{{\left( 5+0.55 \right)}^{\circ }}\] = \[5.55{}^\circ \]
On substituting these converted angles into the question given we get,
y = \[\sin 33.83{}^\circ +\cos 5.55{}^\circ \]
we can get the sine value of 33.83 degrees and the cosine value of 5.55 degrees from the sine table and cosine table respectively.
From the sine table we get the value of \[\sin 30{}^\circ \] is equal to 0.5 and then from the cosine table we get the value of \[\cos 5{}^\circ 30'\] equal to 0.9954. Also, the mean difference for 3 minutes that is,\[3'\] = 0.0001.
Then the value cos B = \[\cos 5{}^\circ 33'\] = \[\cos 5{}^\circ 30'\] – cos \[3'\] which is equal to
= 0.9954 – 0.0001
= 0.9953
That is \[\cos 5{}^\circ 33'\] = 0.9953 -------(1)
From sine table, we get \[\sin 33.83{}^\circ \] = 0.5567
That is \[\sin 30{}^\circ 230'\] = 0.5567 --------(2)
Hence, y = \[\sin 30{}^\circ 230'+\cos 5{}^\circ 33'\] = 0.5567 + 0.9953 = 1.5520 is the required answer.
Note: We need to be very careful while doing calculations. We can solve the above question by converting the angles completely into minutes. Using the mean difference value for \[3'\] = 0.0001, we can find the sine and cosine values by addition and subtraction respectively.
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