
In \[\Delta ABC\], if \[a = 3\], b = 4, c = 5, then find \[\sin 2B\].
A. \[\dfrac{4}{5}\]
B. \[\dfrac{3}{{20}}\]
C. \[\dfrac{{24}}{{25}}\]
D. \[\dfrac{1}{{50}}\]
Answer
161.4k+ views
Hint: First we will find \[\cos B\] using cosine law. Then we will use trigonometric identity to calculate the value of \[\sin B\]. Then put the value of \[\cos B\] and \[\sin B\] in the double angle formula of sine to get the required solution.
Formula used:
Cosine formula:
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
Trigonometry identity:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Double angle formula of sine:
\[\sin 2\theta = 2\sin \theta \cos \theta \]
Complete step by step solution:
We will find the value of \[\cos B\] using the cosine law \[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
Substitute \[a = 3\], b = 4, c = 5 in \[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{4^2} = {3^2} + {5^2} - 2 \cdot 3 \cdot 5\cos B\]
\[ \Rightarrow 16 = 9 + 25 - 30\cos B\]
\[ \Rightarrow 16 - 25 - 9 = - 30\cos B\]
\[ \Rightarrow - 18 = - 30\cos B\]
Divide both sides by -30
\[ \Rightarrow \cos B = \dfrac{{ - 18}}{{ - 30}}\]
\[ \Rightarrow \cos B = \dfrac{3}{5}\]
Now we will calculate the value of \[\sin B\] by using trigonometry identity.
Substitute \[\cos B = \dfrac{3}{5}\] in \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[{\sin ^2}\theta + {\left( {\dfrac{3}{5}} \right)^2} = 1\]
\[ \Rightarrow {\sin ^2}\theta + \dfrac{9}{{25}} = 1\]
Subtract \[\dfrac{9}{{25}}\] from both sides
\[ \Rightarrow {\sin ^2}\theta + \dfrac{9}{{25}} - \dfrac{9}{{25}} = 1 - \dfrac{9}{{25}}\]
\[ \Rightarrow {\sin ^2}\theta = \dfrac{{16}}{{25}}\]
Take square root on both sides:
\[ \Rightarrow \sin \theta = \dfrac{4}{5}\]
Now we will substitute the value of \[\cos B\] and \[\sin B\] in the double angle formula \[\sin 2\theta = 2\sin \theta \cos \theta \]:
\[\sin 2B = 2\sin B\cos B\]
\[ \Rightarrow \sin 2B = 2 \cdot \dfrac{4}{5} \cdot \dfrac{3}{5}\]
\[ \Rightarrow \sin 2B = \dfrac{{24}}{{25}}\]
Hence option C is the correct option.
Note: Students often make a common mistake to calculate the value of \[\sin B\]. They used sine law to find it but to apply sine law we need an angle of the triangle. The correct way to find the \[\sin B\] is using the trigonometry identity.
Formula used:
Cosine formula:
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
Trigonometry identity:
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Double angle formula of sine:
\[\sin 2\theta = 2\sin \theta \cos \theta \]
Complete step by step solution:
We will find the value of \[\cos B\] using the cosine law \[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
Substitute \[a = 3\], b = 4, c = 5 in \[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{4^2} = {3^2} + {5^2} - 2 \cdot 3 \cdot 5\cos B\]
\[ \Rightarrow 16 = 9 + 25 - 30\cos B\]
\[ \Rightarrow 16 - 25 - 9 = - 30\cos B\]
\[ \Rightarrow - 18 = - 30\cos B\]
Divide both sides by -30
\[ \Rightarrow \cos B = \dfrac{{ - 18}}{{ - 30}}\]
\[ \Rightarrow \cos B = \dfrac{3}{5}\]
Now we will calculate the value of \[\sin B\] by using trigonometry identity.
Substitute \[\cos B = \dfrac{3}{5}\] in \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[{\sin ^2}\theta + {\left( {\dfrac{3}{5}} \right)^2} = 1\]
\[ \Rightarrow {\sin ^2}\theta + \dfrac{9}{{25}} = 1\]
Subtract \[\dfrac{9}{{25}}\] from both sides
\[ \Rightarrow {\sin ^2}\theta + \dfrac{9}{{25}} - \dfrac{9}{{25}} = 1 - \dfrac{9}{{25}}\]
\[ \Rightarrow {\sin ^2}\theta = \dfrac{{16}}{{25}}\]
Take square root on both sides:
\[ \Rightarrow \sin \theta = \dfrac{4}{5}\]
Now we will substitute the value of \[\cos B\] and \[\sin B\] in the double angle formula \[\sin 2\theta = 2\sin \theta \cos \theta \]:
\[\sin 2B = 2\sin B\cos B\]
\[ \Rightarrow \sin 2B = 2 \cdot \dfrac{4}{5} \cdot \dfrac{3}{5}\]
\[ \Rightarrow \sin 2B = \dfrac{{24}}{{25}}\]
Hence option C is the correct option.
Note: Students often make a common mistake to calculate the value of \[\sin B\]. They used sine law to find it but to apply sine law we need an angle of the triangle. The correct way to find the \[\sin B\] is using the trigonometry identity.
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