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How do you solve the triangle of $A={{76}^{\circ }}$, $a=34,b=21$?

Last updated date: 22nd Feb 2024
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IVSAT 2024
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Hint: To solve the triangle having angle and side information as $A={{76}^{\circ }}$, $a=34,b=21$, we are going to use the triangle sine properties which says that: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$. In this formula, $a,b\And c$ are the sides of the triangle opposite to angles A, B and C respectively. In this way, we will find all the measurements of the three sides $a,b\And c$ and all the three angles $A,B\And C$.

Complete step-by-step answer:
In the above problem, we have given a triangle in which measurement of angle A, and length of the sides $a\And b$ is given as follows:
$A={{76}^{\circ }}$, $a=34,b=21$
Now, let us draw triangle ABC with sides $a,b\And c$ opposite to angles $A,B\And C$ respectively.
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We know there is a sine relation for finding the solutions of triangle which is as follows:
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
Now, we are taking the first two fractions in the above expression and we get,
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}$
Substituting the values of $a=34,b=21$ and the angle A in the above equation we get,
$\Rightarrow \dfrac{34}{\sin {{76}^{\circ }}}=\dfrac{21}{\sin B}$
Cross multiplying the above equation we get,
$\Rightarrow 34\left( \sin B \right)=21\left( \sin {{76}^{\circ }} \right)$
We know the value of $\sin {{76}^{\circ }}$ which is equal to 0.9703.
$\sin {{76}^{\circ }}=0.9703$
Substituting the above value of sine in the above equation we get,
  & \Rightarrow 34\left( \sin B \right)=21\left( 0.9703 \right) \\
 & \Rightarrow 34\left( \sin B \right)=20.3763 \\
Dividing 34 on both the sides of the above equation we get,
  & \Rightarrow \sin B=\dfrac{20.3763}{34} \\
 & \Rightarrow \sin B=0.5993 \\
Now, rounding off the above value of $\sin B$ we get,
$\Rightarrow \sin B=0.6$
Removing decimal in the above equation by converting the above number 0.6 into fraction we get,
$\sin B=\dfrac{6}{10}=\dfrac{3}{5}$
And we know that $\sin B=\dfrac{3}{5}$ when the value of angle B is ${{37}^{\circ }}$. Now, we have got the value of angle B.
As we know that sum of all the angles of triangle is ${{180}^{\circ }}$ so adding all the angles A, B and C and equating them to ${{180}^{\circ }}$ we get,
  & \Rightarrow A+B+C={{180}^{\circ }} \\
 & \Rightarrow {{76}^{\circ }}+{{37}^{\circ }}+C={{180}^{\circ }} \\
  & \Rightarrow {{113}^{\circ }}+C={{180}^{\circ }} \\
 & \Rightarrow C={{180}^{\circ }}-{{113}^{\circ }} \\
 & \Rightarrow C={{67}^{\circ }} \\
Hence, we have found the angle C also. Now, finding the third side $''c''$ of the above triangle by using the sine relation we get,
$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
Using last two fractions of the above equation we get,
$\Rightarrow \dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
Substituting the value of $b=21$ and angles B and C from the above we get,
$\Rightarrow \dfrac{21}{\sin {{37}^{\circ }}}=\dfrac{c}{\sin {{67}^{\circ }}}$
We know that the value of $\sin {{37}^{\circ }}=\dfrac{3}{5}$ and $\sin {{67}^{\circ }}=0.9205$ so substituting these value of sine in the above we get,
  & \Rightarrow \dfrac{21}{\dfrac{3}{5}}=\dfrac{c}{0.9205} \\
 & \Rightarrow \dfrac{21\times 5}{3}=\dfrac{c}{0.9205} \\
In the L.H.S of the above equation 3 will divide 21 by 7 times.
$\Rightarrow 7\times 5=\dfrac{c}{0.9205}$
Cross multiplying above equation we get,
  & \Rightarrow 7\times 5\times 0.9205=c \\
 & \Rightarrow 32.2175=c \\
Rounding off the above value of side “c” we get,
$\Rightarrow c=32.22$
Hence, we have found the remaining third side also.

Note: In the above problem, we have asked to solve the given triangle. You might be wondering what it is that we have to solve so whenever you are asked to find the solutions of a triangle or to solve the triangle so basically you have to find all the sides and the three angles of the triangle and you have given some of the sides and angles of the triangle.
Just like sine formula we have shown above there exists cosine formula also which is equal to:
  & \cos A=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}; \\
 & \cos B=\dfrac{{{c}^{2}}+{{a}^{2}}-{{b}^{2}}}{2ca}; \\
 & \cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab} \\
So, you can use the above formulas also.