Answer

Verified

338.4k+ views

**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.

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,

$\begin{align}

& \Rightarrow 34\left( \sin B \right)=21\left( 0.9703 \right) \\

& \Rightarrow 34\left( \sin B \right)=20.3763 \\

\end{align}$

Dividing 34 on both the sides of the above equation we get,

$\begin{align}

& \Rightarrow \sin B=\dfrac{20.3763}{34} \\

& \Rightarrow \sin B=0.5993 \\

\end{align}$

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,

$\begin{align}

& \Rightarrow A+B+C={{180}^{\circ }} \\

& \Rightarrow {{76}^{\circ }}+{{37}^{\circ }}+C={{180}^{\circ }} \\

\end{align}$

$\begin{align}

& \Rightarrow {{113}^{\circ }}+C={{180}^{\circ }} \\

& \Rightarrow C={{180}^{\circ }}-{{113}^{\circ }} \\

& \Rightarrow C={{67}^{\circ }} \\

\end{align}$

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,

$\begin{align}

& \Rightarrow \dfrac{21}{\dfrac{3}{5}}=\dfrac{c}{0.9205} \\

& \Rightarrow \dfrac{21\times 5}{3}=\dfrac{c}{0.9205} \\

\end{align}$

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,

$\begin{align}

& \Rightarrow 7\times 5\times 0.9205=c \\

& \Rightarrow 32.2175=c \\

\end{align}$

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:

$\begin{align}

& \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} \\

\end{align}$

So, you can use the above formulas also.

Recently Updated Pages

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

What are the possible quantum number for the last outermost class 11 chemistry CBSE

Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE

What happens when entropy reaches maximum class 11 chemistry JEE_Main

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Write a stanza wise summary of money madness class 11 english CBSE

Which places in India experience sunrise first and class 9 social science CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Which neighbouring country does not share a boundary class 9 social science CBSE

What is Whales collective noun class 10 english CBSE