Question

How do you know how many triangles are created given $A = 61$ , $a = 23$ , $b = 24$ ?

Answer 1

In this question, we need to find how many triangles can be formed when $A = 61$ , $a = 23$ and $b = 24$ . With the use of laws of sine function and laws of cosine function we can find the number of triangles that can be formed . Sine , cosine and tangent are the basic trigonometric functions . Sine function is nothing but a ratio of the opposite side of a right angle to the hypotenuse of the right angle. Similarly, cosine function is nothing but a ratio of the adjacent side of a right angle to the hypotenuse of the right angle . By using cosine law, we can find the side of the triangle. Then by using sine law, we can find the angle of the triangle.

Formula used:
Law of sine:
$\dfrac{a}{{\operatorname {sin}A}} = \dfrac{b}{{\operatorname{sin}B}} = \dfrac{c}{{\operatorname{sin}C}}$
Where $a,\ b,\ c$ are the sides of the triangle and $A,\ B,C$ is the angle of the triangle
Law of cosine:
$a^{2} = b^{2} + c^{2} – 2bc\ \cos\left( A \right)$
Where $a,\ b,\ c$ are the sides of the triangle and $A$ is the angle of the triangle.

Complete step by step answer:
Given, $A = 61$ , $a = 23$ and $b = 24$. From the laws of cosine, we can find the value of side $c$.
$\ \Rightarrow \ a^{2} = b^{2} + c^{2} – 2bc\ \cos\left( A \right)$
By substituting the known values,
We get,
$\Rightarrow \ \left( 23 \right)^{2} = \left( 24 \right)^{2} + c^{2} – 2\left( 24 \right)\left( c \right)\cos\left( 61 \right)\$
Here the value of $\cos(61^{o}) = 0.485$
On simplifying,
We get
$\Rightarrow \ 529 = 576 + c^{2} – 48c \times \left( 0.485 \right)$

By rewriting the terms,
We get,
$c^{2} – 23.28c + \left( 576 – 529 \right) = 0$
On simplifying,
We get,
$c^{2} – 23.28c + 47 = 0$
This equation is in the form of $ax^{2}+ bx + c = 0$
Thus by using quadratic formula,
$x = \dfrac{- b \pm \sqrt{b^{2} – 4ac}}{2a}$
Where $a,\ b,\ c$ are constants which are not equal to $0$ and $x$ is the unknown.
Thus we get,
c = \dfrac{\left\\{ - 23.28 \pm \sqrt{\left( 23.28 \right)^{2} – 4\left( 1 \right)\left( 47 \right)} \right\\}}{2\left( 1 \right)}

On simplifying,
We get,
c = \dfrac{\left\\{ - 23.38 \pm \sqrt{546.6 – 188} \right\\}}{2}
On further simplifying,
We get,
$c = \dfrac{- 23.38 \pm 18.81}{2}$
$\Rightarrow c = \dfrac{\left( - 23.37 + 18.81 \right)}{2}\ and\ c = \dfrac{\left( - 23.37 – 18.81 \right)}{2}$
On simplifying,
We get,
$c = 21.045$ and $2.235$
Hence we get the value of side $c$

Case 1:
From the laws of sine, we can find the value of angle $C$ when $c = 21.045$ .
$\dfrac{a}{{\operatorname {sin}A}} = \dfrac{b}{{\operatorname{sin}B}} = \dfrac{c}{{\operatorname{sin}C}}$
First we can find the value of angle $B$,
$\Rightarrow \dfrac{a}{{\operatorname{sin}A}} = \dfrac{b}{{\operatorname{sin}B}}$
By substituting the known values,
We get,
$\Rightarrow \dfrac{23}{\sin\left( 61^{o} \right)} = \dfrac{24}{{\operatorname{sin}B}}$
By cross multiplying,
We get
$\operatorname{sin}B = \dfrac{\left( 24 \times \sin\left( 61 \right) \right)}{23}$
Here the value of $\sin(61^{o})\ = 0.875$
By substituting the values and simplifying,
We get,
$\operatorname{sin}B = \dfrac{\left( 24 \times 0.875 \right)}{23}$

Thus we get $\operatorname{sin}B = 0.193$
$\Rightarrow \ B = \sin^{- 1}(0.193)$
On simplifying,
We get,
$B = 65.93^{o}$
We know that the sum of three angles of the triangle is $180^{o}$
That is $A + B + C = 180^{o}$
From this we can find the value of $C$.
$61^{o} + 65.93^{o} + C^{o}= 180^{o}$
$\Rightarrow \ C = 180^{o} – 61^{o} – 65.93^{o}$
On simplifying,
We get,
$C = 53.07^{o}$

Case 2: We can find the next angle if $C$ when $c = 2.235$. First we can find the value of angle $B$. We know that $\sin\theta=\sin(\pi-\theta)$
Therefore,
$\operatorname{sin}B = \sin(\pi – B)$
We have find the value of $B = 65.93^{o}$
$\Rightarrow \ B = (180^{o} – 65.93^{o})$
Thus we get, $B = 114.07^{o}$

We know that the sum of three angles of the triangle is $180^{o}$
That is $A + B + C = 180^{o}$
From this we can find the value of $C$.
$61^{o} + 114.07^{o} + C = 180^{o}$
$\Rightarrow \ C = 180^{o} – 61^{o} – 114.07^{o}$
On simplifying,
We get,
$C = 4.93^{o}$

Therefore two triangles can be formed one with angles $A = 61^{o}\ ,\ B = 65.94^{o}\ ,\ C = 53.07^{o}$ and sides$\ a = 23\ ,\ b = 24$ and $c = 21.054$ and the another triangle with angles $A = 61^{o}\ ,\ B = 114.07^{o}\ ,\ C = 4.93^{o}$ and sides $a = 23\ ,\ b = 24$ and $c = 2.235$.

Note: Mathematically , a triangle has three sides, three angles and three vertices . The sum of all internal angles of a triangle is always equal to $180^{o}$. This is known as the property of the triangle. The concept used in this problem is trigonometric identities and laws of trigonometry . We use both the laws of sine and cosine to find the sides and angles of the triangle. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the algebraic formula with the use of trigonometric functions . Trigonometric functions are also known as circular functions or geometrical functions.