Question

How do you solve the triangles when the sides of $a=125,b=200$ and $\angle A={{110}^{\circ }}$. Find the other sides and measurement of angles.

Answer 1

Hint: We first use the theorem for properties of triangles \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\]. We also use the relation of angles that sum of all the angles of a triangle is ${{180}^{\circ }}$. We put the values in the relation to get the remaining angles and the sides.

Complete step by step solution:
We have been given the values of two sides’ length and measurement of one angle.
We are going to use the relation for properties of triangles between angles and sides of a general triangle.
The relation gives that for $\Delta ABC$, we have \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\] where $a,b,c$ are the lengths of the sides and $A,B,C$ are corresponding opposite angles of the sides $a,b,c$ respectively.
We have been given that $a=125,b=200$ and $\angle A={{110}^{\circ }}$.
We put these values in the equation \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\] to get \[\dfrac{125}{\sin \left( {{110}^{\circ }} \right)}=\dfrac{200}{\sin B}=\dfrac{c}{\sin C}\]
We get \[\sin \left( {{110}^{\circ }} \right)=0.9396\].
We get the first linear equation of \[\dfrac{125}{0.9396}=\dfrac{200}{\sin B}\].
Solving the equation, we get \[\sin B=\dfrac{200\times 0.9396}{125}=1.50336\].

We know that the highest value possible for ratio sin is 1. But in this case we get \[\sin B=1.50336\] which is not possible.
Therefore, this triangle doesn’t exist.

Note: The individual ratios in the relation \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\] is equal to the $2R$ where $R$ is the circum-radius of the triangle. So, the relation is \[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R\]. If We know two sides and the angle between them, use the cosine rule and plug in the values for the remaining sides and the angles.