Question

 In a \[\Delta ABC\], if \[{{\sin }^{2}}A+{{\sin }^{2}}B={{\sin }^{2}}C\], then show that the triangle is right-angled.

Answer 1

Hint: For the above question, we will have to know about the right-angled triangle. A right-angled triangle is a triangle in which one angle is a right angle. The relation between the sides and the angles of a right-angle triangle is the basis for trigonometry. The side opposite the right angle is called the hypotenuse and the other two sides are known as base and perpendicular. In the above question, we will use the sine rule of a triangle and also the Pythagoras Theorem. Pythagoras Theorem states that the sum of the squares of the base and perpendicular is equal to the square of the hypotenuse in any right angled triangle.

Complete step by step answer:
We have been given that in a \[\Delta ABC\], \[{{\sin }^{2}}A+{{\sin }^{2}}B={{\sin }^{2}}C\]
We know the sine rule of the triangle is
\[\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\]
Let us suppose the ratio of the sine of the angle to the side is a constant ‘k’.
\[\Rightarrow \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}=k\]
\[\Rightarrow \sin A=ak,\text{ }\sin B=bk\text{ and }\sin C=ck\]
After substituting these values in the given condition, we get,
\[{{\left( ak \right)}^{2}}+{{\left( bk \right)}^{2}}={{\left( ck \right)}^{2}}\]
\[\Rightarrow {{a}^{2}}{{k}^{2}}+{{b}^{2}}{{k}^{2}}={{c}^{2}}{{k}^{2}}\]
Taking \[{{k}^{2}}\] as common and dividing by \[{{k}^{2}}\] to both the sides of the equality, we get,
\[\dfrac{{{k}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)}{{{k}^{2}}}-\dfrac{{{k}^{2}}{{c}^{2}}}{{{k}^{2}}}\]
\[\Rightarrow {{a}^{2}}+{{b}^{2}}={{c}^{2}}\]
Hence, it satisfies the condition of the Pythagoras Theorem.
Therefore, the given triangle is a right-angled triangle at angle C.

Note: We can also solve the given question by only using sine rule and cosine rule. In this method, we will show that the value of cos C is equal to 0 which means the angle C is \[{{90}^{o}}\] , and thus the triangle is right-angled.