Question

In an acute-angled triangle ABC, if $\sin \left( A+B-C \right)=\dfrac{1}{2}$ and $\cos \left( B+C-A \right)=\dfrac{1}{\sqrt{2}}$, then find the measure of each angle of the triangle.

Answer 1

Hint: Find the acute angle solution of the equations $\sin x=\dfrac{1}{2}$ and $\cos x=\dfrac{1}{\sqrt{2}}$. Write equations based on the data given in the question. Solve those equations to calculate the measure of all the angles.

Complete step-by-step answer:
We know that in an acute-angled triangle ABC, we have $\sin \left( A+B-C \right)=\dfrac{1}{2}$ and $\cos \left( B+C-A \right)=\dfrac{1}{\sqrt{2}}$. We have to calculate the measure of each angle of the triangle.

We will first calculate the acute angle solution of the equations $\sin x=\dfrac{1}{2}$ and $\cos x=\dfrac{1}{\sqrt{2}}$.
We know that $\sin {{30}^{\circ }}=\dfrac{1}{2}$ and $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}}$. We also know that $\sin \left( A+B-C \right)=\dfrac{1}{2}$ and $\cos \left( B+C-A \right)=\dfrac{1}{\sqrt{2}}$.
Thus, we have $A+B-C={{30}^{\circ }}.....\left( 1 \right)$ and $B+C-A={{45}^{\circ }}.....\left( 2 \right)$.
We also know that the sum of all angles of a triangle is ${{180}^{\circ }}$. Thus, we have $A+B+C={{180}^{\circ }}.....\left( 3 \right)$.
We will now simplify all the equations.
Subtracting equation (1) from equation (3), we have $\left( A+B+C \right)-\left( A+B-C \right)={{180}^{\circ }}-{{30}^{\circ }}$.
Thus, we have $2C={{150}^{\circ }}$. Rearranging the terms of the previous equation, we have $C=\dfrac{{{150}^{\circ }}}{2}={{75}^{\circ }}.....\left( 4 \right)$.
Substituting equation (4) in equation (2), we have $B+{{75}^{\circ }}-A={{45}^{\circ }}$. Rearranging the terms of the above equation, we have $B-A={{45}^{\circ }}-{{75}^{\circ }}=-{{30}^{\circ }}.....\left( 5 \right)$.
Similarly, substituting equation (4) in equation (3), we have ${{75}^{\circ }}+B+A={{180}^{\circ }}$. Rearranging the terms of the above equation, we have $A+B={{180}^{\circ }}-{{75}^{\circ }}={{105}^{\circ }}.....\left( 6 \right)$.
We will now simplify equations (5) and (5). Adding equation (5) and (6), we have $B-A+\left( A+B \right)={{105}^{\circ }}-{{30}^{\circ }}$. Thus, we have $2B={{75}^{\circ }}\Rightarrow B=\dfrac{{{75}^{\circ }}}{2}={{37.5}^{\circ }}$.
Substituting $B={{37.5}^{\circ }}$ in equation (6), we have $A+{{37.5}^{\circ }}={{105}^{\circ }}$. Thus, we have $A={{105}^{\circ }}-{{37.5}^{\circ }}={{67.5}^{\circ }}$.
Hence, the measure of all the angles of the triangle is $\angle A={{67.5}^{\circ }},\angle B={{37.5}^{\circ }},\angle C={{75}^{\circ }}$.

Note: We can’t solve this question without using the fact that the sum of all interior angles of a triangle is ${{180}^{\circ }}$. If we don’t use this fact, we will get an incorrect measure of the angles of the triangle. We must write the measures of all the angles in degrees or radians.