Question

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

Answer 1

Hint: Find the acute angle solution of the equations $\sin x={\displaystyle \frac{1}{2}}$ and $\cos x={\displaystyle \frac{1}{\sqrt{2}}}$. 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 $\sin (A+B-C)={\displaystyle \frac{1}{2}}$ and $\cos (B+C-A)={\displaystyle \frac{1}{\sqrt{2}}}$. We have to calculate the measure of each angle of the triangle.

We will first calculate the acute angle solution of the equations $\sin x={\displaystyle \frac{1}{2}}$ and $\cos x={\displaystyle \frac{1}{\sqrt{2}}}$.
We know that $\sin {30}^{\circ }={\displaystyle \frac{1}{2}}$ and $\cos {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}}$. We also know that $\sin (A+B-C)={\displaystyle \frac{1}{2}}$ and $\cos (B+C-A)={\displaystyle \frac{1}{\sqrt{2}}}$.
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 $(A+B+C)-(A+B-C)={180}^{\circ }-{30}^{\circ }$.
Thus, we have $2C={150}^{\circ }$. Rearranging the terms of the previous equation, we have $C={\displaystyle \frac{{150}^{\circ }}{2}}={75}^{\circ }.....\left(4\right)$.
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 $B-A+(A+B)={105}^{\circ }-{30}^{\circ }$. Thus, we have $2B={75}^{\circ }\Rightarrow B={\displaystyle \frac{{75}^{\circ }}{2}}={37.5}^{\circ }$.
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 $\mathrm{\angle }A={67.5}^{\circ },\mathrm{\angle }B={37.5}^{\circ },\mathrm{\angle }C={75}^{\circ }$.

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.