Question

Pls answer the b. What is the three various values of a and b

Answer 1

Given:

The right triangle is,

We know that, the sum of angles of triangle is 180 degree.

As it is right triangle the measure of one angle is 90 degree.

$\begin{gathered} \angle a+\angle b+90^{\circ}=180^{\circ} \\ \angle a+\angle b=180^{\circ}-90^{\circ} \\ \angle a+\angle b=90^{\circ} \end{gathered}$

All that values which satisfy the above equation will be the possible values of a and b.

The three possible values of a and b will be,

$\begin{gathered} \angle a=50^{\circ},\angle b=40^{\circ} \\ \angle a=30^{\circ},\angle b=60^{\circ} \\ \angle a=60^{\circ},\angle b=30^{\circ} \end{gathered}$

Using these three values complete the chart,

1)

$\begin{gathered} \angle a=50^{\circ},\angle b=40^{\circ} \\ \angle a+\angle b=90^{\circ} \\ \sin (a+b)=\sin (90^{\circ^{}})=1 \\ \sin (a)=\sin (50^{\circ})=0.766\text{ }.\ldots\text{(up to 3 decimal place)} \\ \sin (b)=\sin (40^{\circ})=0.643\ldots\text{.(up to 3 decimal places)} \\ \sin (a)+\sin (b)=0.776+0.643=1.419 \end{gathered}$

2)

$\begin{gathered} \angle a=30^{\circ},\angle b=60^{\circ} \\ \angle a+\angle b=90^{\circ} \\ \sin (a+b)=\sin (90^{\circ^{}})=1 \\ \sin (a)=\sin (30^{\circ})=\frac{1}{2} \\ \sin (b)=\sin (60^{\circ})=\frac{\sqrt[]{3}}{2} \\ \sin (a)+\sin (b)=\frac{1}{2}+\frac{\sqrt[]{3}}{2}=\frac{1+\sqrt[]{3}}{2} \end{gathered}$

3)

$\begin{gathered} \angle a=60^{\circ},\angle b=30^{\circ} \\ \angle a+\angle b=90^{\circ} \\ \sin (a+b)=\sin (90^{\circ^{}})=1 \\ \sin (a)=\sin (60^{\circ})=\frac{\sqrt[]{3}}{2} \\ \sin (b)=\sin (30^{\circ})=\frac{1}{2} \\ \sin (a)+\sin (b)=\frac{\sqrt[]{3}}{2}+\frac{1}{2}=\frac{\sqrt[]{3}+1}{2} \end{gathered}$