Chapter 9.9, Problem 19PSC

a

To determine

To verify:The given expression, ${(\sin \angle A)}^2+{(\cos \angle A)}^2=1$ under the given conditions.

Explanation of Solution

Given information:The following information has been given

Right triangle ABC

  $\angle C={90}^0$

AB is the hypotenuse

We know that basic definition of trigonometric ratios states that

  $\sin \angle A=\frac{BC}{AB}$ , and

  $\cos \angle A=\frac{AC}{AB}$

Thus, substituting this value in our original equation, we get

  $\frac{B{C}^2}{A{B}^2}+\frac{A{C}^2}{A{B}^2}=1\to B{C}^2+A{C}^2=A{B}^2$

Now, know that $B{C}^2+A{C}^2=A{B}^2$ holds true due to Pythagoras theorem.

Hence, verification is successful

b

To determine

To show:The statement $\frac{a}{\sin \angle A}=\frac{b}{\sin \angle B}$ ,using definitions of trigonometric ratios

Explanation of Solution

Given information:The following information has been given

Right triangle ABC

  $\angle C={90}^0$

We know from the basic definition of trigonometric ratios that

  $\sin \angle A=\frac{a}{c}\to \frac{a}{\sin \angle A}=c$ , and

  $\sin \angle B=\frac{b}{c}\to \frac{b}{\sin \angle B}=c$

Now, substituting these values in our original equation, we get

  $c=c$

Hence, the statements are verified to be true

c

To determine

To verify:The statement $\frac{\sin \angle A}{\cos \angle A}=\tan \angle A$ using definitions of trigonometric ratios

Explanation of Solution

Given information:The following information has been given

Right triangle ABC

  $\angle C={90}^0$

We know from the basic definition of trigonometric ratios that

  $\sin \angle A=\frac{a}{c}$

  $\cos \angle A=\frac{b}{c}$

  $\tan \angle A=\frac{a}{b}$

Now, substituting these values in our original equation, we get

  $\frac{a}{b}=\frac{a}{b}$

Hence, the verification is correct

d

To determine

To show:The statement $\sin \angle A=\cos (\angle {90}^0-A)$ using definitions of trigonometric ratios

Explanation of Solution

Given information:The following information has been given

Right triangle ABC

  $\angle C={90}^0$

We know that for this right triangle,

  $\angle {90}^0-A=B$

We know from the basic definition of trigonometric ratios that

  $\sin \angle A=\frac{a}{c}$ , and

  $\cos (\angle {90}^0-A)=\cos \angle B=\frac{a}{c}$

Thus, substituting the values in our original equation, we get

  $\frac{a}{c}=\frac{a}{c}$

Hence, the verification is successful