Chapter 9.9, Problem 13PSB

a

To determine

To find:The value of $\tan \angle ACD$

Answer to Problem 13PSB

The value of $\tan \angle ACD=\frac{2}{3}$

Explanation of Solution

Given information:The following information has been given

  $CD=6$

  $BD=9$

Formula used:Basic similarity has been equated

Calculation:We know that $\angle ACD=\angle CBD$ , hence we can say that

  $\tan \angle ACD=\tan \angle CBD$

Hence, we get

  $\tan \angle CBD=\frac{2}{3}$

Hence, we can say that as $\angle ACD=\angle CBD$ ,

  $\tan \angle ACD=\frac{2}{3}$

b

To determine

To find:The value of $\sin \angle A$

Answer to Problem 13PSB

The value of $\sin \angle A=\frac{3}{\sqrt{13}}$

Explanation of Solution

Given information:The following information has been given

  $CD=6$

  $BD=9$

Formula used:Basic trigonometric identities have been used

Calculation:We apply the Pythagoras Theorem in right triangle BCD, and we get

  $6^2+9^2=B{C}^2$

Hence, we get

  $BC=3\sqrt{13}$

Now, we apply the sine formula in the triangle RET, we get

  $\sin \angle A=\sin \angle BCD=\frac{9}{3\sqrt{13}}$

Hence, we get

  $\sin \angle A=\frac{3}{\sqrt{13}}$