Chapter 8.6, Problem 19WE

a.

To determine

To find: the lengths of the side AB and AC of the ΔABC

Answer to Problem 19WE

The lengths of the side AB is $16$

The length of the side AC is $16$

Explanation of Solution

Given Information: ΔABC with $m\angle B=m\angle C=72$ and BC=10

Formula used:

The law of sines can be used to determine the measure of two sides when the third side and the angles of the triangle are given.

  

For a triangle ABC with length of sides a,b and c and with angles A, B and C, the law of sines can be expressed as,

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Calculation: Consider the triangle ΔABC shown below. Based on the given value of BC and the angles of the triangle, the law of sines can be expressed as,

  $\frac{10}{\sin 36°}=\frac{b}{\sin 72°}=\frac{c}{\sin 72°}$ ........(1)

  

Since, two angles of the triangle are equal, the two corresponding sides are also equal and hence the triangle is isosceles in nature.

Now, considering the first two terms of the equation (1),

  $\begin{array}{l}\frac{10}{\sin 36°}=\frac{b}{\sin 72°}\\ b=\frac{\sin 72°}{\sin 36°}\times 10\end{array}$

Using the trigonometric table of values in P 311 of the textbook, the value of b can be obtained as,

  $\begin{array}{c}b=\frac{0.9511}{0.5878}\times 10\\ =16.18\\ \simeq 16\end{array}$

Thus, the sides of the triangle ABC , AC=AB=16

b.

To determine

To find: the length of bisector of to B$\angle A$C

Answer to Problem 19WE

The length of bisector of $\angle A$ to BC is $15$

Explanation of Solution

Given Information: ΔABC with $m\angle B=m\angle C=72$ , BC=10 with AD being the angle bisector of $\angle A$ to BC

Formula used:

a) The angle bisector theorem for a triangle ABC relates the arms of the triangle to the base as,

  

  $\frac{AB}{AC}=\frac{BD}{BC}$

Having found the ratio of measures of the sides BD and DC their values can be estimated for a given measure of BC.

b) Now, law of sines can be used to find the measure of the side AD by considering triangle ABD or ADC by using any one of the two relations given below.

  $\frac{BD}{\sin \frac{A}{2}}=\frac{AD}{\sin B}$

  $\frac{DC}{\sin \frac{A}{2}}=\frac{AD}{\sin C}$

Calculation: Consider the triangle ΔABC with AD as the angle bisector of $\angle A$ .

  

Applying the angle bisector theorem,

  $\frac{AB}{AC}=\frac{BD}{BC}$

Using the measure of AB and AC from previous part,

  $\begin{array}{l}\frac{16}{16}=\frac{BD}{BC}\\ BD=DC\end{array}$

This gives the measure of the sides BD and DC as

  $\begin{array}{c}BD=DC\\ =\frac{BC}{2}\\ =\frac{10}{2}\\ =5\end{array}$

Now, using the law of sines as in (b),

  $\begin{array}{c}\frac{BD}{\sin \frac{A}{2}}=\frac{AD}{\sin B}\\ \frac{5}{\sin 18°}=\frac{AD}{\sin 72°}\\ AD=\frac{5\sin 72°}{\sin 18°}\end{array}$

Again, with the use of trigonometric values from table in P311, the measure of side AD is obtained as,

  $\begin{array}{c}AD=\frac{5\left(0.9511\right)}{0.3090}\\ =15.38\\ \simeq 15\end{array}$