Chapter 8.6, Problem 18WE

a.

To determine

To calculate:The length of the altitude from $A$

Answer to Problem 18WE

The length of the altitude from $A$ is 12.

Explanation of Solution

Given information:

In $\Delta ABC$ , $AB=AC=13$ and $BC=10$

Calculation:

Let’s draw a tringle first,

  

Now, the length of altitude from A, i.e. AM is,

  $AM=\sqrt{A{C}^2-M{C}^2}=\sqrt{{13}^2-5^2}=12$

b.

To determine

To calculate:Find the measures of three angles of $\Delta ABC$

Answer to Problem 18WE

  $A=45.24°$ and $B=C=67.38°$

Explanation of Solution

Given information:

In $\Delta ABC$ , $AB=AC=13$ and $BC=10$

Calculation:

Let’s draw a tringle first,

  

Now, from the cosine law,

  $\cos A=\frac{b^2+c^2-a^2}{2bc}$

  $\therefore \cos A=\frac{{13}^2+{13}^2-{10}^2}{2\cdot 13\cdot 13}=0.704$

  $\therefore A=45.24°$

Now, a triangle is an isosceles triangle, Therefore, other two angles must be equal,

  $\therefore A+B+C=180°$

  $\therefore 45.24+2B=180°$

  $\therefore B=C=67.38°$

c.

To determine

To calculate:The length of altitude from $C$

Answer to Problem 18WE

The length of altitude from $C$ is 9.23

Explanation of Solution

Given information:

In $\Delta ABC$ , $AB=AC=13$ and $BC=10$

Calculation:

Let’s draw a tringle first,

  

Now, from the part b, $C=67.38°$ , also $\angle BNC=90°$ , $\therefore \angle NBC=180-90-67.38=22.62°$

Now, $\Delta BNC$ is a right-angled triangle.

  $\therefore BC\mathrm{cosNBC}=BN$

  $\therefore 10\cos 22.62°=BN$

  $\therefore BN=9.23$