Help with #2 By using distances show that the points A(8, 3), B(7, –5) and C(4,-3) arevertices of a right triangle. Find the six trigonometric functions of angle A

Answered: By using distances show that the points…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter6: The Trigonometric Functions

Section6.2: Trigonometric Functions Of Angles

Problem 4E

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Help with #2

By using distances show that the points A(8, 3), B(7, –5) and C(4,-3) are
vertices of a right triangle. Find the six trigonometric functions of angle A

Expert Solution

Step 1

Given vertices of triangle are $A (8, 3), B (7, - 5) and C (4, - 3)$ .

It is known that the distance between two points $(x_{1}, y_{1}) a n d (x_{2}, y_{2})$ is given by $d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$ .

Compute AB as follows.

$\begin{array}{rcl} A B & = & \sqrt{{(7 - 8)}^{2} + {(- 5 - 3)}^{2}} \\ = & \sqrt{1 + 64} \\ = & \sqrt{65} \end{array}$

Compute BC as follows.

$\begin{array}{rcl} B C & = & \sqrt{{(4 - 7)}^{2} + {(- 3 + 5)}^{2}} \\ = & \sqrt{9 + 4} \\ = & \sqrt{13} \end{array}$

Step 2

Now compute AC as follows.

$\begin{array}{rcl} A C & = & \sqrt{{(4 - 8)}^{2} + {(- 3 - 3)}^{2}} \\ = & \sqrt{16 + 36} \\ = & \sqrt{52} \end{array}$

Note that 65= 13+52.

That is, $A B^{2} = B C^{2} + A C^{2}$ .

Thus, the triangle formed by vertices $A (8, 3), B (7, - 5) and C (4, - 3)$ is a right angle triangle.

Draw the rough diagram as shown below.

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