Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 2 - Section 2.1 - Definition II: Right Triangle Trigonometry - 2.1 Problem Set - Page 63: 70

Answer

$\sin\angle D=\frac{\sqrt3}{3}$ $\cos\angle D= \frac{\sqrt6}{3}$

Work Step by Step

It's obvious that the triangle $\triangle EGD$ is the right triangle, where $\angle G$ is the right angle. We have to find sine and cosine of the angle $EDG$, let's note it by angle $D$. By the definition sine is equal to the ratio of the side opposite a given angle (in a right-angled triangle) to the hypotenuse. $\sin\angle D= \dfrac{opposite }{hypotenuse}$. In our case $\sin\angle D= \dfrac{EG}{ED}.$ Similarly, cosine is equal to the ratio of the side adjacent to an acute angle (in a right-angled triangle) to the hypotenuse. $\cos\angle D= \dfrac{adjacent}{hypotenuse}=\dfrac{DG}{ED}$. We just have to find $ED$, $DG$ because we know that $EG = 3$ inches. For that we will use Pythagorean theorem. The triangle $\triangle GCD$ is the right triangle, where $\angle C$ is the right angle. So $DG^{2}=GC^{2}+CD^{2}.$ For our cube $GC=CD= 3 $ inches. And DG is equal $DG=3\sqrt{2} $ inches. The next right triangle is our well known $\triangle EGD$. Similarly, $ED^2=DG^2+EG^2=(3\sqrt2)^2+3^2=3\cdot9.$ So $ED=3\sqrt3$ inches. Now we have everything to find sine and cosine. $$\sin\angle D=\dfrac{EG}{ED}=\dfrac{3}{3\sqrt3}=\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{3}$$ $$\cos\angle D=\dfrac{DG}{ED}=\dfrac{3\sqrt2}{3\sqrt3}=\dfrac{\sqrt2}{\sqrt3}=\dfrac{\sqrt6}{3}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.