
Concept explainers
a.
To write: Why sine of an acute
a.

Explanation of Solution
Sine of any angle
As hypotenuse is the longest side,
So,
Sine of any angle
Hence,
The value of sine of an acute angle is always less than one.
b.
To write: Why cosine of an acute angle is always less than one.
b.

Explanation of Solution
Cosine of the angle
As hypotenuse is the longest side,
So,
Cosine of any angle
Hence,
Cosine of an acute angle is always less than one
Chapter 8 Solutions
McDougal Littell Jurgensen Geometry: Student Edition Geometry
Additional Math Textbook Solutions
Calculus for Business, Economics, Life Sciences, and Social Sciences (14th Edition)
Elementary Statistics
A Problem Solving Approach To Mathematics For Elementary School Teachers (13th Edition)
A First Course in Probability (10th Edition)
Basic Business Statistics, Student Value Edition
University Calculus: Early Transcendentals (4th Edition)
- 1/6/25, 3:55 PM Question: 14 Similar right triangles EFG and HIJ are shown. re of 120 √65 adjacent E hypotenuse adjaca H hypotenuse Item Bank | DnA Er:nollesup .es/prist Sisupe ed 12um jerit out i al F 4 G I oppe J 18009 90 ODPO ysma brs & eaus ps sd jon yem What is the value of tan J? ed on yem O broppo 4 ○ A. √65 Qx oppoEF Adj art saused taupe ed for yem 4 ○ B. √65 29 asipnisht riod 916 zelprisht rad √65 4 O ○ C. 4 √65 O D. VIS 9 OD elimiz 916 aelonsider saused supsarrow_forwardFind all anglesarrow_forwardFind U V . 10 U V T 64° Write your answer as an integer or as a decimal rounded to the nearest tenth. U V = Entregararrow_forward
- Find the area of a square whose diagonal is 10arrow_forwardDecomposition geometry: Mary is making a decorative yard space with dimensions as shaded in green (ΔOAB).Mary would like to cover the yard space with artificial turf (plastic grass-like rug). Mary reasoned that she could draw a rectangle around the figure so that the point O was at a vertex of the rectangle and that points A and B were on sides of the rectangle. Then she reasoned that the three smaller triangles resulting could be subtracted from the area of the rectangle. Mary determined that she would need 28 square meters of artificial turf to cover the green shaded yard space pictured exactly.arrow_forward7. 11 m 12.7 m 14 m S V=B₁+ B2(h) 9.5 m 16 m h+s 2 na 62-19 = 37 +, M h² = Bu-29arrow_forward
- Elementary Geometry For College Students, 7eGeometryISBN:9781337614085Author:Alexander, Daniel C.; Koeberlein, Geralyn M.Publisher:Cengage,Elementary Geometry for College StudentsGeometryISBN:9781285195698Author:Daniel C. Alexander, Geralyn M. KoeberleinPublisher:Cengage Learning

