Answered: 1) Fill in the blanks. Show your work…

Trigonometry (11th Edition)

11th Edition

ISBN:9780134217437

Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels

Chapter1: Trigonometric Functions

Section: Chapter Questions

Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.

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Question

1) Fill in the blanks. Show your work if necessary.

a) Given a 30o -60o -90o triangle and the side opposite 60o is 5. The side opposite 30o is and the hypotenuse is .

b) Given a 45o -45o -90o triangle and the side hypotenuse is 7. The shorter sides (legs) are each .

Polygon with three sides, three angles, and three vertices. Based on the properties of each side, the types of triangles are scalene (triangle with three three different lengths and three different angles), isosceles (angle with two equal sides and two equal angles), and equilateral (three equal sides and three angles of 60°). The types of angles are acute (less than 90°); obtuse (greater than 90°); and right (90°).

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To Determine:

1) Fill in the blanks. Show your work if necessary.

a) Given a 30o -60o -90o triangle and the side opposite 60o is 5. The side opposite 30o is ? and the hypotenuse is ? .

b) Given a 45o -45o -90o triangle and the side hypotenuse is 7. The shorter sides (legs) are each ?

Given: we have a triangle whose angles are given

Part a) a) Given a 30o -60o -90o triangle and the side opposite 60o is 5.

Explanation: see in the figure below we have three angles given and a side now we will use the formula to find the remaining sides of the triangles.

$\sin 60^{\circ} = \frac{p e r p e n d i c u l a r}{h y p o t n e u s e} = \frac{5}{A C} \frac{\sqrt{3}}{2} = \frac{5}{A C} A C = \frac{10}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{10 \sqrt{3}}{3}$

by using pythagorean theorem we have

${(h y p o t e n u s e)}^{2} = (b a s e)^{2} + {(p e r p e n d i c u l a r)}^{2} {(\frac{10 \sqrt{3}}{3})}^{2} = {(b a s e)}^{2} + {(5)}^{2} \frac{100 \times 3}{9} = 25 + {(B C)}^{2} {(B C)}^{2} = \frac{100}{3} - 25 = \frac{100 - 75}{3} = \frac{25}{3} B C = \frac{\sqrt{25}}{\sqrt{3}} = \frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5 \sqrt{3}}{3}$

so we have base= $\frac{5 \sqrt{3}}{3}$ and hypotenuse = $\frac{10 \sqrt{3}}{3}$

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