BaANote: Triangle may not be drawn to scale.Suppose a = 11 and A = 65 degrees.%3DFind:b =C =B =degrees BaAbNote: Triangle may not be drawn to scale.Suppose a =4 and b = 9.Find an exact value or give at least two decimal places:sin(A) =cos(A) =tan(A) =sec(A) =csc(A) =%3Dcot(A) =%D

(1) By the law of sines, 11sin65=bsinB=csin90 Then, c=12.137. By Pythagoras theorem,…

Answered: B a A b Note: Triangle may not be drawn…

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN:9780547587776

Author:HOLT MCDOUGAL

Publisher:HOLT MCDOUGAL

Chapter9: Real Numbers And Right Triangles

Section9.8: The Sine And Cosine Ratios

Problem 15E

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Question

B
a
A
Note: Triangle may not be drawn to scale.
Suppose a = 11 and A = 65 degrees.
%3D
Find:
b =
C =
B =
degrees

B
a
A
b
Note: Triangle may not be drawn to scale.
Suppose a =
4 and b = 9.
Find an exact value or give at least two decimal places:
sin(A) =
cos(A) =
tan(A) =
sec(A) =
csc(A) =
%3D
cot(A) =
%D

Expert Solution

Step 1

(1)

By the law of sines,

$\frac{11}{\sin 65} = \frac{b}{\sin B} = \frac{c}{\sin 90}$

Then, $c = 12.137$ .

By Pythagoras theorem,

$\begin{array}{rcl} b & = & \sqrt{c^{2} - a^{2}} \\ = & \sqrt{147.3 - 121} \\ = & 5.129 \end{array}$ .

Therefore, $b = 5.129$ .

Since sum of interior angles of a triangle is 180.

Therefore,

$\begin{array}{rcl} B & = & 180 - 90 - 65 \\ = & 25 \end{array}$

Hence,

$b = 5.129$

$c = 12.137$

$B = 25$ degree

Step 2

(2)

Given $a = 4 and b = 9$ .

By Pythagoras theorem,

$\begin{array}{rcl} c & = & \sqrt{a^{2} + b^{2}} \\ = & \sqrt{16 + 81} \\ = & 9.85 \end{array}$ .

Therefore, $c = 9.85$ .

By the trigonometric ratios,

$\begin{array}{rcl} \sin A & = & \frac{opposite side}{hypotenuse} \\ = & \frac{a}{c} \end{array}$

Therefore, $\sin A = 0.406$

$\begin{array}{rcl} \cos A & = & \frac{adjacent side}{hypotenuse} \\ = & \frac{b}{c} \end{array}$

Therefore, $\cos A = 0.914$

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B a A b Note: Triangle may not be drawn to scale. Suppose a = 11 and A = 65 degrees. Find: b = C = B = degrees