Savvas Learning Co Geometry Student Edition Chapter 8 Right Triangles and Trigonometry Exercise
Page 487 Exercise 1 Problem 1
Given: 0.5858 = \(\frac{24}{x}\)
To Find: We have to solve the given expression.
Using algebraic identities.
Given
0.5858 = \(\frac{24}{x}\)
x = \(\frac{24}{0.5858}\)
0.5858
∴ x = 40.9696
The value of x is 40.9696.
Page 487 Exercise 2 Problem 2
Given: 0.8572 = \(\frac{5271}{x}\)
To Find – We have to solve the given expression.
Using algebraic identities.
Given
0.8572 = \(\frac{5271}{x}\)
0.8572x = 5271
x = \(\frac{5271}{0.8572}\)
∴ x = 6149.0900
The value of x is 6149.0900.
Page 487 Exercise 3 Problem 3
Given: 0.5 = \(\frac{x}{3x+5}\)
To Find – We have to solve the given expression.
Using algebraic identities.
Given , 0.5 = \(\frac{x}{3x+5}\)
1.5x + 2.5 = x
1.5x − x + 2.5 = 0
0.5x + 2.5 = 0
0.5x = −2.5
x = −2.5
0.5
∴ x = −5
The value of x is −5.
Page 487 Exercise 4 Problem 4
Given: \(\overline{C D}\) ∥ \(\overline{A B}\)
To Find – We have to prove the given pair of triangles as similar.
Using the theorems and postulates of similarities.
Given in the diagram \(\overline{C D}\) ∥ \(\overline{A B}\)
Therefore
\(\overline{C E}\) = \(\overline{E B}\)
\(\overline{A E}\) = \(\overline{E D}\)
Therefore, the given pair of triangles are similar.
We have proved that the given pair of triangles are similar.
Page 487 Exercise 5 Problem 5
Given: A pair of triangles.
To Find – We have to prove the given pair of triangles as similar.
Using the theorems and postulates of similarities.
From the given diagram
\(\frac{21}{27}\) = \(\frac{14}{18}\)
∴ \(\frac{7}{9}\) = \(\frac{7}{9}\)
And , \(\frac{21}{9.6}\) = \(\frac{14}{6.4}\)
∴ \(\frac{7}{3.2}\) = \(\frac{7}{3.2}\)
And , \(\frac{27}{9.6}\) = \(\frac{18}{6.4}\)
∴ \(\frac{9}{3.2}\) = \(\frac{9}{3.2}\)
Therefore, the given pair of triangles are similar.
By proving the ratio of the corresponding sides of the triangles are same we have proved that the given pair of triangles are similar.
Page 487 Exercise 6 Problem 6
Given: \(\overline{J K}\) ⊥ \(\overline{M L}\)
To Find – We have to prove the given pair of triangles as similar.
Using the theorems and postulates of similarities.
In the given diagram given that
Given \(\overline{J K}\) ⊥ \(\overline{M L}\)
From the diagram we get
tan M = \(\frac{15}{21}\)
M = tan-1(\(\frac{5}{7}\))
∠M = 35.5°
tan L = \(\frac{20}{28}\)
L = tan-1(\(\frac{5}{7}\))
∴ ∠L = 35.5°
Therefore ∠M = ∠L
The given pair of triangles are similar by ASA property of similarity.
By ASA property of similarity we have proved that the given pair of triangles are similar.
Page 487 Exercise 7 Problem 7
Given: A diagram of a ΔABCb with right ∠C and altitude \(\overline{C D}\)
To Find – We have to find the value of x.
Using the rules of the right angles.
From the given diagram we get
BD = 16
DA = 9
tan∠B = tan∠ACD
⇒ \(\frac{16}{x} – {9}{x}\)
⇒ \(\frac{7}{x}\) = 0
∴ x = 7
In ΔABC the value of x is 7.
Page 487 Exercise 8 Problem 8
Given: A diagram of aΔABC with right ∠C and altitude \(\overline{C D}\)
To Find –We have to find the value of x.
Using the rules of the right angles.
From the given diagram we get
AB = 32 [Given]
BC = 16 [Given]
∠D = 90° [Given]
Cos∠ABC = cos∠DBC
\(\frac{16}{32-x}\) = \(\frac{x}{16}\)
32x − x2 = 256
x2 − 32x + 256 = 0
x2 − 16x − 16x + 256 = 0
x(x − 16) − 16(x − 16) = 0
(x−16)(x−16) = 0
∴ x = 16
In ΔABC the value of x is 16.
Page 487 Exercise 9 Problem 9
Given: A diagram of a ΔABC with right ∠C and altitude \(\overline{C D}\)
To Find – We have to find the value of x.
Using the rules of the right angles.
From the diagram we get
AD = 4 [Given]
DB = 9 [Given]
∠D = 90°[Given]
∠C = 90° [Given]
cos∠DAC = cos∠BAC
\(\frac{4}{x}\) = \(\frac{x}{13}\)
x2 = 52
∴ x = 7.211
In ΔABC the value of x is 7.211.
Page 487 Exercise 10 Problem 10
Given: The triangle
To find – The value of x.
We will use geometric form and calculate the value of x
The geometric means is as follows:
\(\frac{d}{b}\) = \(\frac{b}{e}\)
\(\frac{x+16}{15}\) = \(\frac{15}{x}\)
x(x + 16) = 225x
x2 + 16x − 225x = 0
x = 0,209
The value of x in △ABC with right ∠C and altitude is 209 \(\overline{C D}\)
Page 487 Exercise 11 Problem 11
We have to describe how might you describe an angle of elevation in geometry.
We observe that the angle of elevation is the angle between the horizontal line of sight and the line of sight up to an object.
An angle of elevation is the angle between the horizontal to a line of sight to the object when looking up.
In geometry it is the angle of elevation is the angle between the horizontal line of sight and the line of sight up to an object.
An angle of elevation in geometry is the angle of elevation is the angle between the horizontal line of sight and the line of sight up to an object.
Page 487 Exercise 12 Problem 12
We have to describe how to calculate the magnitude of a line segment in the coordinate plane.
For a line segment AB containing the points, A,B the magnitude is the length of the line segment.
|AB| = \(\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\)
The magnitude of a line segment in the coordinate plane can be calculated by the distance between the two points containing the formula and taking the magnitude
Page 487 Exercise 13 Problem 13
We have to what does the prefix indicate in these words and what geometric figure do you think is associated with the phrase trigonometric ratio.
We observe that tri represents three in quantity so so the geometric figure associated with the phrase trigonometric ratio is a triangle.
The geometric figure is a triangle.
The prefix tri indicates three and the geometric figure do you think is associated with the phrase trigonometric ratio is triangle.