a Answer ∠ A = 60 ° Explanation Given information: ∠ B = 30 ° , ∠ C = 90 ° Formula used: Sum of all interior angle of a right angle triangle is ∠ A + ∠ B + ∠ C = 180 ° Calculation: Substituting the values of angle B and C in the formula ∠ A + 30 ° + 90 ° = 180 ° Calculating the value of angle A ∠ A = 60 ° b. To determine To calculate length of A C using sine ratio Answer 55 2 Explanation Given information: h y p o t e n u s e = A B = 55 Formula used: sin θ = p e r p e n d i c u l a r h y p o t e n u s e Calculation: The angle of sine ratio is 30 ° Substituting it in the sine expression sin 30 ° = 1 2 Substituting value of hypotenuse and sine ratio in the formula 1 2 = A C 55 Using cross multiplication method value of AC is A C = 55 2 c. To determine To calculate length of B C using cosine ratio Answer 47.6 f t Explanation Given information: h y p o t e n u s e = A B = 55 Formula used: cos θ = b a s e h y p o t e n u s e Calculation: The angle of cosine ratio is 30 ° Substituting it in the cosine expression cos 30 ° = 3 2 Substituting value of hypotenuse and cosine ratio in the formula 3 2 = B C 55 Using cross multiplication method B C = 55 × 3 2 Calculating value of the above expression B C = 55 × 1.73 = 47.6 f t d. To determine To prove : To verify length of A C and B C using tangent ratio Explanation Given information : A C = 55 2 , B C = 55 × 3 2 Formula used : tan θ = p e r p e n d i c u l a r b a s e Proof: Using formula, tan 30 ° = A C B C Substituting values of AC and BC in the formula 1 3 = 55 2 55 × 3 2 Calculating 1 3 = 55 2 × 2 55 3 Finally, 1 3 = 1 3 Hence, using tangent ratio the length of AC and BC are verified

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN: 9780547587776

Author: HOLT MCDOUGAL

Publisher: HOLT MCDOUGAL

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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°).

Chapter 10, Problem 16PTTS

To determine

To measure ∠A

Expert Solution

Answer to Problem 16PTTS

∠A=60°

Explanation of Solution

Given information:

∠B=30° , ∠C=90°

Formula used:

Sum of all interior angle of a right angle triangle is

∠A+∠B+∠C=180°

Calculation:

Substituting the values of angle B and C in the formula

∠A+30°+90°=180°

Calculating the value of angle A

∠A=60°

To determine

To calculate length of AC using sine ratio

Expert Solution

Answer to Problem 16PTTS

552

Explanation of Solution

Given information:

hypotenuse=AB=55

Formula used:

sinθ=perpendicularhypotenuse

Calculation:

The angle of sine ratio is

30°

Substituting it in the sine expression

sin30°=12

Substituting value of hypotenuse and sine ratio in the formula

12=AC55

Using cross multiplication method value of AC is

AC=552

To determine

To calculate length of BC using cosine ratio

Expert Solution

Answer to Problem 16PTTS

47.6ft

Explanation of Solution

Given information:

hypotenuse=AB=55

Formula used:

cosθ=basehypotenuse

Calculation: The angle of cosine ratio is

30°

Substituting it in the cosine expression

cos30°=32

Substituting value of hypotenuse and cosine ratio in the formula

32=BC55

Using cross multiplication method

BC=55×32

Calculating value of the above expression

BC=55×1.73=47.6 ft

To determine

To prove :

To verify length of AC and BC using tangent ratio

Expert Solution

Explanation of Solution

Given information :

AC=552 , BC=55×32

Formula used :

tanθ=perpendicularbase

Proof:

Using formula,

tan30°=ACBC

Substituting values of AC and BC in the formula

13=55255×32

Calculating

13=552×2553

Finally,

13=13

Hence, using tangent ratio the length of AC and BC are verified

Chapter 10, Problem 15PTTS

Chapter 10, Problem 1CP

Chapter 10 Solutions

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

Ch. 10.1 - Prob. 1C Ch. 10.1 - Prob. 2C Ch. 10.1 - Prob. 3C Ch. 10.1 - Prob. 1E Ch. 10.1 - Prob. 2E Ch. 10.1 - Prob. 3E Ch. 10.1 - Prob. 4E Ch. 10.1 - Prob. 5E Ch. 10.1 - Prob. 6E Ch. 10.1 - Prob. 7E

Ch. 10.1 - Prob. 8E Ch. 10.1 - Prob. 9E Ch. 10.1 - Prob. 10E Ch. 10.1 - Prob. 11E Ch. 10.1 - Prob. 12E Ch. 10.1 - Prob. 13E Ch. 10.1 - Prob. 14E Ch. 10.1 - Prob. 15E Ch. 10.1 - Prob. 16E Ch. 10.1 - Prob. 17E Ch. 10.1 - Prob. 18E Ch. 10.1 - Prob. 19E Ch. 10.1 - Prob. 20E Ch. 10.1 - Prob. 21E Ch. 10.1 - Prob. 22E Ch. 10.1 - Prob. 23E Ch. 10.1 - Prob. 24E Ch. 10.1 - Prob. 25E Ch. 10.1 - Prob. 26E Ch. 10.1 - Prob. 27E Ch. 10.1 - Prob. 28E Ch. 10.1 - Prob. 29E Ch. 10.1 - Prob. 30E Ch. 10.1 - Prob. 31E Ch. 10.1 - Prob. 32E Ch. 10.2 - Prob. 1C Ch. 10.2 - Prob. 2C Ch. 10.2 - Prob. 3C Ch. 10.2 - Prob. 1E Ch. 10.2 - Prob. 2E Ch. 10.2 - Prob. 3E Ch. 10.2 - Prob. 4E Ch. 10.2 - Prob. 5E Ch. 10.2 - Prob. 6E Ch. 10.2 - Prob. 7E Ch. 10.2 - Prob. 8E Ch. 10.2 - Prob. 9E Ch. 10.2 - Prob. 10E Ch. 10.2 - Prob. 11E Ch. 10.2 - Prob. 12E Ch. 10.2 - Prob. 13E Ch. 10.2 - Prob. 14E Ch. 10.2 - Prob. 15E Ch. 10.2 - Prob. 16E Ch. 10.2 - Prob. 17E Ch. 10.2 - Prob. 18E Ch. 10.2 - Prob. 19E Ch. 10.2 - Prob. 20E Ch. 10.2 - 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