7. determine whether the triangle is acute, right, or obtuse for each of the following triangles with given sides.
a. 9, 12, 17
b. 20, 21, 28
c. 15, 36, 39
8. find the value of x.
(image of a right triangle with segments: a to d is 3, d to c is 9, right angle at d and b)
9. find the value of x.
(image of a right triangle with segments: a to d is 5, d to c is 15, right angle at d and b)
10. the tangent of ∠b is ______.
(image of a right triangle with bc=45, ac=28, hypotenuse ab=53, right angle at c)
11. find the value of x.
(image of a right triangle with angle 59°, side 17)
12. write cos b.
(image of a right triangle with bc=45, ac=28, hypotenuse ab=53, right angle at c; handwritten: 45/53)
13. what is x to the nearest hundredth?
(image of a right triangle with angle 61°, side 17)
14. find y. round the result to the nearest hundredth.
(image of a right triangle with angle 61°, side 17)

Question

Accepted Answer

### Problem 7A: Determine Triangle Type (9, 12, 17)
# Explanation:
## Step1: Identify the longest side
Longest side $ c = 17 $, other sides $ a = 9 $, $ b = 12 $.
## Step2: Apply Pythagorean Inequality
Check $ a^2 + b^2 $ vs $ c^2 $.
$ a^2 + b^2 = 9^2 + 12^2 = 81 + 144 = 225 $
$ c^2 = 17^2 = 289 $
Since $ a^2 + b^2 < c^2 $, the triangle is obtuse.
# Answer: Obtuse

### Problem 7B: Determine Triangle Type (20, 21, 28)
# Explanation:
## Step1: Identify the longest side
Longest side $ c = 28 $, other sides $ a = 20 $, $ b = 21 $.
## Step2: Apply Pythagorean Inequality
Check $ a^2 + b^2 $ vs $ c^2 $.
$ a^2 + b^2 = 20^2 + 21^2 = 400 + 441 = 841 $
$ c^2 = 28^2 = 784 $
Since $ a^2 + b^2 > c^2 $, the triangle is acute.
# Answer: Acute

### Problem 7C: Determine Triangle Type (15, 36, 39)
# Explanation:
## Step1: Identify the longest side
Longest side $ c = 39 $, other sides $ a = 15 $, $ b = 36 $.
## Step2: Apply Pythagorean Inequality
Check $ a^2 + b^2 $ vs $ c^2 $.
$ a^2 + b^2 = 15^2 + 36^2 = 225 + 1296 = 1521 $
$ c^2 = 39^2 = 1521 $
Since $ a^2 + b^2 = c^2 $, the triangle is right.
# Answer: Right

### Problem 8: Find $ x $ (Right Triangle Altitude)
# Explanation:
## Step1: Recall Geometric Mean Theorem
In a right triangle, the leg is the geometric mean of the adjacent segment and the hypotenuse segment.
For leg $ AB = x $, segment $ AD = 3 $, hypotenuse segment $ AC = 3 + 9 = 12 $.
Thus, $ x^2 = AD \times AC $.
## Step2: Solve for $ x $
$ x^2 = 3 \times 12 = 36 $
$ x = \sqrt{36} = 6 $
# Answer: 6

### Problem 9: Find $ x $ (Right Triangle Leg)
# Explanation:
## Step1: Recall Geometric Mean Theorem
For leg $ BC = x $, segment $ CD = 15 $, hypotenuse segment $ AC = 5 + 15 = 20 $.
Thus, $ x^2 = CD \times AC $.
## Step2: Solve for $ x $
$ x^2 = 15 \times 20 = 300 $
$ x = \sqrt{300} = 10\sqrt{3} \approx 17.32 $ (or exact form $ 10\sqrt{3} $)
# Answer: $ 10\sqrt{3} $ (or ≈17.32)

### Problem 10: Tangent of $ \angle B $
# Explanation:
## Step1: Recall Tangent Definition
$ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} $ for $ \angle B $.
In $ \triangle ABC $, $ \angle C = 90^\circ $, opposite side to $ \angle B $ is $ AC = 28 $, adjacent side is $ BC = 45 $.
## Step2: Compute $ \tan B $
$ \tan B = \frac{AC}{BC} = \frac{28}{45} $
# Answer: $ \frac{28}{45} $

### Problem 11: Find $ x $ (Right Triangle Side)
# Explanation:
## Step1: Identify Triangle Type
Right triangle with angle $ 59^\circ $, hypotenuse $ 17 $, $ x $ is adjacent to $ 59^\circ $.
Use $ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} $.
## Step2: Solve for $ x $
$ \cos 59^\circ = \frac{x}{17} $
$ x = 17 \cos 59^\circ \approx 17 \times 0.5150 \approx 8.755 \approx 8.76 $
# Answer: ≈8.76

### Problem 12: $ \cos B $
# Explanation:
## Step1: Recall Cosine Definition
$ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} $ for $ \angle B $.
In $ \triangle ABC $, $ \angle C = 90^\circ $, adjacent side to $ \angle B $ is $ BC = 45 $, hypotenuse $ AB = 53 $.
## Step2: Compute $ \cos B $
$ \cos B = \frac{BC}{AB} = \frac{45}{53} $
# Answer: $ \frac{45}{53} $

### Problem 13: Find $ x $ (Right Triangle Side)
# Explanation:
## Step1: Identify Triangle Type
Right triangle with angle $ 61^\circ $, adjacent side $ 17 $, $ x $ is hypotenuse.
Use $ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} $.
## Step2: Solve for $ x $
$ \cos 61^\circ = \frac{17}{x} $
$ x = \frac{17}{\cos 61^\circ} \approx \frac{17}{0.4848} \approx 35.07 $
# Answer: ≈35.07

### Problem 14: Find $ y $ (Right Triangle Side)
# Explanation:
## Step1: Identify Triangle Type
Right triangle with angle $ 61^\circ $, adjacent side $ y $, opposite side $ 17 $.
Use $ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} $.
## Step2: Solve for $ y $
$ \tan 61^\circ = \frac{17}{y} $
$ y = \frac{17}{\tan 61^\circ} \approx \frac{17}{1.8040} \approx 9.42 $
# Answer: ≈9.42

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 7A: Determine Triangle Type (9, 12, 17)

Step1: Identify the longest side

Step2: Apply Pythagorean Inequality

Step1: Identify the longest side

Step2: Apply Pythagorean Inequality

Step1: Identify the longest side

Step2: Apply Pythagorean Inequality

Answer:

Problem 7B: Determine Triangle Type (20, 21, 28)