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Solutions for Session 8, Part C
See solutions for Problems: C1 | C2 | C3 | C4 | C5 | C6 | C7
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Problem C1 | |
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The triangle with hypotenuse 4: By similarity with the original triangle, we can say:
5/4 = 3/a, so a = 12/5.
The triangle with hypotenuse 10: By similarity with the original triangle, we can say:
5/10 = 3/a, so a = 30/5 = 6.
The triangle with hypotenuse 1: By similarity with the original triangle, we can say:
5/1 = 3/a, so a = 3/5.
<< back to Problem C1
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Problem C2 | |
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The triangle with hypotenuse 4: By similarity with the original triangle, we can say:
5/4 = 4/b, so b = 16/5.
The triangle with hypotenuse 10: By similarity with the original triangle, we can say:
5/10 = 4/b, so b = 40/5 = 8.
The triangle with hypotenuse 1: By similarity with the original triangle, we can say:
5/1 = 4/b, so b = 4/5.
<< back to Problem C2
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Problem C3 | |
a. | sin A = 3/5 and cos A = 4/5 |
b. | sin B = 4/5 and cos B = 3/5 |
<< back to Problem C3
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Problem C4 | |
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We calculate the hypotenuse with the Pythagorean theorem and find that it is 13.
a. | sin A = 12/13, cos A = 5/13 |
b. | sin B = 5/13, cos B = 12/13 |
c. | tan A = 12/5, tan B = 5/12 |
<< back to Problem C4
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Problem C5 | |
a. | sin 30° = cos 60° = 1/2
sin 60° = cos 30° =  /2
tan 30° = 1/
tan 60° =
These values are constant for any triangle with angles 30°-60°-90°.
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b. | sin 45° = cos 45° = 1/
tan 45° = 1
These values are constant for any triangle with angles 45°-45°-90°.
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<< back to Problem C5
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Problem C6 | |
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Angles A and B are adjacent, so sin A = cos B. The hypotenuse is the same for both angles, but the roles of "adjacent side" and "opposite side" switch. The side opposite angle B is adjacent to angle A, and vice versa.
<< back to Problem C6
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Problem C7 | |
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Let x be the length of the ramp. Then we have a right triangle with hypotenuse x, shorter leg 2, and the angle opposite to the shorter leg of 10°. Since sin 10° = 2/x, we have x = 2/sin 10° = 2/0.17  11.765 feet.
<< back to Problem C7
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