Question

math 10 foundations
trigonometry
what is the length of the missing side of the following right - angle triangles? show all work!

1. a = 3; b = 4; c =?
2. a = 6; b = 8; c =?
3. a = 12; b =?; c = 22
4. a = 9; b =?; c = 13
5. a =?; b = 20; c = 23
6. a =?; b = 10; c = 17
7. to get from point a to point b you must avoid walking through a pond. to avoid the pond, you must walk 34 meters south and 41 meters east. to the nearest meter, how many meters would be saved if it were possible to walk through the pond?

Explanation:

Step1: Recall Pythagorean theorem

For a right - triangle with sides \(a\), \(b\) and hypotenuse \(c\), \(c^{2}=a^{2}+b^{2}\) (when \(c\) is unknown) or \(a^{2}=c^{2}-b^{2}\) (when \(a\) is unknown) or \(b^{2}=c^{2}-a^{2}\) (when \(b\) is unknown).

Step2: Solve problem 10

Given \(a = 3\), \(b = 4\), find \(c\). Using \(c^{2}=a^{2}+b^{2}\), we have \(c^{2}=3^{2}+4^{2}=9 + 16=25\), so \(c=\sqrt{25}=5\).

Step3: Solve problem 11

Given \(a = 6\), \(b = 8\), find \(c\). Using \(c^{2}=a^{2}+b^{2}\), \(c^{2}=6^{2}+8^{2}=36 + 64 = 100\), so \(c=\sqrt{100}=10\).

Step4: Solve problem 12

Given \(a = 12\), \(c = 22\), find \(b\). Using \(b^{2}=c^{2}-a^{2}\), \(b^{2}=22^{2}-12^{2}=(22 + 12)(22 - 12)=34\times10 = 340\), so \(b=\sqrt{340}\approx18.44\).

Step5: Solve problem 13

Given \(a = 9\), \(c = 13\), find \(b\). Using \(b^{2}=c^{2}-a^{2}\), \(b^{2}=13^{2}-9^{2}=(13 + 9)(13 - 9)=22\times4 = 88\), so \(b=\sqrt{88}\approx9.38\).

Step6: Solve problem 14

Given \(b = 20\), \(c = 23\), find \(a\). Using \(a^{2}=c^{2}-b^{2}\), \(a^{2}=23^{2}-20^{2}=(23 + 20)(23 - 20)=43\times3 = 129\), so \(a=\sqrt{129}\approx11.36\).

Step7: Solve problem 15

Given \(b = 10\), \(c = 17\), find \(a\). Using \(a^{2}=c^{2}-b^{2}\), \(a^{2}=17^{2}-10^{2}=(17 + 10)(17 - 10)=27\times7 = 189\), so \(a=\sqrt{189}\approx13.75\).

Step8: Solve problem 16

The non - pond path length is \(34 + 41=75\) meters. Using the Pythagorean theorem to find the straight - line (through the pond) distance \(d\), \(d^{2}=34^{2}+41^{2}=1156+1681 = 2837\), so \(d=\sqrt{2837}\approx53.26\) meters. The distance saved is \(75 - 53.26\approx22\) meters.

Answer:

1. \(c = 5\)
2. \(c = 10\)
3. \(b\approx18.44\)
4. \(b\approx9.38\)
5. \(a\approx11.36\)
6. \(a\approx13.75\)
7. Approximately 22 meters