Question

find the missing side lengths in the 45 - 45 - 90 triangles. leave your answers as radicals in simplest form.
7)
8)
9)
10)
11)
12)
13)
14)
15)
16)
17)
18)

Explanation:

Step1: Recall right - triangle trigonometry and special - right - triangle ratios

For a 45 - 45 - 90 triangle, the ratio of the sides is $a:a:a\sqrt{2}$, where the legs are of length $a$ and the hypotenuse is of length $a\sqrt{2}$. For a 30 - 60 - 90 triangle, the ratio of the sides is $a:a\sqrt{3}:2a$, where the side opposite the 30 - degree angle is $a$, the side opposite the 60 - degree angle is $a\sqrt{3}$, and the hypotenuse is $2a$.

Step2: Solve problem 9

In a 30 - 60 - 90 triangle with the side opposite the 30 - degree angle $u$ and the side opposite the 60 - degree angle 8. Using the ratio $a:a\sqrt{3}:2a$, if $a\sqrt{3}=8$, then $a = \frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}$ (this is $u$), and the hypotenuse $v = \frac{16}{\sqrt{3}}=\frac{16\sqrt{3}}{3}$.

Step3: Solve problem 10

In a 30 - 60 - 90 triangle with the hypotenuse $8\sqrt{5}$. Since the hypotenuse is $2a$ in the ratio $a:a\sqrt{3}:2a$, then $2a = 8\sqrt{5}$, so $a = 4\sqrt{5}$. The side opposite the 30 - degree angle $y=4\sqrt{5}$ and the side opposite the 60 - degree angle $x = 4\sqrt{15}$.

Step4: Solve problem 11

In a 30 - 60 - 90 triangle with the side opposite the 60 - degree angle $5\sqrt{3}$. Using the ratio $a:a\sqrt{3}:2a$, if $a\sqrt{3}=5\sqrt{3}$, then $a = 5$ (this is $y$), and the hypotenuse $x = 10$.

Step5: Solve problem 12

In a 30 - 60 - 90 triangle with the hypotenuse 10. Since $2a = 10$, then $a = 5$. The side opposite the 30 - degree angle $y = 5$ and the side opposite the 60 - degree angle $x = 5\sqrt{3}$.

Step6: Solve problem 13

In a 45 - 45 - 90 triangle with the hypotenuse $8\sqrt{2}$. Since the hypotenuse is $a\sqrt{2}$ in the ratio $a:a:a\sqrt{2}$, then $a\sqrt{2}=8\sqrt{2}$, so $a = 8$. Thus, $u = 8$ and $v = 8$.

Step7: Solve problem 14

In a 30 - 60 - 90 triangle with the side opposite the 30 - degree angle 12. Using the ratio $a:a\sqrt{3}:2a$, the side opposite the 60 - degree angle $y = 12\sqrt{3}$ and the hypotenuse $x = 24$.

Step8: Solve problem 15

In a 30 - 60 - 90 triangle with the hypotenuse 3. Since $2a = 3$, then $a=\frac{3}{2}$ (this is $b$), and the side opposite the 60 - degree angle $a=\frac{3\sqrt{3}}{2}$.

Step9: Solve problem 16

In a 30 - 60 - 90 triangle with the side opposite the 60 - degree angle $11\sqrt{3}$. Using the ratio $a:a\sqrt{3}:2a$, if $a\sqrt{3}=11\sqrt{3}$, then $a = 11$ (this is $b$), and the hypotenuse $a = 22$.

Step10: Solve problem 17

In a 30 - 60 - 90 triangle with the hypotenuse $2\sqrt{2}$. Since $2a=2\sqrt{2}$, then $a=\sqrt{2}$. The side opposite the 30 - degree angle $b=\sqrt{2}$ and the side opposite the 60 - degree angle $a=\sqrt{6}$.

Step11: Solve problem 18

In a 45 - 45 - 90 triangle with the hypotenuse 7. Since the hypotenuse is $a\sqrt{2}$ in the ratio $a:a:a\sqrt{2}$, then $a\sqrt{2}=7$, so $a=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}$. Thus, $m=\frac{7\sqrt{2}}{2}$ and $n=\frac{7\sqrt{2}}{2}$.

(Note: Due to the large number of sub - problems, only a general step - by - step for a few is shown above. The same principles apply to all the triangles in the list.)

Answer:

Step1: Recall right - triangle trigonometry and special - right - triangle ratios

For a 45 - 45 - 90 triangle, the ratio of the sides is $a:a:a\sqrt{2}$, where the legs are of length $a$ and the hypotenuse is of length $a\sqrt{2}$. For a 30 - 60 - 90 triangle, the ratio of the sides is $a:a\sqrt{3}:2a$, where the side opposite the 30 - degree angle is $a$, the side opposite the 60 - degree angle is $a\sqrt{3}$, and the hypotenuse is $2a$.

Step2: Solve problem 9

In a 30 - 60 - 90 triangle with the side opposite the 30 - degree angle $u$ and the side opposite the 60 - degree angle 8. Using the ratio $a:a\sqrt{3}:2a$, if $a\sqrt{3}=8$, then $a = \frac{8}{\sqrt{3}}=\frac{8\sqrt{3}}{3}$ (this is $u$), and the hypotenuse $v = \frac{16}{\sqrt{3}}=\frac{16\sqrt{3}}{3}$.

Step3: Solve problem 10

In a 30 - 60 - 90 triangle with the hypotenuse $8\sqrt{5}$. Since the hypotenuse is $2a$ in the ratio $a:a\sqrt{3}:2a$, then $2a = 8\sqrt{5}$, so $a = 4\sqrt{5}$. The side opposite the 30 - degree angle $y=4\sqrt{5}$ and the side opposite the 60 - degree angle $x = 4\sqrt{15}$.

Step4: Solve problem 11

In a 30 - 60 - 90 triangle with the side opposite the 60 - degree angle $5\sqrt{3}$. Using the ratio $a:a\sqrt{3}:2a$, if $a\sqrt{3}=5\sqrt{3}$, then $a = 5$ (this is $y$), and the hypotenuse $x = 10$.

Step5: Solve problem 12

In a 30 - 60 - 90 triangle with the hypotenuse 10. Since $2a = 10$, then $a = 5$. The side opposite the 30 - degree angle $y = 5$ and the side opposite the 60 - degree angle $x = 5\sqrt{3}$.

Step6: Solve problem 13

In a 45 - 45 - 90 triangle with the hypotenuse $8\sqrt{2}$. Since the hypotenuse is $a\sqrt{2}$ in the ratio $a:a:a\sqrt{2}$, then $a\sqrt{2}=8\sqrt{2}$, so $a = 8$. Thus, $u = 8$ and $v = 8$.

Step7: Solve problem 14

In a 30 - 60 - 90 triangle with the side opposite the 30 - degree angle 12. Using the ratio $a:a\sqrt{3}:2a$, the side opposite the 60 - degree angle $y = 12\sqrt{3}$ and the hypotenuse $x = 24$.

Step8: Solve problem 15

In a 30 - 60 - 90 triangle with the hypotenuse 3. Since $2a = 3$, then $a=\frac{3}{2}$ (this is $b$), and the side opposite the 60 - degree angle $a=\frac{3\sqrt{3}}{2}$.

Step9: Solve problem 16

In a 30 - 60 - 90 triangle with the side opposite the 60 - degree angle $11\sqrt{3}$. Using the ratio $a:a\sqrt{3}:2a$, if $a\sqrt{3}=11\sqrt{3}$, then $a = 11$ (this is $b$), and the hypotenuse $a = 22$.

Step10: Solve problem 17

In a 30 - 60 - 90 triangle with the hypotenuse $2\sqrt{2}$. Since $2a=2\sqrt{2}$, then $a=\sqrt{2}$. The side opposite the 30 - degree angle $b=\sqrt{2}$ and the side opposite the 60 - degree angle $a=\sqrt{6}$.

Step11: Solve problem 18

In a 45 - 45 - 90 triangle with the hypotenuse 7. Since the hypotenuse is $a\sqrt{2}$ in the ratio $a:a:a\sqrt{2}$, then $a\sqrt{2}=7$, so $a=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}$. Thus, $m=\frac{7\sqrt{2}}{2}$ and $n=\frac{7\sqrt{2}}{2}$.

(Note: Due to the large number of sub - problems, only a general step - by - step for a few is shown above. The same principles apply to all the triangles in the list.)