38° z SohCah Toa 10’ y. β SohCah Toa 15cm x 24cm.

9
38° z z = 16.24 ft Soh Cah Toa 10 y ¿ 90 38 = 52 ° y = 12.80 ft 10 2 + 12.80 2 = 16.24 2 100 + 163.84 = 263.74

Transcript of 38° z SohCah Toa 10’ y. β SohCah Toa 15cm x 24cm.

38°

z z = 16.24 ft

Soh Cah Toa

10’

y

¿90−38=52°

y = 12.80 ft102 + 12.802 = 16.242 100 + 163.84 = 263.74 √

β = 51.3˚

Soh Cah Toa

15cm

x

β  ¿ 90−51.3=36.7 °

24cm

x2 = 351x = 3

37˚ x = x = 25.76cm

Soh Cah Toax

α ¿90−37=53 °

15.5cm

y2 = 426.3276y = 20.57 cm

y

13.5

°x x = 95.53 in

Soh Cah Toa22.3”

y

¿90−13.5=76.5 °y = 92.87 in

22.32 + 92.872 = 95.532 9122.13 ≈ 9125.98 √

15

.5

m

= 30˚

Soh Cah Toa

31 mx

¿90−30=60 °

28

Given regular square pyramid with slant height of 50ft. The perimeter of the Base is 112ft

Find the altitude.

14

x50

Family! (7 – 24 – 25)

x = 48ft

12

9

20

Find the diagonal of a regular rectangular prism with dimensions of 9, 12, and 20

Find the altitude.

202 + 152 = x2

Family! (3 – 4 – 5)

x = 25

• √2 • √2

= 42√2 2

21√2

42

21√245

I have the hyp, so to get the legs, divide by √2

42 √2

= 21√2

45

Use special right ∆ rules to solve the triangle. Answers in

simplified radical form

12√3

60

18

6√3

30

I have the long leg, so get short leg first by dividing by √3

Then, from the short leg to get the hyp, multiply by 2

• √3 • √3

= 18√3 3

18 √3

= 6√3

Use special right ∆ rules to solve the triangle. Answers in

simplified radical form