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shusha [124]
3 years ago
9

Can someone help me with number 3 please??!!

Mathematics
1 answer:
ipn [44]3 years ago
7 0
There are 2 way to solve this.

one using Pythagoras theorem and 2nd using trigonometry

so lets solve it by both

using Pythagoras theorem we know

base^2 + perpendicular^2 = hypotanes^2

6^2 + x^2 = 12^2

36 + x^2 = 144

x^2 = 144- 36 = 108

x = √(108) = √( 2×2×3×3×3)

= (2×3) √ (3) = 6 √3

so answer is option 2

bow lets use trigonometry

we know
sin theta = perpendicular / hypotanes
sin 60 = x /12
x = 12 × sin 60
we kNow sin 60 = √3/ 2
so
x = 12×√3 /2 = 6√3
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Please help with question 15 and 16!!!
kykrilka [37]
15. [4] insufficient information. only given SA
16. There are 3 right triangles, 3 Pythagorean theorem equations 
x²+b²= 16²
12²+a² = x²
a²+4² = b²

combine equations using substitution to get in terms of only x
substitute b² for (a² +x²) in first equation
x²+a²+4²=16²
substitute a² for (x² - 12²)
x²+x²-12²+4² = 16²
solve for x
2x² - 144 + 16 = 256
2x² = 256 + 144 - 16
2x² = 384
x² = 384/2
x² = 192
x = 13.85

rounded to nearest tenth, x = 13.9




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explanation:

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