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Answer |
#1 Finde the area of the following tringles: (use Logarithms)
A) a=12.7, b=21.5, c=28.6.
B) c=126, a=45.81, and b=61.03694
C) An isosceles triangle in which each of the equal sides is 14.72 in. and the vertex angle is 47.467
(i need it in very detailed thanx) |
| Answer |
For problems 1 and 2, use the formula to find the area S of an oblique triangle:
S = [s (s-a)(s-b)(s-c)]^(1/2) where s = (1/2)(a + b + c)
Applying rules of logs:
S = [s (s-a)(s-b)(s-c)]^(1/2)-->
logS=(1/2)[log(s(s-a)(s-b)(s-c)] -->*{you may want to just multiply the terms out, rather than go the next extra step; in the solutions, I used the next step.}
logS=(1/2)[logs + log(s-a) + log(s-b) + log(s-c)]
--------
A) s =(1/2)(12.7+21.5+28.6)=31.4
s-a=31.4-12.7=18.7
s-b=31.4-21.5=9.9
s-c=31.4-28.6=2.8
logS=(1/2)(log31.4+log18.7+log9.9+log2.8) -->logS = (1/2)(2.801049343) -->logS=1.400524671-->S=Inv log(1.400524671) = 10^(1.400524671) =25.14922876
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B) This triangle doesn\'t exist... If you add sides a and b, you will see that the sum is less than the length of side c, so this triangle can\'t be formed. I am wondering if you meant one of those variables to represent an angle?
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C) You have two sides and the included angle. So, to find the area you would use the formula:
S=absinC/2 -->
logS = log(absinC/2) -->
logS=log a + logb +log(sinC) - log 2
(I assumed using degrees rather than radians for the angle; if not, the value for the sine of the angle will change.)
logS = log(14.72)+log(14.72)+log(sin(47.467)) - log2 -->
logS=1.902187169 -->
S = 10^(1.902187169) -->
S=79.83386752
I hope this was helpful! |
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