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n200080 [17]
3 years ago
12

Suppose an isosceles triangle has two sides of length 42 and 35, and that the angle between these two sides is 120 degree. what

is the length of the third side of the triangle? a. 43.65 b.77 c.58.04 d.66.78
Mathematics
1 answer:
Citrus2011 [14]3 years ago
3 0
Suppose that a=42, b=35, c is unknown, and C = 120°

c^2 = a^2 + b^2 -2ab(cosC)

c^2 = 42^2 + 35^2 -(2*42*35*cos120)

(cos120=-0.5)

c^2 = 1764 + 1225 -(2940*cos120)

c^2 = 2989 -(2940*-0.5)

c^2 = 2989+1470

c^2=4459

c= \sqrt{c^2} 

c=\sqrt{4459}
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8 0
3 years ago
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Find the missing side length XI. round your answer to the nearest tenth.​
lianna [129]

Answer: XI = 10.35 mm

Step-by-step explanation:

Considering the given triangle XIF, to determine XI, we would apply the sine rule. It is expressed as

a/SinA = b/SinB = c/SinC

Where a, b and c are the length of each side of the triangle and angle A, Angle B and angle C are the corresponding angles of the triangle. Likening it to the given triangle, the expression becomes

XI/SinF = XF/SinI = FI/SinX

The sum of the angles in a triangle is 180°. It means that

F + 63 + 52 = 180

F = 180 - (63 + 52)

F = 65°

Therefore

XI/Sin 65 = 9/Sin 52

Cross multiplying, it becomes

XISin52 = 9Sin65

0.788XI = 9 × 0.906

0.788XI = 8.154

XI = 8.154/0.788

XI = 10.35

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