62.2k views
2 votes
In triangle




ABCA, B, C,


=
7
AB=7A, B, equals, 7 and


=
10
BC=10B, C, equals, 10. Which of the following cannot be the length of


ACA, C ?

User Jary
by
7.7k points

1 Answer

2 votes
Let's use the Law of Cosines to solve this problem.

We know that:

AC^2 = AB^2 + BC^2 - 2(AB)(BC)cos(angle B)

Since angle B is obtuse, cos(angle B) is negative. Therefore, we have:

AC^2 = (7A)^2 + (10B)^2 + 2(7A)(10B)cos(angle B)

AC^2 = 49A^2 + 100B^2 + 140ABcos(angle B)

We are asked to find which of the following cannot be the length of AC. Let's check each option:

A) 10
If AC = 10, then AC^2 = 100. We can substitute this into our equation:

100 = 49A^2 + 100B^2 + 140ABcos(angle B)

We know that A, B, and cos(angle B) are all positive, so the left-hand side of the equation is less than the right-hand side. Therefore, 10 can't be the length of AC.

B) 13
If AC = 13, then AC^2 = 169. We can substitute this into our equation:

169 = 49A^2 + 100B^2 + 140ABcos(angle B)

We know that A, B, and cos(angle B) are all positive, so the left-hand side of the equation is greater than the right-hand side. Therefore, 13 can be the length of AC.

C) 15
If AC = 15, then AC^2 = 225. We can substitute this into our equation:

225 = 49A^2 + 100B^2 + 140ABcos(angle B)

We know that A, B, and cos(angle B) are all positive, so the left-hand side of the equation is greater than the right-hand side. Therefore, 15 can be the length of AC.

D) 17
If AC = 17, then AC^2 = 289. We can substitute this into our equation:

289 = 49A^2 + 100B^2 + 140ABcos(angle B)

We know that A, B, and cos(angle B) are all positive, so the left-hand side of the equation is greater than the right-hand side. Therefore, 17 can be the length of AC.

Therefore, the answer is A) 10.
User Ilia Reshetnikov
by
8.4k points
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