Answer:
a) a = 7.37, b = 15.13, C = 67°
b) 1 triangle
Step-by-step explanation:
<h3>a)</h3>
Two angles and one side are given. That means the triangle is uniquely determined, and the remaining sides can be found from the Law of Sines.
The third angle is ...
C = 180° -A -B = 180° -29° -84° = 67°
Then the Law of Sines tells you ...
a/sin(A) = b/sin(B) = c/sin(C)
a = sin(A)/sin(C)·c = sin(29°)/sin(67°)·14 ≈ 7.37
b = sin(B)/sin(C)·c = sin(84°)/sin(67°)·14 ≈ 15.13
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<h3>b)</h3>
Ordinarily, when the given angle (B = 30°) is opposite the shorter of the given sides (b = 10 < a = 20), it means there are two possible solutions to the triangle.
However, when the sine of the angle is exactly equal to the ratio of the given sides: sin(30°) = 10/20 = 1/2, the larger angle can only be 90°. That is, the one triangle that can be formed is a right triangle.
The Law of Sines tells you this.
sin(A)/a = sin(B)/b
sin(A) = (a/b)sin(B) . . . . . . multiply by 'a'
A = arcsin(a/b·sin(B)) = arcsin(20/10·sin(π/6)) = arcsin(1)
A = 90°
There are two solutions for this problem. It depends if order matters or not. As asked in the problem, we have to show both solutions.
The first solution is for the condition that order does not matter. This is a combination problem:
nCr = n!/r!(n-r)!
where n=10 because there are 10 total students, and r=2 because you choose 2 students at a time
nCr = 10!/2!(10-2)! = 45 ways
The second solution is for the condition that order matters. This is a permutation problem.
nPr = n!/(n-r)! = 10!/(10-2)! = 90 ways
It’s D, when lines are perpendicular, the slopes of the lines are opposite reciprocals.
Answer:
-1.5
Step-by-step explanation:
8÷12 is 1.5 and 4÷12 is 3 so 1.5-3 is -1.5