1)In triangle ABC, C=60 degrees, a=12, and b=5. Find c.
A)109.0
B)10.4
C)11.8
D)15.1
I chose B
2)Which triangle should be solved by beginning with the Law of Cosines?
A)A=115 degrees, a=19, b=13
B)A=62 degrees, B=15 degrees, b=10
C)B=48 degrees, a=22, b=5
D)A=50 degrees, b=20, c=18
I chose A
3)P(-9/40,40/41) is located on the unit circle. Find sin theta.
A)40/41
B)-9/41
C)-9/40
D)-40/9
I chose A
4)Find the exact value of cos(-420)
A)-1/2
B)1/2
C)square root of 3 over 2
D)square root of -3 over 2
I chose B
5)Write the equation sin y= x in the form of an inverse function.
A)y=Sin-1x
B)x=Sin-1y
C)y=Sin-1x
D)y=Sinx
I chose A
2,3 are wrong.
On 5, what is the difference between a, c?
the only diff is c has sin-1x and a has Sin-1x
how is 2 not A? my book says to solve an oblique triangle: two angles and any side, and two sides and an angle opposite one of them is solved by Law of Sines. Two angles and their included angle or three sides are solved by Law of Cosines.
<Two angles> and their included angle or three sides? I think that should be <two sides>, then angle C is the one included by sides a and b. That's why answer A is wrong.
help with laws of sines
1) To find c in triangle ABC, we can use the Law of Cosines. The formula is: c^2 = a^2 + b^2 - 2ab*cos(C), where a, b, and C are given. Plugging in the values, we get: c^2 = 12^2 + 5^2 - 2(12)(5)*cos(60). Simplifying this equation, c^2 = 144 + 25 - 120*cos(60). On evaluating cos(60) which is equal to 0.5, we get c^2 = 169 - 120(0.5), c^2 = 169 - 60, c^2 = 109. Now to find c, we take the square root of 109, which is approximately 10.4. Thus, the answer is B) 10.4.
2) The Law of Cosines should be used to solve triangles where we are given two sides and the included angle. Looking at the given options, A) A=115 degrees, a=19, b=13 fits this criteria. To solve such a triangle, we use the formula: c^2 = a^2 + b^2 - 2ab*cos(C), where C is the included angle. So, in this case, we use c^2 = 19^2 + 13^2 - 2(19)(13)*cos(115). Evaluating cos(115), we get c^2 = 361 + 169 - 494cos(115). Therefore, we solve for c by taking the square root of the value obtained which is approximately 21.59. Thus, the answer is A) A=115 degrees, a=19, b=13.
3) To find sin theta for point P(-9/40, 40/41) on the unit circle, we can use the Pythagorean identity sin^2 theta + cos^2 theta = 1. Since P is on the unit circle, the distance from the origin to P is 1. Therefore, we can write (-9/40)^2 + (40/41)^2 = 1. Simplifying this equation, we get 81/1600 + 1600/1681 = 1. Adding these fractions results in (81 + 1600)/1600 = 1681/1600 = 1. Thus, sin theta is equal to the square root of the fraction 40/41. So the answer is A) 40/41.
4) To find the exact value of cos(-420 degrees), we can use the unit circle. We know that cos is the x-coordinate on the unit circle, and -420 degrees is equivalent to -420 + 360 = -60 degrees. On the unit circle, -60 degrees is in the fourth quadrant. In the fourth quadrant, cos is positive. Looking at the options, the only positive value for cos is B) 1/2. Thus, the answer is B) 1/2.
5) To write the equation sin y = x in the form of an inverse function, we can solve for y. We take the inverse sine of both sides of the equation. Applying the inverse sine function, we have sin^(-1)(sin y) = sin^(-1)(x). This simplifies to y = sin^(-1)(x). Therefore, the answer is A) y = sin^(-1)(x).