Apply the Law of Cosines to the triangle whose vertices are the golfer, the hole, and the ball.
B
b
Apply the Law of Cosines to the triangle whose one angle measure is equal to θ.
A
a
About 163.4 yards
B
b
About 3.5^(∘).
Practice makes perfect
a
We are asked to determine the distance x from the golfer's ball to the hole. To do so, let's first consider the given diagram.
We can see that x is the length of one of the sides of the triangle whose vertices are the golfer, the hole, and the ball. We are given two side length of this triangle and their included angle. Therefore, we will use the Law of Cosines to find x.
Law of Cosines
If â–³ ABC has sides of length a, b, and c, then the following are true. a^2=b^2+c^2-2bccosA b^2=a^2+c^2-2accosB c^2=a^2+b^2-2abcosC
Let's apply the Law of Cosines to our triangle.
x^2= 260^2+ 400^2-2( 260)( 400)cos15^(∘)
Now we will solve the above equation for the distance x from the golfer's ball to the hole.
The distance from the golfer's ball to the hole is about 163.4 yards.
b
Assume the golfer is able to hit the ball precisely the distance found in Part A. Recall that this distance is about 163.4 yards. We are asked to find the maximum angle θ by which the ball can be off target in order to land no more than 10 yards from the hole.
To find the value of θ, we assume that the ball is exactly 10 yards off target. We can see that θ is an angle of a triangle, and we know all side lengths of the triangle. Again, we will use the Law of Cosines to find θ. Let's apply the law to our triangle.
10^2 = 163.4^2 + 163.4^2 - 2( 163.4 )( 163.4 )cosθ
Let's solve the above equation for θ.
Now we will use the inverse cosine ratio.
cosθ=53 299.12/53 399.12 [0.7em]
⇕ [0.6em]
θ=cos^(- 1)(53 299.12/53 399.12)
Finally, let's estimate the value of θ using a calculator.
θ=cos^(- 1)(53 299.12/53 399.12)≈ 3.5^(∘)
The maximum angle by which the ball can be off target in order to land no more than 10 yards from the hole is about 3.5^(∘).
