Joan is playing golf - Leaving Cert Mathematics - Question 9 - 2015

Using the Law of Sines in triangle ATX:

1. Calculate |TH| = 385 m, |AT| = 190 m, and angle |∠ATH| = 18°.

2. Using sine law:

$\frac{|AH|}{\sin(18°)} = \frac{190}{\sin(A)}$

Calculating |AH| gives:

$|AH| = 212.57 m \Rightarrow |AH| \approx 213 m$

Set h = -6r² + 22t + 8. To find height at time t = 0:

$h(0) = -6(0)² + 22(0) + 8 = 8 m$

Thus, the height of K above OB is 8 m.

To find the angle of elevation |∠OBK|:

1. When the ball is at B, we have the height at K = 8 m and distance from B to O as 152 m.

2. Therefore, using tangent:

$\tan(\theta) = \frac{8}{152} \Rightarrow \theta = \arctan(\frac{8}{152}) \approx 3.01°$

Thus, the angle of elevation from B to K is approximately 3°.

|CD| can be expressed as:

$|CD| = 25$ (constant height of tree)

And for d, using trigonometric relations:

$d = 2h$

To find h, we can write:

$d² + |CD|² = 25² \Rightarrow (2h)² + 25² = 25² \Rightarrow 4h² + 625 - 625 = 0$

From which:

$4h² = 0 \Rightarrow h = 10 m$. Therefore, h is 10 m.