Stay onside

Let $t$ be the time it takes for the pass to occur. In this time, the receiver runs a distance $7t$ , in S.I. units. The component of the velocity of the ball that is parallel to the trajectories of the runners is $10 \cos \theta$ , where $\theta$ is the desired angle.

Now, the two horizontal distances run the ball and the receiver must be the same, so: $10t \cos \theta = 7t \implies \cos \theta = \dfrac {7}{10}.$ Thus, taking the arccosine, we have $\theta = \boxed {45.57^\circ}$ .

Let the ball reach the other player $P$ in time $t$ . In this time, $P$ would travel the distance $7t$ ; the ball would travel $5 \text{ m}$ perpendicular to $P\text{'s}$ trajectory and $7t$ along $P\text{'s}$ trajectory. Therefore, the distance $l$ that the ball travels can be expressed in $t$ by Pythagoras from a right triangle to be $\sqrt{25 + 49t^2}$ .

Since we know the velocity $u$ of the ball, we can determine $t$ from $u = \frac{l}{t}$ as $10 = \frac{ \sqrt{25 + 49t^2}}{t}$ . Solving this for positive $t$ values gives $t = \sqrt{\frac{25}{51}}$ .

Now we can find the angle from the same right triangle we expressed $l$ from, since $\tan{\left(\frac{\pi}{2} - \alpha\right)} = \frac{7t}{5} = \frac{7\cdot \sqrt{\frac{25}{51}}}{5} = 0.9802$ . Hence, $\alpha = \frac{\pi}{2} - \arctan{0.9802} = \boxed{45.57^{\circ}}$

Since the ball and the teamate both move at a constant rate, you know that the distance each moves will be the same ratio of 7/10. The angle you want the complement of sin(7/10), which is approximately 45.57.

The ball and the player both have got to reach a certain point after a certain time $t$ . Let us call this point $O$ and the points of where the player with the ball and his teammate are with $P_{1}$ and $P_{2}$ accordingly.

Now, let us form a triangle with sides $\overline{P_{1}P_{2}}=5,\overline{P_{1}O}=c$ and $\overline{P_{2}O}=b$ ( $\overline {P_{1}P_{2}}$ is actually the distance between the two players, which we know). This triangle is also a right one, with the right angle being $\angle P_{1}P_{2}O$ .

Now, we know for a fact that because $v=\frac{s}{t}$ , where $v$ is the velocity, $s$ is the length traveled and $t$ is the time, the two missing sides of the triangle can be expressed as

$b=7t$ and $c=10t$

The angle we are actually looking for is $\angle P_{2}OP_{1}$ , which we will call $\angle \alpha$ for better illustrative purposes. From the given triangle, we know that

$\cos\alpha =\frac{b}{c} \Rightarrow \cos\alpha=\frac{7}{10}$

From this point on, we simply find $\arccos \frac{7}{10}$ , which is $45,568$ .

Thus, the answer is $45,57$ .

Let * \­( x \­) * be the displacement of the ball respect the trajectory of the players. Let $\alpha$ be the angle we want to know. We know the ball's speed is 10 m/s. Part of that speed must go move 5 m perpendicularly to their trajectory. That will be \­(10 \times \sin \alpha \­). By the other hand, the speed parallel to their trajectory will be the same speed as the players, 7 m/s. So \­(7 = \10 \times \cos \alpha \­). We can solve the angle now. Note that \­(10 \times \sin \alpha \­) is not necessary. Finally, we have \­(\frac{7}{10} = \cos \alpha \­), so \­(alpha = 45.573 \­)

It was a guess lol

consider two players. velocity of both the players are 7m/s. velocity of ball when he passes the ball = 10 m/s. now applying trigonometry ; sin (x) = 7/10. which gives x = 44.427004. bt according to ques wee should subtract our ans from 90 because we want angle with respect to their trajectory. therefore ans is 90- x i.e 45.572996

Let $t$ be the amount of time it takes for the soccer ball to transfer between the two players. In that time, the ball travels $10t \; \text{m}$ and the receiving player travels $7t \; \text{m}.$ The angle $\theta$ between the trajectory of the defender and the trajectory of the ball therefore has cosine $7/10$ (draw the diagram to see this.) Thus, we have $\theta = \arccos{(7/10)} \approx \boxed{45.6^{\circ}}.$