Answer:
![cos\theta=0.6](https://img.qammunity.org/2019/formulas/physics/middle-school/zesme8f7yade7euhpax5lrkjh1qwqil9by.png)
Step-by-step explanation:
In the given triangle, perpendicular distance is 12 units and the hypotenuse of the triangle is 15 units. Let b is the base of the triangle. It can be calculated using the Pythagoras theorem as :
![H^2=P^2+B^2](https://img.qammunity.org/2019/formulas/physics/middle-school/ys18qsn4kcyngj0urxgajgy825wpykn0wm.png)
![B=√(H^2-P^2)](https://img.qammunity.org/2019/formulas/physics/middle-school/5mn673t526qndt73ib7o79hx5v7fuc5oak.png)
![B=√(15^2-12^2)](https://img.qammunity.org/2019/formulas/physics/middle-school/2a2oyu2np26h8dbklupl2bu5cbrckkmyvz.png)
B = 9 units
Now we need to find the value of
. Using trigonometric identities as :
![cos\theta=(B)/(H)](https://img.qammunity.org/2019/formulas/physics/middle-school/oga9835n1laz83ce6a39c37wxujtfclxeo.png)
![cos\theta=(9)/(15)](https://img.qammunity.org/2019/formulas/physics/middle-school/1e12k7ozgsdv4s96rzmyh3n5das1dk0ils.png)
![cos\theta=0.6](https://img.qammunity.org/2019/formulas/physics/middle-school/zesme8f7yade7euhpax5lrkjh1qwqil9by.png)
So, the value of
is 0.6. Hence, this is the required solution.