Question

On a road trip you and your family stop a truck stop to take a break and to let your puppy Fido stretch. You notice that they have a triangular dog park that is fenced in on two sides. The third side of the field is formed by a creek. If the fences measure 150 feet and 98 feet, and the side along the creek is 172 feet, what are the measures of the angles made by the dog park?

Answer 1

cosine theorem can be used to calculate any of these three angles
 cosα=(150^2+172^2-98^2)/2*150*172

Answer 2

Answer:

Angles are 34.59°, 85.08° and 60.33°

Step-by-step explanation:

Let ABC is a triangle, ( that show the dog park)

In which,

AB = 150 feet

BC = 98 feet

CA = 172 feet,

By the cosine law,

$BC^2=AB^2+AC^2-2(AB)(AC)cos A$

$2(AB)(AC)cos A=AB^2+AC^2-BC^2$

$\implies cos A=\frac{AB^2+AC^2-BC^2}{2(AB)(AC)}-----(1)$

Similarly,

$\implies cos B=\frac{AB^2+BC^2-AC^2}{2(AB)(BC)}-----(2)$

$\implies cos C=\frac{BC^2+AC^2-AB^2}{2(BC)(AC)}-----(3)$

By substituting the values in equation (1),

$cos A=\frac{150^2+172^2-98^2}{2\times 150\times 172}$

$=\frac{22500+29584-9604}{51600}$

$=\frac{42480}{51600}$

$\approx 0.8233$

$\implies m\angle A\approx 34.59^{\circ}$

Similarly,

From equation (2) and (3),

m∠B ≈ 85.0°, m∠C ≈ 60.33°