Question

A Cat, sitting in the top of a tree, spots a dog and a Firefighter both on the flat ground below. From the cats point of view , the dog is 10 meters south from the base of the tree, at an angle of depression of 65 degrees, and the Firefighter is some distance east of the tree at an angle of depression of 50 degrees. How far is the Firefighter from the dog?

Answer 1

This is the concept of application of trigonometry;
The height of the tree will be given by:
tan theta=opposite/adjacent
theta=65
opposite=h
adjacent= 10
thus;
tan 65=h/10
h=10 tan 65
h=21.45 m

The distance between the dog and the man is x;
the distance of the man to the tree is (x+10) m
this will be given by:
tan 50= 21.45/(10+x)
getting the reciprocal of the expression we get;
1/tan 50=(10+x)/21.45
cross multiplying the above expression we get;
21.45=(10+x)tan 50
21.45=(10+x)1.2
21.45=12+1.2x
1.2x=21.45-12
1.2x=9.45
x=9.45/1.2
x=7.875
We conclude that the distance of the Firefighter and the dog is x=7.875 m

Answer 2

Answer:

15 meters

Step-by-step explanation:

Denote points: C - cat, B - base of the tree, D - dog, F - Firefighter.

Consider triangle CBD. This triangle is right triangle with right angle CBD and m∠BCD=65°, DB=10 m. Then

$\dfrac{BC}{BD}=\tan 65^{\circ},\\ \\BC=BD\cdot \tan 65^{\circ}\approx 10\cdot 2.14=21.4\ m.$

Consider right triangle CBF. In this triangle angle CBF is right angle and m∠CFB=50°, then

$\dfrac{BC}{BF}=\tan 50^{\circ},\\ \\BF=\dfrac{BC}{\tan 50^{\circ}}\approx \dfrac{21.4}{1.91}\approx 11.2\ m.$

Consider right triangle BDF. In this triangle angle FBD is right and by the Pythagorean theorem,

$DF^2=BF^2+BD^2,\\ \\DF^2=(11.2)^2+10^2,\\ \\DF^2=225.44,\\ \\DF\approx 15\ m.$