Trigonometry help? (Word Problem) ?

Given:

27km=distance from Toronto Airport to Detroit Airpot. Let's call this a

18 km = distance from Toronto Airport to destination of 2nd plane. Let's call this b.

65 deg = angle between the courses of the two planes.

Find: distance between the two planes. Let's call this c.

Solution: If you draw a sketch of these distances you will form a triangle. The two sides, a and b, intersect at a point representing Toronto Airport while side c, connecting the endpoints of a and b, represents the distance between the two planes whey they reached their destinations.

Since we want c, we use cosine law to solve for it.

c^2 = a^2 + b^2 - 2abcosC, where C = 65 deg

c = sqrt[a^2 + b^2 - 2abcosC]

c = sqrt[27^2 + 18^2 - 2(27)(18)cos65deg]

c = 25.3 km answer

Hope this helps.

teddy boy

Let airports be T , D and X

In ∆TDX , TD = 27 , TX = 18 and /_T = 65°

Using Cosine rule :-

t ² = 27² + 18² - (2 + 27 x 18) cos 65°

t = 25.3 km

Planes are 25.3 km apart.

From the ship, the angle of elevation of a point B at the top of a cliff is 21 degrees 13 minutes. After the ship was sailed 2500 feet directly toward B, its angle of elevation is found to be 47 degrees 17 minutes. find the height of the cliff

Here you've got to use the Trigonometric ratio Tan Tan 30 = Height at which the kite is flying / 50M Height of the kite is Tan 30 x 50 M = 28.86 M

law of cosines -> you have 2 legs of the triangle, you can calculate the third.

c² = a² + b² - 2abcosC

PS there are lots of anomalies in that question, but I'm biting my tongue.

they are sqrt{18^2 + 27^2 -- 2*18*27cos(65)} km apart = 25.34 km

just draw out a triangle with all of the numbers in it, and you'll see

25km