What is the length of leg s of the triangle below? 90° 45° E 72 45°

Mathematics

1 answer:

Julli [10]3 years ago

8 0

Answer:

s = 6

Step-by-step explanation:

It's equal to the other leg, since it's an equilateral triangle, where two angles are 45°

Answered by GAUTHMATH

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the 2nd term in a geometric sequence is 20 the 4th term in the same sequence is 45/4 or 11.25 what is the common ratio?

lbvjy [14]

Let x be the common ratio and y be the 3rd term

11.25/x=y This means the fourth term divided by the common ratio will get the third term
20x=y This means the second term multiplied by the common ratio will get the third term

11.25/x=20x Set the two y's equal to each other, and solve for x!
20x^2=11.25
x^2=.5625
x=+-√.5625
x=.75 or -.75

5 0

3 years ago

Read 2 more answers

A flagpole stands perpendicular to the ground. It is supported by a 20-foot cable that runs from the top of the flagpole to the

3241004551 [841]

Answer:

h = 16 feet

Step-by-step explanation:

Pythagoras´ theorem establishes, that in a right triangle square of the hypothenuse (c²) is the sum of leg ( b²) plus leg (a²)

Then

c² = a² + b²

where according to the problem statement

(20)² = h² + (12)² ( h is the height of the flagpole )

h² = 400 - 144

h² = 256

h = √ 256

h = 16 feet

7 0

3 years ago

ozzi

Answer:

90 km, N 46° E

Step-by-step explanation:

A jet flies due North for a distance of 50 km and then on a bearing of N 70° E for a further 60 km. Find the distance and bearing of the jet from its starting point.

Look at the diagram I drew of this scenario. You can see the jet flies North for 50 km, and then turns at a 70° angle to fly another 60 km. We want to find the distance from the starting point, SP, to angle C (labeled).

This will be the jet's distance from its starting point.

In order to find the bearing of the jet from its starting point, we will need to find the angle formed between distances b and c, labeled angle A.

The Law of Cosines will allow us to use two known sides and one known angle to solve for the sides opposite of the known angle.

In this case, the known angle is 110° (angle B) so we will use the Law of Cosines respective to B.

b² = a² + c² - 2ac cosB

Substitute the known values into the equation and solve for b, the distance from the starting point (A) to the endpoint (C).

b² = (60)² + (50)² - 2(60)(50) cos(110°)
b² = 6100 -(-2052.12086)
b² = 8152.12086
b = 90.28909602
b ≈ 90 km

The distance of the jet from its starting point is 90 km. Now we can use this b value in order to calculate angle A, the bearing of the jet.

The Law of Cosines with respect to A:

a² = b² + c² - 2bc cosA

Substitute the known values into the equation and solve for A, the bearing from the starting point (clockwise of North).

(60)² = (90.28909602)² + (50)² - 2(90.28909602)(50) cosA
3600 = 8152.12086 - 6528.909602 cosA
-4552.12086 = -6528.909602 cosA
0.6972252853 = cosA
A = cos⁻¹(0.6972252853)
A = 45.79519
A ≈ 46°

The bearing of the jet from its starting point is N 46° E. This means that it is facing northeast at an angle of 46° clockwise from the North.