1. 20 squared + 48 squared = 2704
square root of 2704 is 52. Answer: 52 inches.
2. -18 squared + 82 squared = 6400
square root of 6400 is 80 Answer: 80 inches
We will use trigonometric identities to solve this. I will use θ (theta) for the angle.
First of all, we know that cotθ = 1/tanθ. This is a trigonometric identity.
We can replace cotθ in the expression with 1/tanθ.
Simplify: 1/tanθ * tanθ = tanθ/tanθ = 1
So now, we have:
Next, we also know that secθ = 1/cosθ. This is another trigonometric identity.
We can replace secθ with 1/cosθ in our expression.
Simplify:
Our third trigonometric identity that we will use is tanθ = sinθ/cosθ.
We can replace sinθ/cosθ with tanθ.
Now we have as our final answer:
Hope this helps!
Answer:
15 ft
Step-by-step explanation:
12^2 + 9^2 = c^2
144 + 81 = c^2
225 = c^2
c = 15
Answer:
The third side is 5.7 centimeters.
Step-by-step explanation:
The hypothenuse is the largest side of a right triangle. So in this question, we know that the hypothenuse is 7.
One of the sides is 4, and the other is x.
Applying Pythagoras' Theorem:
Since it is a length measure, we only take the positive value
The third side is 5.7 centimeters.
Gideon is painting all sides thus we find the overall area of the square pyramid given in the picture. A square pyramid is a polyhedron that consists of a square and 4 triangles. To solve the area that Gideon will have to paint, we can calculate the area of the square and the area of the triangles then add these two values of area. We proceed as follows:
Area of the square = s^2 where s is the measure of the side
Area of the square = 4^2 = 16 in^2
Area of the triangle= 1/2 bh where b is the base of the triangle and h is height of the triangle
Area of the triangle = 1/2 (4) (5) = 10 in^2
We multiply the area of the triangle to four in order to obtain the total area of the triangles.
Total area of the triangles = 4 x 10 = 40 in^2
We then add the two areas,
Total Area = 16 + 40 = 56 in^2
Thus the total area that Gideon will have to paint for one square pyramid is 56 in^2.
