The sun subtend an angle at 35degree from the centre of the earth whose distant from the centre of the earth is 382,100km. Find
the diameter of the sun
2 answers:
Examine the illustration provided below:
Solution:
240951.33km
Detailed Explanation:
To tackle this type of problem, the first step is to recreate the diagram as depicted in the accompanying file (Figure 1).
Upon observing the diagram, it becomes apparent that triangle ABC is established by bisecting the 35-degree angle (the bisection of this angle divides it into two equal segments, resulting in two angles of 17.5 degrees each).
Given that triangle ABC is a right triangle, we can employ trigonometric ratios to ascertain the sun's diameter.
In triangle ABC, BC serves as the perpendicular side (opposite to ), while AC is the base (adjacent side).
=17.5 degrees
AC equals 382100km
Substituting the values yields:
Diameter=d=2*Radius
Since BC represents the sun's radius from its center, we derive the diameter by multiplying it by 2.
The diameter of the sun is d=2*120475.66km
You might be interested in
Given:
A quadratic function has a line of symmetry positioned at x = –3.5 with one root located at –9.
To find:
The second root.
Solution:
It is understood that the line of symmetry splits the quadratic function's graph into two identical halves. Hence, both roots are equidistant from this line.
This implies that the line of symmetry passes through the midpoint of the two roots.
Let the other root be denoted as x.
Multiply both sides by 2.
Add 9 to both sides.
Consequently, the other zero of the quadratic function is concluded to be 2.
Solution:
We know, g(x) = f(x) + k --------(1)
It is given that f(x) = (x+2)
and g(x) = (x+5)
Substituting f(x) and g(x) into equation (1):
→ (x+5) =
(x+2) + k
→ = k
→ k =
Thus, the value of k is 1.
The solutions to the equation are
Explanation:
Given the equation
, we aim to find the solutions of the equation.
Let us use and
Thus, the equation transforms into
Factoring the equation results in:
Inserting back and allows us to solve for x.
Initially, we will substitute
Consequently, we find;
Similarly, by substituting , we derive;
Thus, the solutions to the equation are