Ido not understand why the rt triangel was in ratio of 1:√3:2 in Sue's question
where in triangle abc c isthert angle and sin a √3/2 - find csc b-please explain where the 1 comes from
(We know that in rt triangle
Sin=P/H)
given sinA=_/3/2
SO P=_/3 , H=2
ACB is rt triangle
by pythagoras theorem
H^2=P^2+B^2
B^2=H^2-P^2
B^2=(2)^2-(_/3)^2
B^2=4-3
B^2=1
B=_/1
B=1
(where P=perpendicular,
B=base,H=hypotenuse)
thank you.
In Sue's question, we have a right triangle ABC where angle C is the right angle. We are given that sin(A) = √3/2.
To understand why the ratio of the sides of the right triangle is 1:√3:2, we can use the trigonometric ratios of a right triangle.
In a right triangle, there are three trigonometric ratios: sine (sin), cosine (cos), and tangent (tan). These ratios are defined using the sides of the triangle.
For the given triangle ABC, let's label the sides as follows:
- The side opposite angle A is side a.
- The side opposite angle B is side b.
- The side adjacent to angle A is side c.
Now, let's calculate the trigonometric ratios for the given triangle:
- We are given that sin(A) = √3/2. By definition, sin(A) is equal to the ratio of the side opposite angle A (a) to the hypotenuse (c) of the right triangle. Therefore, we can write: a/c = √3/2.
To simplify this ratio, we can rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of √3/2, which is 2/√3. This gives us: (a/c)*(2/√3) = (√3/2)*(2/√3).
Simplifying further, we get: (2a)/(c√3) = √3/√3 = 1.
So, the ratio of side a to side c is 2a/c = 1.
In other words, in Sue's question, the ratio of the sides of the right triangle is 1:√3:2.