Thus L.H.S = R.H.S that is 2/√3cosx + sinx = sec(Π/6-x) is proved
We have to prove that
2/√3cosx + sinx = sec(Π/6-x)
To prove this we will solve the right-hand side of the equation which is
R.H.S = sec(Π/6-x)
= 1/cos(Π/6-x)
[As secƟ = 1/cosƟ)
= 1/[cos Π/6cosx + sin Π/6sinx]
[As cos (X-Y) = cosXcosY + sinXsinY , which is a trigonometry identity where X = Π/6 and Y = x]
= 1/[√3/2cosx + 1/2sinx]
= 1/(√3cosx + sinx]/2
= 2/√3cosx + sinx
R.H.S = L.H.S
Hence 2/√3cosx + sinx = sec(Π/6-x) is proved
Learn more about trigonometry here : brainly.com/question/7331447
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Hello,
Using Thalès 's theorem:
Let's h the height of the monument.
If you don't know what is the Thalès's theorem,
imagine that the similar triangles.
Answer:
x = 3.85
Step-by-step explanation:
Given equation:
6ˣ = 1,000
now,
on taking log both sides, we get
⇒ log(6ˣ) = log(1,000)
or
⇒ log(6ˣ) = log(10³)
now we know the property of log function that
log(aᵇ) = b × log(a)
thus, applying the above property, we get
⇒ x × log(6) = 3log(10)
or
⇒ x × log(2×3) = 3log(10)
now,
we have another property of log function as":
log(A) = log(A) + log(B)
therefore,
x × [log(2) + log(3)] = 3log(10)
also,
log(10) = 1
log(2) = 0.3010
log(3) = 0.4771
Thus,
⇒ x × [0.3010 + 0.4771 ] = 3 × 1
or
⇒ x × 0.7781 = 3
or
⇒ x = 3.85
