Inverse Trigonometric Functions: Real-World Scenarios (6)

hard • 3 min read
Tags Inverse Trigonometric Functions

Question

A surveyor stands at a horizontal distance of 50 m50 \, \text{m} from the base of a tower. The angle of elevation to the top of the tower is given by sin⁡−1(3/5)+cos⁡−1(4/5)\sin^{-1}(3/5) + \cos^{-1}(4/5). Find this combined angle and the height of the tower.

Solution — Step by Step

For any xx in [−1,1][-1, 1]:

sin⁡−1(x)+cos⁡−1(x)=π2\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}

But here the arguments are different: 3/53/5 and 4/54/5. So we cannot use the identity directly.

Let α=sin⁡−1(3/5)\alpha = \sin^{-1}(3/5) and β=cos⁡−1(4/5)\beta = \cos^{-1}(4/5).

Then sin⁡α=3/5\sin\alpha = 3/5, so cos⁡α=4/5\cos\alpha = 4/5 (taking principal value, α\alpha in first quadrant).

And cos⁡β=4/5\cos\beta = 4/5, so sin⁡β=3/5\sin\beta = 3/5.

So α=β\alpha = \beta! Both angles are the same — the angle whose sine is 3/53/5 and cosine is 4/54/5.

α+β=2α\alpha + \beta = 2\alpha, where sin⁡α=3/5\sin\alpha = 3/5 and cos⁡α=4/5\cos\alpha = 4/5.

sin⁡(2α)=2sin⁡αcos⁡α=2⋅(3/5)(4/5)=24/25\sin(2\alpha) = 2\sin\alpha\cos\alpha = 2 \cdot (3/5)(4/5) = 24/25.

cos⁡(2α)=cos⁡2α−sin⁡2α=16/25−9/25=7/25\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = 16/25 - 9/25 = 7/25.

So combined angle θ=2α\theta = 2\alpha, with tan⁡θ=(24/25)/(7/25)=24/7\tan\theta = (24/25)/(7/25) = 24/7.

tan⁡θ=h/50\tan\theta = h/50, so h=50tan⁡θ=50⋅(24/7)=1200/7≈171.4 mh = 50 \tan\theta = 50 \cdot (24/7) = 1200/7 \approx 171.4 \, \text{m}.

Combined angle: θ=tan⁡−1(24/7)\theta = \tan^{-1}(24/7). Tower height ≈171.4 m\approx 171.4 \, \text{m}.

Why This Works

The trick is recognising that "sin⁡−1(3/5)\sin^{-1}(3/5)" and "cos⁡−1(4/5)\cos^{-1}(4/5)" both name the same angle (in the first quadrant). Reducing a confusing combination to a familiar quantity (2α2\alpha) makes the problem trivial.

The 3-4-5 right triangle is one of the most common “free” tools in inverse trig — whenever we see 3/53/5, 4/54/5, 24/2524/25, or 7/257/25, the underlying angle is built from this triangle.

Pattern hint: sin⁡−1(a/c)+cos⁡−1(b/c)\sin^{-1}(a/c) + \cos^{-1}(b/c) where a2+b2=c2a^2 + b^2 = c^2 → both terms name the same first-quadrant angle. Add them, get 2α2\alpha, use double-angle formulas.

Alternative Method — tan⁡−1\tan^{-1} Conversion

Convert each term to tan⁡−1\tan^{-1}:

  • sin⁡−1(3/5)=tan⁡−1(3/4)\sin^{-1}(3/5) = \tan^{-1}(3/4) (from 3-4-5 triangle).
  • cos⁡−1(4/5)=tan⁡−1(3/4)\cos^{-1}(4/5) = \tan^{-1}(3/4) (same triangle).

Sum: 2tan⁡−1(3/4)2\tan^{-1}(3/4).

Use tan⁡(2A)=2tan⁡A/(1−tan⁡2A)=2(3/4)/(1−9/16)=(3/2)/(7/16)=24/7\tan(2A) = 2\tan A/(1 - \tan^2 A) = 2(3/4)/(1 - 9/16) = (3/2)/(7/16) = 24/7. Same result.

Common Mistake

Students often add the arguments instead of the angles: sin⁡−1(3/5)+cos⁡−1(4/5)≠sin⁡−1(3/5+4/5)\sin^{-1}(3/5) + \cos^{-1}(4/5) \ne \sin^{-1}(3/5 + 4/5). Inverse trig functions are not linear in their arguments.

Another classic: forgetting that sin⁡−1(x)+cos⁡−1(x)=π/2\sin^{-1}(x) + \cos^{-1}(x) = \pi/2 ONLY when both have the same argument. Different arguments require the conversion approach.

JEE Main 2024 used sin⁡−1(5/13)+cos⁡−1(12/13)\sin^{-1}(5/13) + \cos^{-1}(12/13) — the 5-12-13 triangle. Same template. CBSE Class 12 boards regularly include 4-mark inverse trig problems built around 3-4-5 or 5-12-13.

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