Trigonometry · real student question

Solve for x: sqrt(x^2 + 26.8^2 + 100^2) = x / cos(20 degrees).

Question

Solve for xx:

x2+26.82+1002=xcos20\sqrt{x^2+26.8^2+100^2}=\frac{x}{\cos 20^\circ}

Step-by-step solution

  1. Note the domain before squaring. The left side is a square root, so it is non-negative; the right side is x/cos20x/\cos 20^\circ with cos20>0\cos 20^\circ>0. Hence any solution must satisfy x0x\ge 0, and a negative root produced later would have to be discarded as extraneous.

  2. Square both sides and collect the constant. Squaring is legitimate here because both sides are non-negative:

    x2+26.82+1002=x2cos220    26.82+1002=x2(1cos2201)x^2+26.8^2+100^2=\frac{x^2}{\cos^2 20^\circ}\;\Longrightarrow\;26.8^2+100^2=x^2\left(\frac{1}{\cos^2 20^\circ}-1\right)

    The constant is K=26.82+10000=10718.24K=26.8^2+10000=10718.24.

  3. Use the Pythagorean identity to simplify the bracket. Since sec2θ1=tan2θ\sec^2\theta-1=\tan^2\theta:

    1cos2201=tan220\frac{1}{\cos^2 20^\circ}-1=\tan^2 20^\circ

    so the equation becomes x2tan220=10718.24x^2\tan^2 20^\circ=10718.24. Recognising this identity is what turns a messy trigonometric equation into a one-line solve.

  4. Solve for xx, keeping only the positive root.

    x=10718.24tan20=103.52890.3639702284.44x=\frac{\sqrt{10718.24}}{\tan 20^\circ}=\frac{103.5289}{0.3639702}\approx 284.44

    The negative root x-x satisfies the squared equation but not the original one, since it would make the right-hand side negative while the left side stays positive.

  5. Verify by substituting back. With x284.44x\approx 284.44 the left side is x2+10718.24\sqrt{x^2+10718.24} and the right side is x/cos20x/\cos 20^\circ; both evaluate to the same number to seven significant figures, confirming the root and ruling out an extraneous solution.

Answer

x=10718.24tan20284.44x=\frac{\sqrt{10718.24}}{\tan 20^\circ}\approx 284.44

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