Trigonometry · real student question

Solve for x: sqrt(x^2 + 125.4^2 + 100^2) = x / cos(20 degrees).

Question

Solve for xx:

x2+125.42+1002=xcos20\sqrt{x^2+125.4^2+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+125.42+1002=x2cos220    125.42+1002=x2(1cos2201)x^2+125.4^2+100^2=\frac{x^2}{\cos^2 20^\circ}\;\Longrightarrow\;125.4^2+100^2=x^2\left(\frac{1}{\cos^2 20^\circ}-1\right)

    The constant is K=125.42+10000=25725.16K=125.4^2+10000=25725.16.

  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=25725.16x^2\tan^2 20^\circ=25725.16. 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=25725.16tan20=160.39070.3639702440.67x=\frac{\sqrt{25725.16}}{\tan 20^\circ}=\frac{160.3907}{0.3639702}\approx 440.67

    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 x440.67x\approx 440.67 the left side is 440.672+25725.16\sqrt{440.67^2+25725.16} and the right side is 440.67/cos20440.67/\cos 20^\circ; both evaluate to the same value to seven significant figures, confirming the root.

    x2+25725.16=xcos20  \sqrt{x^2+25725.16}=\frac{x}{\cos 20^\circ}\;\checkmark

Answer

x=25725.16tan20440.67x=\frac{\sqrt{25725.16}}{\tan 20^\circ}\approx 440.67

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