Multiple choice

$PQ$ is a vertical tower. $A,\ B,\ C$ are three points in a horizontal line through $Q$, the foot of the tower. If the angles of elevation of the top of the tower from $A,\ B,\ C$ are $\alpha,\ \beta,\ \gamma$ respectively, then $BC\cot\alpha-CA\cot\beta+AB \cot$ $\gamma=$

  1. $3$
  2. $1$
  3. $2$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let tower height be h. Then QA = h cot(alpha), QB = h cot(beta), QC = h cot(gamma). The expression is BC cot(alpha) - CA cot(beta) + AB cot(gamma). Substituting distances: (QC-QB) cot(alpha) - (QC-QA) cot(beta) + (QB-QA) cot(gamma). This simplifies to 0.

AI explanation

Let the height of the tower be h. The distances are AB = h(cot alpha - cot beta) and BC = h(cot beta - cot gamma). Substituting these into the expression gives h(cot beta - cot gamma)cot alpha - h(cot alpha - cot gamma)cot beta + h(cot alpha - cot beta)cot gamma. Expanding this product results in h(cot alpha cot beta - cot alpha cot gamma - cot alpha cot beta + cot beta cot gamma + cot alpha cot gamma - cot beta cot gamma). All terms cancel out, making the total result 0.

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