IIT-JEE 1980Question 1
(i) $PQ$ is a vertical tower. $P$ is the foot and $Q$ is the top of the tower. $A,B,C$ are three points in the horizontal plane through $P$. The angles of elevation of $Q$ from $A,B,C$ are equal, and each is equal to $\theta$. The sides of the triangle $ABC$ are $a,b,c$; and the area of the triangle $ABC$ is $\Delta$. Show that the height of the tower is $\dfrac{abc\tan\theta}{4\Delta}$.
(ii) $AB$ is a vertical pole. The end $A$ is on the level ground. $C$ is the middle point of $AB$. $P$ is a point on the level ground. The portion $CB$ subtends an angle $\beta$ at $P$. If $AP=n\,AB$, then show that $\tan\beta=\dfrac{n}{2n^{2}+1}$.
Trigonometric Ratio & Identities
Answerproved
Solution
Concept. (i) Equal elevations put the foot at equal distances from $A,B,C$: it is the circumcentre. (ii) $\beta$ is the difference of the angles subtended by $AB$ and $AC$. Formulas used. (1) $R=\dfrac{abc}{4\Delta}$(2) $\tan(X-Y)=\dfrac{\tan X-\tan Y}{1+\tan X\tan Y}$Part (i). Each of $PA,PB,PC$ equals $h\cot\theta$, so $P$ is the circumcentre and $h\cot\theta=R$. By (1), $$h=\frac{abc\tan\theta}{4\Delta}.$$ Part (ii). Let $AB=1$, so $AC=\frac12$ and $AP=n$. The angles at $P$ subtended by $AB$ and $AC$ have tangents $\frac1n$ and $\frac{1}{2n}$. By (2), $$\tan\beta=\frac{\frac1n-\frac{1}{2n}}{1+\frac{1}{2n^2}}=\frac{\frac{1}{2n}}{\frac{2n^2+1}{2n^2}}=\frac{n}{2n^2+1}.$$ Hence proved.