Aspire Faculty ID #19144 · Topic: NIMCET 2026 · Just now
NIMCET 2026

An engineer standing at a point $P$ wishes to determine the width of a new rectangular pond. She finds the distance to the western-most point $A$ of the pond from $P$ to be $60$ m, while the distance to the northern-most point $B$ of the pond from $P$ is $80$ m. If the angle between the two lines of sight at $P$ is $60^\circ$, then the width $AB$ in metres of the pond is, where $AB$ is not parallel to line of North-South:

Solution

Given,

$PA=60$

$PB=80$

$\angle APB=60^\circ$

Using cosine rule in triangle $APB$,

$AB^2=PA^2+PB^2-2(PA)(PB)\cos 60^\circ$

$AB^2=60^2+80^2-2(60)(80)\cdot \frac{1}{2}$

$AB^2=3600+6400-4800$

$AB^2=5200$

$AB=\sqrt{5200}$

$AB=\sqrt{400\cdot 13}$

$AB=20\sqrt{13}$

Previous 10 Questions — NIMCET 2026

Nearest first

Next 10 Questions — NIMCET 2026

Ascending by ID
Ask Your Question or Put Your Review.

loading...