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Home » 11. A ladder rests against a tree on one side of a street. The foot of the ladder makes an angle of with the ground. When the ladder is turned over to rest against another tree on the other side of the street it makes an angle of with the ground. If the length of the ladder is     , find the width of the street.
11. A ladder rests against a tree on one side of a street. The foot of the ladder makes an angle of {{50}^{\circ }} with the ground. When the ladder is turned over to rest against another tree on the other side of the street it makes an angle of {{40}^{\circ }} with the ground. If the length of the ladder is

    \[60m\]

, find the width of the street.
Class 10 | Frank | Heights and Distances | ICSE | Maths
11. A ladder rests against a tree on one side of a street. The foot of the ladder makes an angle of {{50}^{\circ }} with the ground. When the ladder is turned over to rest against another tree on the other side of the street it makes an angle of {{40}^{\circ }} with the ground. If the length of the ladder is

    \[60m\]

, find the width of the street.

As per given information in the question,

Let us assume two trees are denoted by PQ and RS and O be a point on the street PR, which is the distance between the two trees.

Then,

We Assume that OS and OQ represent the ladder which at rest on both trees.

Then,

In ΔORS,

    \[\cos {{50}^{\circ }}=\frac{OR}{OS}\]

Now as we know,\cos {{50}^{\circ }}=0.6428

    \[0.6428=\frac{OR}{60}\]

    \[OR=60\times 0.6428\]

    \[OR=38.568\]

Now, consider ΔPQO,

\cos {{40}^{\circ }}=\frac{PQ}{QO}

As we know, \cos {{40}^{\circ }}=0.7660

0.7660=\frac{PO}{60}

PO=0.7660\times 60

PO=45.96

Therefore, PR=PO+OR

PR=45.96+38.56

PR=84.52m

Hence the width of the street PR is 84.52m.

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