Question

a ladder leans against a wall. it makes an angle of 45 with the ground and reaches tje height of 10 meter on the wall . find the length of yhe ladder and also the diatance between the lower end of the ladder and the base of the wall

Answer 1

AC is the length of the ladder and
BC is the distance of the lower end of
ladder from the wall

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{Length\:of\:ladder=14.1\:m}}}$

$\green{\therefore{\text{Distance\:between\:wall\:and\:foot\:of\:ladder=10\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a ladder leans against a wall. it makes an angle of 45 with the ground and reaches the height of 10 meter on the wall.

• We have to find the length of ladder and distance between wall and foot of ladder.

$\green{\underline \bold{Given :}} \\ : \implies \text{Height\:of\:wall= 10\: feet} \\ \\ : \implies \text{Angle\:of\:elevation=}45^{\circ}\\\\ \red{\underline \bold{To \: Find:}} \\ : \implies \text{Length\:of\:ladder= ?}\\\\ :\implies \text{Distance\:between\:wall\:and\:foot\:of\:tower=?}$

• Accroding to given question :

$\bold{In \: \triangle \: ABC} \\ : \implies sin \: \theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \\ \\ : \implies sin\: 45^{\circ}= \frac{AB}{AC} \\ \\ : \implies \frac{1}{\sqrt{2}}= \frac{10}{AC} \\ \\ : \implies AC= 10\sqrt{2} \ \\ \\ \green{ : \implies \text{AC=14.1\:m}}$

$\bold{again \: in \: \triangle \: abc} \\ : \implies cos \: \theta \degree = \frac{\text{Base}}{\text{Hypotenuse}} \\ \\ \implies cos \: 45^{ \circ} = \frac{BC}{10 \sqrt{2} } \\ \\ : \implies \frac{1}{ \sqrt{2} } = \frac{BC}{10 \sqrt{2} } \\ \\ \green{ : \implies \text{BC = 10 \: m}}$