Question

From a window 20m high above the ground in a street, the angle of elevation and depression of the top and the foot of another house opposite side of the street are 60° and 45° respectively. Find the height of opposite house​

Answer 1

Answer:

Step-by-step explanation:

Let AP 60 m be the height of the window above the ground.(AP = QC)

CD = h m be the height of the house on the opposite side of the Street

GIVEN:

∠QPD = 60°(angle of elevation of the top of D of house CD)

∠QPC = 45°  (angle of depression of the foot C of the house CD)

QD = CD - CQ

QD = CD - AP        [CQ = AP]                                this is an example for your question

QD =  (h - 60) m

In ∆PQC,

tan 45° = QC/PQ = P/B

1 = 60/PQ

PQ = 60 m

In ∆PQD ,

tan 60° = QD/PQ = P/B

√3 = (h-60)/60

60√3 = (h-60)

60√3 +60 = h

60(√3+1) = h

Hence, the height of the opposite house is 60(√3+1) m.

Answer 2

ANSWER:-

Given:

From a window 20m high above the ground in a street, the angle of elevation and depression of the top and the foot of another house opposite side of the street are 60° & 45° respectively.

To find:

Find the height of opposite house.

Solution:

Let the height of the opposite house be DC= h m

In right ∆ADE,

$tan60 \degree = \frac{DE}{AE} \\ \\ = > \sqrt{3} = \frac{h - 20}{AE} \\ \\ = > AE = \frac{h - 20}{ \sqrt{3} } ................(1)$

In right ∆ACE,

$tan45 \degree = \frac{CE}{AE} \\ \\ = > 1 = \frac{20}{AE} \\ \\ = > AE = 20 \degree..............(2)$

Now,

Comparing equation (1) & (2) we get;

$\frac{h - 20}{ \sqrt{3} } = 20 \\ \\ = > h - 20 = 20 \sqrt{3} \\ \\ = > h = 20 \sqrt{3} + 20 \\ \\ = > h = 20(1 + \sqrt{3} ) \\ \\ = > h = 20(1 + 1.732) \\ \\ = > h = 20 \times 2.732 \\ \\ = > h = 54.64m$

Thus,

Height of the opposite house is 54.64m.

Width of the street

=) AE= BC= 20m

Hope it helps ☺️