Solution for Two Triangles

4. From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take 3 = 1.732)

Instructions

Solution : AB denotes the height of the building, BD the flagstaff and P the given point. Note that there are two right triangles PAB and PAD. We are required to find the length of the flagstaff, i.e. DB and the distance of the building from the point P, i.e. PA.

Since, we know the height of the building AB, we will first consider the right triangle PAB.

We have tan 30°cot 30° = ABAP

133 = /AP

Therefore, AP = 103310

i.e. the distance of the building from P is 103 m = m. (upto two decimal places)

Next, let us suppose DB = x m. Then AD = 10+x10x m.

Now, in right triangle PAD, tan 45° = ADAPAPAD = 10+x10310310+x

Therefore, = 10+x103

i.e., x = 103−1103+1 = (upto two decimal places)

So, the length of the flagstaff is 7.32 m.

5. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

Instructions

Solution : AB is the tower and BC is the length of the shadow when the Sun’s altitude is 60°, i.e., the angle of elevation of the top of the tower from the tip of the shadow is 60° and DB is the length of the shadow, when the angle of elevation is °.

Now, let AB be h m and BC be x m. According to the question, DB is m longer than BC.

So, DB = (40 + x) m

Now, we have two right triangles ABC and ABDBCD.

In triangle ABC, tan 60° = ABBCBCAB

313 = hxxh (1)

In triangle ABD, tan 30° = ABBDBDAB

133 = hx+40x+40h (2)

From (1), we have h = x33x

Putting this value in (2), we get x33 = x + 40, i.e. = x + 40

Thus, x =

So, h = 203203 [From (1)]

Therefore, the height of the tower is 203 m.

6. The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multistoreyed building and the distance between the two buildings.

Instructions

Solution : PC denotes the multistoryed building and AB denotes the 8 m tall building. We are interested to determine the height of the multi-storeyed building, i.e. PC and the distance between the two buildings i.e. .

Look at the figure carefully. Observe that PB is a to the parallel lines PQ and BD.

Therefore, ∠ QPB and ∠ PBD are alternate interioralternate exterior angles and thus are .

So ∠ PBD = °. Similarly, ∠ PAC = °.

In right triangle PBD, we have:

PDBDBDPD = tan 30° = 133

BD = 313PD

In right triangle PAC, we have:

PCACACPC = tan 45° =

PC =

Also, PC = PD +

Therefore, PD + DC =

Since, AC = BD and DC = AB = m, we get:

PD + 8 = BD = 3

This gives PD = 83−1 = 83+13−13+1 = 43+143−1

So, the height of the multi-storeyed building is 43+1+8 = 43+343−3 and the distance between the two buildings is also 43+343−3 m.

Example 7

7. From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river.

Solution : A and B represent points on the bank on opposite sides of the river, so that AB is the width of the river. P is a point on the bridge at a height of 3 m, i.e. DP = m.

We are interested to determine the width of the river, which is the length of the side ABAP of the triangle APB.

Now, AB = AD +

In right triangle APD, ∠ A = °.

So, tan 30°sin 30° = PDADADPD

13 = 3ADAD3

AD = 3333 m

Also in right triangle PBD, ∠ B = °. So, BD = PD = m.

Now, AB = BD + AD = 3+33 = 31+333 m.

Therefore, the width of the river is 31+3 m.

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