Distance AB. Now in ΔABC, tan30

→1/√3 = 50/AB

→ AB=50√3 m

Let us see one more example where the horizontal distance is given and the height of the object is asked.

AB = 32m. In ΔABC, tan30

→ √3 = BC/AB → BC=32√3 m

Hence the height of the building is 32√3m.

Next there are problems where a person observes two ships and the angles of depression/elevation are given and we have to find the distance between the ships etc.**Example 3:** From the top of a light house 80m high, two ships are observed. The angles of elevation of these ships as seen from the light house are found to be 45^{0} and 30^{0}. Find the distance between the two ships.
**Sol:** Here AB is the light house and the ships are at points

C and D. Let CD = x

In ΔABC, tan45^{0} =AB/AC

→ 1 = 80/AC

→ AC=80m

Hence AD = 80 + x

In ΔBAD, tan30^{0} =AB/AD

→ 1/√3 = 80/(80+x)

→ (80+x) = 80√3

→ x=80(√3-1) = 80 x 0.732 =58.56m

Hence the distance between the ships is 58.56 m.
**Example 4:** On the roof of a building 20 m high, a boy is flying a kite. To keep that kite flying a thread of length 75 m is released. If the thread makes an angle of 45^{0} with the horizontal, then at what height from the ground is the kite flying?
**Sol:** Let AB is the building of height 20 m and BC is the string.

In ΔBDC, sin45^{0} =CD/BC

→ 1/√2 = CD/BC

→ 1/√2 = CD/75

→ CD = 75/√2 = 43.3m

Hence the distance above the ground = 20 + 43.3 = 63.3 m
**Example 5: **Two poles of height 30 meters and 10 meters are erected on the ground. The top of the poles are tied by a wire of length 40 m. Find the angle made by the wire with the horizontal.
**Sol: ** Let AB and CD are the poles of heights 30 m and 10 m respectively.

Let the angle made by the wire with horizontal is θ.

In ΔDEB, sinθ = BE/BD =20/40 = 1/2

→ sinθ = 30^{0}
**Example 6:** The angles of depression of the top and the bottom of a tree as seen from the top of a tower 90 meters high are 30 and 60 respectively. Find the height of the tree.
**Sol:** Let AB is the tower of height 90 m and CD is the tree.

In ΔABC, tan60^{0} =AB/AC

→ √3 = 90/AC

→ AC = 30√3

Hence DE = AC = 30√3m

In ΔDEB, tan30^{0} =BE/DE

→ 1/√3 = BE/30√3

→BE = 30m

Hence CD = AE = 90 – 30 = 60m.
**Example 7:** The upper part of a tree is broken and the top touches the ground making an angle of 30°. The distance of the point where the top touches the ground to the base of the tree is 72 m. What was the height of the tree?
**Sol:** Let AC is the tree which broke at point B so that the top C touches the ground. Here AC = 72 m

In ΔABC, tan30^{0} = AB/AC

→ 1/√3 = AB/72

→ AB = 72/√3 = 24√3m

Again in ΔABC, sin30^{0} = AB/AC

→ 1/2 = 24√3/BC

→ BC = 48√3m

Hence the length of the tree = AB + BC = 24√3+48√3 = 72√3 = 124.7m
**Example 8:** Two aero planes flying in opposite directions are at points A and B, 3000 m high from the ground. When the pilots look at a point P on the ground, the angles of depression made by them are 300 and 600 respectively. Find the distance between the two aero planes.
**Sol:** The planes are at point A & B.

In ΔADP, AD/DP =tan 30^{0}

→ 3000/DP =1/√3 → DP = 3000√3m

Again in ΔBPC, tan 60^{0}=3000/PC

→ √3 = 3000/PC → PC = 3000/√3

The distance between the planes = 3000√3 + 3000/√3 = 12000/√3 =6928.4 m
**Example 9:** When a man looks from the foot and the top of a tower at the roof of a building, the angles of elevation and depression are of 27^{0} and 63^{0} respectively. If the height of the building is 40m, find the height of the tower. (tan 63^{0}=2)
**Sol:** Let AB is the tower and CD is the building of height 40m.

In ΔACD, AC/DC = cot27^{0} = cot(90-63)

→ AC/40 = tan63^{0} = 2

→ AC=80m

Now DE=AC=80m

In ΔBED, tan 63^{0}= BE/DE

→2=BE/80 →BE=160m ∴Height of the tower = AE + EB = 40 + 160 = 200m
**Example 10:** There is a cliff 80 m high on the bank of a lake. A man is sitting on the cliff. He looks at a bird in the sky and the angle of elevation is formed to be of 30^{0}. When he looks at the shadow of the bird in the lake, the angle of depression is 60^{0}. Find the height of the bird from the lake surface.
**Sol:** Let AB is the cliff and the man is sitting at B. Let C is the position of the bird and D is its shadow.

In ΔADB, tan 60^{0} = AB/AD

→ √3 = 80/AD → AD=80/√3 m

Now BE = AD = 80/√3 m

In ΔBEC, tan 30^{0}= CE/BE

→ 1/√3 = CE/(80/√3) → CE=80/3 m

∴ The height of the bird = DC = DE +CE = 80 + 80/3 m = 320/3 m

Next there are problems where a person observes two ships and the angles of depression/elevation are given and we have to find the distance between the ships etc.

C and D. Let CD = x

In ΔABC, tan45

→ 1 = 80/AC

→ AC=80m

Hence AD = 80 + x

In ΔBAD, tan30

→ 1/√3 = 80/(80+x)

→ (80+x) = 80√3

→ x=80(√3-1) = 80 x 0.732 =58.56m

Hence the distance between the ships is 58.56 m.

In ΔBDC, sin45

→ 1/√2 = CD/BC

→ 1/√2 = CD/75

→ CD = 75/√2 = 43.3m

Hence the distance above the ground = 20 + 43.3 = 63.3 m

Let the angle made by the wire with horizontal is θ.

In ΔDEB, sinθ = BE/BD =20/40 = 1/2

→ sinθ = 30

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In ΔABC, tan60

→ √3 = 90/AC

→ AC = 30√3

Hence DE = AC = 30√3m

In ΔDEB, tan30

→ 1/√3 = BE/30√3

→BE = 30m

Hence CD = AE = 90 – 30 = 60m.

In ΔABC, tan30

→ 1/√3 = AB/72

→ AB = 72/√3 = 24√3m

Again in ΔABC, sin30

→ 1/2 = 24√3/BC

→ BC = 48√3m

Hence the length of the tree = AB + BC = 24√3+48√3 = 72√3 = 124.7m

In ΔADP, AD/DP =tan 30

→ 3000/DP =1/√3 → DP = 3000√3m

Again in ΔBPC, tan 60

→ √3 = 3000/PC → PC = 3000/√3

The distance between the planes = 3000√3 + 3000/√3 = 12000/√3 =6928.4 m

In ΔACD, AC/DC = cot27

→ AC/40 = tan63

→ AC=80m

Now DE=AC=80m

In ΔBED, tan 63

→2=BE/80 →BE=160m ∴Height of the tower = AE + EB = 40 + 160 = 200m

In ΔADB, tan 60

→ √3 = 80/AD → AD=80/√3 m

Now BE = AD = 80/√3 m

In ΔBEC, tan 30

→ 1/√3 = CE/(80/√3) → CE=80/3 m

∴ The height of the bird = DC = DE +CE = 80 + 80/3 m = 320/3 m