Let AB be the building of height 7 m. Let CD be the cable tower of height H. Let BC be the distance between the building and the tower. From point A, the angle of elevation to the top of the tower (D) is 60^°, and the angle of depression to the foot of the tower (C) is 45^°. We draw a line AE parallel to BC. Thus, AE = BC and EC = AB = 7 m. We need to find the total height of the tower, H = CD = CE + ED = 7 + h, where h = ED.

Let AB be the building of height 7 m. Let CD be the cable tower of height H. Let BC be the distance between the building and the tower. From point A, the angle of elevation to the top of the tower (D) is 60^°, and the angle of depression to the foot of the tower (C) is 45^°. We draw a line AE parallel to BC. Thus, AE = BC and EC = AB = 7 m. We need to find the total height of the tower, H = CD = CE + ED = 7 + h, where h = ED.

Consider the right-angled triangle ABC. The angle of depression from A to C is 45^°. Since AE BC, the alternate interior angle ACB is also 45^°. Using the tangent function in ABC, we have tan 45^° = (AB)/(BC). Since AB = 7 m and tan 45^° = 1, we find that BC = 7 m. Also, AE = BC = 7 m.

Consider the right-angled triangle ABC. The angle of depression from A to C is 45^°. Since AE BC, the alternate interior angle ACB is also 45^°. Using the tangent function in ABC, we have tan 45^° = (AB)/(BC). Since AB = 7 m and tan 45^° = 1, we find that BC = 7 m. Also, AE = BC = 7 m.

Now consider the right-angled triangle AED. The angle of elevation from A to D is 60^°. We know AE = 7 m from the previous step. Using the tangent function, tan 60^° = (ED)/(AE). Since tan 60^° = √3, we can find h = ED.

Now consider the right-angled triangle AED. The angle of elevation from A to D is 60^°. We know AE = 7 m from the previous step. Using the tangent function, tan 60^° = (ED)/(AE). Since tan 60^° = √3, we can find h = ED.

The total height of the cable tower CD is the sum of CE and ED. We know CE = 7 m and we just calculated ED = h = 7√3 m. Substitute these values to find the total height H.

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60 degrees and the angle of depression of its foot is 45 degrees. Determine the height of the tower.

Answer: The height of the tower is $7(1 + \sqrt{3})\text{ m}$ .

Step-by-step solution

Step 1: Define variables and set up the first triangle

Let $AB$ be the building of height 7 m. Let $CD$ be the cable tower of height $H$ . Let $BC$ be the distance between the building and the tower. From point $A$ , the angle of elevation to the top of the tower ( $D$ ) is $60^\circ$ , and the angle of depression to the foot of the tower ( $C$ ) is $45^\circ$ . We draw a line $AE$ parallel to $BC$ . Thus, $AE = BC$ and $EC = AB = 7$ m. We need to find the total height of the tower, $H = CD = CE + ED = 7 + h$ , where $h = ED$ .

Step 2: Calculate the distance to the tower using the angle of depression

Consider the right-angled triangle $\triangle ABC$ . The angle of depression from $A$ to $C$ is $45^\circ$ . Since $AE \parallel BC$ , the alternate interior angle $\angle ACB$ is also $45^\circ$ . Using the tangent function in $\triangle ABC$ , we have $\tan 45^\circ = \frac{AB}{BC}$ . Since $AB = 7$ m and $\tan 45^\circ = 1$ , we find that $BC = 7$ m. Also, $AE = BC = 7$ m.

Step 3: Calculate the height of the tower above the building

Now consider the right-angled triangle $\triangle AED$ . The angle of elevation from $A$ to $D$ is $60^\circ$ . We know $AE = 7$ m from the previous step. Using the tangent function, $\tan 60^\circ = \frac{ED}{AE}$ . Since $\tan 60^\circ = \sqrt{3}$ , we can find $h = ED$ .

Step 4: Determine the total height of the tower

The total height of the cable tower $CD$ is the sum of $CE$ and $ED$ . We know $CE = 7$ m and we just calculated $ED = h = 7\sqrt{3}$ m. Substitute these values to find the total height $H$ .

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