Question

Please help to solve the questions with diagrams!! thank you!!​

Answer 1

5. distance from B = 19.5 m, height of clock tower = 9.1 m

6.(i) 30.2 m

6.(ii) 71.1 m

Explanation:

Question 5

The given scenario can be modeled as two right triangles that have the same height (see attachment 1).

Define the variables

Let x = the distance between point B and the foot of the clock tower.

Let y = the height of the clock tower.

Use the given information and the tan trigonometric ratio to create equations for the height of the clock tower.

Tan trigonometric ratio

`\sf \tan(\theta)=(O)/(A)`

where:

• 
  `\theta` is the angle
• O is the side opposite the angle
• A is the side adjacent the angle

For triangle (point A):

• θ = 22°
• O = y
• A = 3 + x

Therefore:

`\implies \sf \tan(22^(\circ))=(y)/(3+x)`

`\implies \sf y=(3+x)\tan(22^(\circ))`

For triangle (point B):

• θ = 25°
• O = y
• A = x

Therefore:

`\implies \sf \tan(25^(\circ))=(y)/(x)`

`\implies \sf y=x\tan(25^(\circ))`

Substitute the equation for Triangle A into the equation for Triangle B and solve for x:

`\implies \sf (3+x)\tan(22^(\circ))=x\tan(25^(\circ))`

`\implies \sf 3\tan(22^(\circ))+x\tan(22^(\circ))=x\tan(25^(\circ))`

`\implies \sf 3\tan(22^(\circ))=x\tan(25^(\circ))-x\tan(22^(\circ))`

`\implies \sf 3\tan(22^(\circ))=x[\tan(25^(\circ))-\tan(22^(\circ))]`

`\implies \sf x=(3\tan(22^(\circ)))/(\tan(25^(\circ))-\tan(22^(\circ)))`

`\implies \sf x=19.46131668...m`

Therefore, the distance of B to the foot of the clock tower is 19.5 m (nearest tenth).

To find the height of the clock tower, substitute the found value of x into one of the found equations for y:

`\implies \sf y=x\tan(25^(\circ))`

`\implies \sf y=\left((3\tan(22^(\circ)))/(\tan(25^(\circ))-\tan(22^(\circ)))\right)\tan(25^(\circ))`

`\implies \sf y=(3\tan(22^(\circ))\tan(25^(\circ)))/(\tan(25^(\circ))-\tan(22^(\circ)))`

`\implies \sf y=9.074961005...m`

Therefore, the height of the clock tower is 9.1 m (nearest tenth).

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Question 6

The given scenario can be modeled as two right triangles where the sum of their heights is 92 m (see attachment 2).

Define the variables

Let x = the distance between the foot of the building and the foot of the lamp post.

Let y = the height of the lamp post.

Use the given information and the tan trigonometric ratio to create equations for the height of the lamp post.

Tan trigonometric ratio

`\sf \tan(\theta)=(O)/(A)`

where:

• 
  `\theta` is the angle
• O is the side opposite the angle
• A is the side adjacent the angle

For the angle of elevation triangle:

• θ = 23°
• O = y
• A = x

Therefore:

`\implies \sf \tan(23^(\circ))=(y)/(x)`

`\implies \sf y=x\tan(23^(\circ))`

For the angle of depression triangle:

• θ = 41°
• O = 92 - y
• A = x

Therefore:

`\implies \sf \tan(41^(\circ))=(92-y)/(x)`

`\implies \sf x\tan(41^(\circ))=92-y`

`\implies \sf y=92-x\tan(41^(\circ))`

Substitute the equation for the angle of elevation triangle into the equation for depression triangle and solve for x:

`\implies \sf x\tan(23^(\circ))=92-x\tan(41^(\circ))`

`\implies \sf x\tan(23^(\circ))+x\tan(41^(\circ))=92`

`\implies \sf x[\tan(23^(\circ))+\tan(41^(\circ))]=92`

`\implies \sf x=(92)/(\tan(23^(\circ))+\tan(41^(\circ)))`

`\implies \sf x=71.11047605...m`

Therefore, the distance of the lamp post from the foot of the building is 71.1 m (nearest tenth).

To find the height of the lamp post, substitute the found value of x into one of the found equations for y:

`\implies \sf y=x\tan(23^(\circ))`

`\implies \sf y=\left((92 )/(\tan(23^(\circ))+\tan(41^(\circ)))\right)\tan(23^(\circ))`

`\implies \sf y=(92 \tan(23^(\circ)))/(\tan(23^(\circ))+\tan(41^(\circ)))`

`\implies \sf y=30.18460625...m`

Therefore, the height of the lamp post is 30.2 m (nearest tenth).