Question

point b on the ground is 5 cm from point E at the entrance to Ollie's house. He is 1.8 m tall and is standing at Point D, below the cemsot. He walks towards point B. Find the distance Ollie is from the entrance to his house when he first activates the sensor.

Answer 1

Point B on the ground is 5 cm from point E at the entrance to Ollie's house.

Ollie is at a distance of 2.45 m from the entrance to his house when he first activates the sensor.

The complete question is as follows:

Ollie has installed security lights on the side of his house that is activated by a  sensor. The sensor is located at point C directly above point D. The area covered by the sensor is shown by the shaded region enclosed by triangle ABC. The distance from A to B is 4.5 m, and the distance from B to C is 6m. Angle ACB is 15°.

The objective of this information is:

• To find angle CAB and;
• Find the distance Ollie is from the entrance to his house when he first activates the sensor.

The diagrammatic representation of the information given is shown in the image attached below.

Using  cosine rule to determine angle CAB, we have:

$\mathbf{\dfrac{AB}{Sin \hat {ACB}} = \dfrac{BC}{Sin \hat {CAB}}= \dfrac{CA}{Sin \hat {ABC}}}$

Here:

$\mathbf{\dfrac{AB}{Sin \hat {ACB}} = \dfrac{BC}{Sin \hat {CAB}}}$

$\mathbf{\dfrac{4.5}{Sin \hat {15^0}} = \dfrac{6}{Sin \hat {CAB}}}$

$\mathbf{Sin \hat {CAB} = \dfrac{Sin 15 \times 6}{4.5}}$

$\mathbf{Sin \hat {CAB} = \dfrac{0.2588 \times 6}{4.5}}$

$\mathbf{Sin \hat {CAB} = 0.3451}$

∠CAB = Sin⁻¹ (0.3451)

∠CAB = 20.19⁰

From the diagram attached;

• assuming we have an imaginary position at the base of Ollie Standing point called point F when Ollie first activates the sensor;          

Then, we can say:

∠CBD = ∠GBF

∠GBF = (CAB + ACB)      

(because the exterior angles of a Δ is the sum of the two interior angles.

∠GBF = 15° + 20.19°

∠GBF = 35.19°

Using the trigonometric function for the tangent of an angle.

$\mathbf{Tan \theta = \dfrac{GF}{BF}}$

$\mathbf{Tan \ 35.19 = \dfrac{1.8 \ m }{BF}}$

$\mathbf{BF = \dfrac{1.8 \ m }{Tan \ 35.19}}$

$\mathbf{BF = \dfrac{1.8 \ m }{0.7052}}$

BF = 2.55 m

Finally, the distance of Ollie║FE║ from the entrance of his bouse is:

= 5 - 2.55 m

= 2.45 m

Therefore, we can conclude that Ollie is at a distance of 2.45 m from the entrance to his house when he first activates the sensor.

Learn more about exterior angles here: