ABCDEF is a triangular prism which represents part of a hill. ABCF is the horizontal rectangular base. D is vertically above C. |a) Work out the height CD. (2) b) Tony walks in a straight line from A to D. Find length AC and use it to work out the size of angle CAD to 1 decimal place.

Also given m∠DBC = 6°, we can use the trigonometric ratio

[tex]\tan 6^\circ = \dfrac{\text{side opposite} }{\text{side adjacent} } \\\\\tan 6^\circ = \dfrac{CD}{BC}\\\\\tan 6^\circ = \dfrac{CD}{470\;m}\\\\[/tex]

Cross-multiplying we get:
[tex]\tan 6^\circ \cdot 470 = CD\\\\49.39899 = CD\\\\CD = 49.4\;m[/tex]

Part b)

ACD is a right triangle with AD as hypotenuse, AC and CD as legs of the right triangle

By the Pythagorean Theorem
AD² = AC² + CD² [1]

Since ABCF is a rectangle, AC forms the diagonal of the rectangle, ABC is a right triangle with legs AB and BC

By the Pythagorean Theorem
AC² = AB² + BC² [3]

We are given AB = 360 m, BC = 470 m and substituting these values into equation [2]

AC² = 360² + 470²

AC² = 129,600 + 220,900

AC² = 350,500

AC = √(350,500)

AC = 592.03 m

ΔACD is a right triangle with legs AC and CD

Using the trigonometric ratio for tan (m∠CAD) = CD/AC

we get

tan ∠CAD = 49.4/592.03

tan ∠CAD = 0.08344

m ∠CAD = tan⁻¹(0.08344)

m ∠CAD = 4.7698°

m∠CAD = 4.8° to the nearest 1 decimal place

ABCDEF is a triangular prism which represents part of a hill. ABCF is the horizontal rectangular base. D is vertically above C. |a) Work out the height CD. (2) b) Tony walks in a straight line from A to D. Find length AC and use it to work out the size of angle CAD to 1 decimal place.​

Answer :

Other Questions

ABCDEF is a triangular prism which represents part of a hill. ABCF is the horizontal rectangular base. D is vertically above C. |a) Work out the height CD. (2) b) Tony walks in a straight line from A to D. Find length AC and use it to work out the size of angle CAD to 1 decimal place.