Question

Need help please math questions # 12 and 14

Answer 1

Explanation:

Answer 2

`\large \boxed{\text{ Q12. 0.85 m; Q14. 24 750 ft}}`

Explanation:

Question 12.

If Jeff is using a right triangle, the third angle of the triangle is 45°.

They have an isosceles triangle, so the distance from the foot of the elevator to the base of the cliff is x ft.

We can use Pythagoras' Theorem:

The square on the hypotenuse is equal to the sum of the squares of the other two sides.

d² = x² + x²

1.2² = 2x²

1.44 = 2x²

x² = 0.72

x ≈ 0.85 m

Question 14

Hugo's angle of descent is the same as the angle of elevation from the point of touchdown.

The glidepath distance is the horizontal distance from the point where they start the descent to the point of touchdown (side c in the diagram).

Thus we have a right triangle in which ∠B = 22° and b = 10 000 ft.

`\begin{array}{rcl}\tan B & = & (b)/(c)\\\\\tan 22^(\circ) & = & \frac{\text{10 000 ft}}{c}\\\\0.4040 & = & \frac{\text{10 000 ft}}{c}\\\\c & = & \frac{\text{10 000 ft}}{0.4040}\\\\ & = &\text{24 750 ft}\\\end{array}\\\text{The length of Hugo's glide path is $\large \boxed{\textbf{24 750 ft}}$}`