Right triangles and trigonometry #3 - Q&A

Question 1 A plane, diving with constant speed at an angle of 53.0° with the vertical, releases a projectile at an altitude of 730 m. The projectile hits the ground 5.00 s after release. a) What is the speed of the plane? b) How far does the projectile travel horizontally during its flight? What are the c) horizontal and d) vertical components of its velocity just before striking the ground? Answer 1 The speed of the plane is the initial speed of the projectile which can be obtained from: 𝑦 − 𝑦 0 = 𝑣 0 𝑦 𝑡 − 1 2 𝑔 𝑡 2 y−y 0 ​ =v 0y ​ t− 2 1 ​ gt 2 𝑦 = 𝑦 0 + 𝑣 0 𝑦 𝑡 − 1 2 𝑔 𝑡 2 y=y 0 ​ +v 0y ​ t− 2 1 ​ gt 2 − 1 2 ( 9.8 ) ( 5 ) 2 = − 730 − 2 1 ​ (9.8)(5) 2 =−730 𝑣 0 𝑦 = 121.5   𝑚 𝑠 v 0y ​ =121.5 s m ​ The angle with the horizontal is 90 − 53 = 37 90−53=37 It is downward, it is going to be 𝜃 = − 37 θ=−37 Now that we know the angle and the y-component of the velocity, we can use the relation to get the overall speed: 𝑣 0 = 𝑣 0 𝑦 sin ⁡ ( − 37 ) = 201.5   𝑚 𝑠 v 0 ​ = sin(−37) v 0y ​ ​ =201.5 s m ​ b) 𝑥 = 𝑥 0 + 𝑣 0 𝑥 𝑡 , 𝑥 = 𝑣 0 𝑥 𝑡 x=x 0 ​ +v 0x ​ t,x=v 0x ​ t 𝑥 = 𝑣 0 cos ⁡ 𝜃 = ( 201.5 ) cos ⁡ ( − 37 ) x=v 0 ​ cosθ=(201.5)cos(−37) 𝑥 = 802   𝑚 x=802m c) 𝑣 𝑥 = 𝑣 0 cos ⁡ 𝜃 = ( 201.5 ) ( cos ⁡ ( − 37 ) ) = 160   𝑚 𝑠 v x ​ =v 0 ​ cosθ=(201.5)(cos(−37))=160 s m ​ d) 𝑣 𝑦 = 𝑣 0 𝑦 − 𝑔 𝑡 = − 121.5   𝑚 𝑠 − ( 9.8 ) ( 5 ) = − 170   𝑚 𝑠 v y ​ =v 0y ​ −gt=−121.5 s m ​ −(9.8)(5)=−170 s m ​ Answer 2 To solve the given problem, let's break it down into parts: a) Finding the speed of the plane: The vertical motion of the plane is independent of its horizontal motion. Since the plane is diving at an angle of 53.0 degrees with the vertical and the time of flight for the projectile is given as 5.00 s, we can use the equation for vertical displacement: 𝑦 = 𝑦 0 + 𝑣 0 𝑦 𝑡 − 1 2 𝑔 𝑡 2 y=y 0 ​ +v 0y ​ t− 2 1 ​ gt 2 where 𝑦 y is the vertical displacement, 𝑣 0 𝑦 v 0y ​ is the vertical component of initial velocity, 𝑡 t is the time, and 𝑔 g is the acceleration due to gravity. Here, 𝑦 = − 730   𝑚 y=−730m (taking downward direction as negative), 𝑡 = 5.00   𝑠 t=5.00s, and 𝑔 = − 9.8   𝑚 𝑠 2 g=−9.8 s 2 m ​ . Since the plane is diving at an angle of 53.0 degrees with the vertical, the initial vertical velocity of the projectile can be expressed as: 𝑣 0 𝑦 = 𝑣 0 sin ⁡ 𝜃 v 0y ​ =v 0 ​ sinθ where 𝑣 0 v 0 ​ is the speed of the plane and 𝜃 = 53.0 ° θ=53.0°. By substituting the given values into the equations, we solve for 𝑣 0 v 0 ​ . b) Finding the horizontal distance traveled by the projectile: The horizontal distance traveled by the projectile is unaffected by the vertical motion of the plane. We use the equation for horizontal distance: 𝑥 = 𝑣 0 𝑥 𝑡 x=v 0x ​ t where 𝑥 x is the horizontal distance, 𝑣 0 𝑥 v 0x ​ is the horizontal component of initial velocity, and 𝑡 t is the time of flight. Since there is no acceleration in the horizontal direction, 𝑣 0 𝑥 v 0x ​ remains constant throughout the motion. By substituting the given values into the equation, we can calculate 𝑥 x. c) Finding the horizontal component of velocity just before striking the ground: The horizontal component of velocity remains constant throughout the motion. Thus, the horizontal component of velocity just before striking the ground is equal to the horizontal component of initial velocity. d) Finding the vertical component of velocity just before striking the ground: The vertical component of velocity changes due to the effect of gravity. We can use the equation for vertical velocity: 𝑣 𝑦 = 𝑣 0 𝑦 + 𝑔 𝑡 v y ​ =v 0y ​ +gt where 𝑣 𝑦 v y ​ is the vertical component of velocity, 𝑣 0 𝑦 v 0y ​ is the initial vertical component of velocity, 𝑔 g is the acceleration due to gravity, and 𝑡 t is the time of flight. By substituting the given values into the equation, we can calculate 𝑣 𝑦 v y ​ . Answer 3 Result: a) The speed of the plane is approximately 44.4   𝑚 𝑠 44.4 s m ​ . b) The projectile travels approximately 150 m horizontally during its flight. c) The horizontal component of the velocity just before striking the ground is approximately 44.4   𝑚 𝑠 44.4 s m ​ . d) The vertical component of the velocity just before striking the ground is approximately − 6.02   𝑚 𝑠 −6.02 s m ​ . Solution: a) To find the speed of the plane, we can use the vertical motion of the projectile. The vertical displacement of the projectile is given by the equation: 𝑦 = 𝑣 𝑦 0 𝑡 + 1 2 𝑔 𝑡 2 y=v y0 ​ t+ 2 1 ​ gt 2 where 𝑦 y is the vertical displacement (730 m), 𝑣 𝑦 0 v y0 ​ is the initial vertical velocity of the projectile, 𝑡 t is the time of flight (5.00 s), and 𝑔 g is the acceleration due to gravity ( − 9.8   𝑚 𝑠 2 −9.8 s 2 m ​ ). Since the plane is diving at an angle of 53.0 degrees with the vertical, the initial vertical velocity of the projectile can be determined using trigonometry: 𝑣 𝑦 0 = 𝑣 sin ⁡ ( 𝜃 ) v y0 ​ =vsin(θ) where 𝑣 v is the speed of the plane and 𝜃 θ is the angle of dive (53.0 degrees). Substituting this expression into the first equation and solving for 𝑣 v, we get: 𝑦 = 𝑣 sin ⁡ ( 𝜃 ) 𝑡 − 1 2 𝑔 𝑡 2 y=vsin(θ)t− 2 1 ​ gt 2 Simplifying further, we have: 𝑣 = 𝑦 + 1 2 𝑔 𝑡 2 sin ⁡ ( 𝜃 ) 𝑡 v= sin(θ)t y+ 2 1 ​ gt 2 ​ Substituting the given values, we get: 𝑣 = 730 − 1 2 ( 9.8 ) ( 5.00 ) 2 sin ⁡ ( 53.0 ° ) ( 5.00 ) v= sin(53.0°)(5.00) 730− 2 1 ​ (9.8)(5.00) 2 ​ Calculating the value, we find: 𝑣 ≈ 44.4   𝑚 𝑠 v≈44.4 s m ​ (Rest of Answer 3 continues on in a similar manner...) Question 2 The converse of the Pythagorean theorem is also a true statement: If the sum of the squares of the lengths of two sides of a triangle is equal to the square of the length of the third side, then the triangle is a right triangle. Use the distance formula and the Pythagorean theorem to determine whether the set of points could be vertices of a right triangle: ( − 4 , 3 ) , ( 0 , 5 ) , ( 3 , − 4 ) (−4,3),(0,5),(3,−4) Answer 1 We know that the square of distance between two points ( 𝑥 1 , 𝑦 1 ) (x 1 ​ ,y 1 ​ ) and ( 𝑥 2 , 𝑦 2 ) (x 2 ​ ,y 2 ​ ) is ( 𝑥 2 − 𝑥 1 ) 2 + ( 𝑦 2 − 𝑦 1 ) 2 (x 2 ​ −x 1 ​ ) 2 +(y 2 ​ −y 1 ​ ) 2 and the converse of Pythagoras theorem, we have if sum of square of any two side is equal to square of third side of triangle, then it is right angled triangle. Using distance formula and converse of Pythagoras theorem, we find that ABC is right angled triangle right angled at A. 𝐴 𝐵 2 = ( 0 − ( − 4 ) ) 2 + ( 5 − 3 ) 2 = 16 + 4 = 20 AB 2 =(0−(−4)) 2 +(5−3) 2 =16+4=20 𝐵 𝐶 2 = ( 3 − 0 ) 2 + ( − 4 − 5 ) 2 = 9 + 81 = 90 BC 2 =(3−0) 2 +(−4−5) 2 =9+81=90 𝐶 𝐴 2 = ( 3 − ( − 4 ) ) 2 + ( − 4 − 3 ) 2 = 49 + 49 = 98 CA 2 =(3−(−4)) 2 +(−4−3) 2 =49+49=98 𝐴 𝐵 2 + 𝐴 𝐶 2 ≠ 𝐵 𝐶 2 AB 2 +AC 2  =BC 2 ( 98 ≠ 90 ≠ 20... 𝐴 𝐵 2 + 𝐴 𝐶 2 ≠ 𝐵 𝐶 2 AB 2 +AC 2  =BC 2 , confirming that the points do not form a right triangle. Question 3 What is the perimeter of the triangle below, to the nearest tenth of an inch, if each square on the grid measures 8 inches on each side? Answer 1 We have a right triangle with legs measuring 2 units and 4 units. Using the Pythagorean Theorem, the missing length (hypotenuse), 𝑥 x, is: 2 2 + 4 2 = 𝑥 2 4 + 16 = 𝑥 2 20 = 𝑥 2 𝑥 = 20 2 2 +4 2 =x 2 4+16=x 2 20=x 2 x= 20 ​ The perimeter in terms of the number of units is: 𝑃 = 2 + 4 + 20 = 6 + 20 P=2+4+ 20 ​ =6+ 20 ​ Since 1 unit = 8 inches, the perimeter is: 𝑃 = 8 ( 6 + 20 ) 𝑃 ≈ 83.8   inches P=8(6+ 20 ​ )P≈83.8inches Question 4 The shaded sector above covers 1 2 2 1 ​ of the circle. If the radius of the circle above is 3 cm, what is the area of the sector in terms of 𝜋 π? A. 1.125   cm 2 1.125cm 2 B. 14.13   cm 2 14.13cm 2 C. 4.5   cm 2 4.5cm 2 D. 2.25   cm 2 2.25cm 2 Answer 1 If the sector covers 1 2 2 1 ​ of the circle, then we need to find the area of the circle and multiply this area by 1 2 2 1 ​ : The area of a circle is 𝜋 𝑟 2 πr 2 , where 𝑟 r is the radius. If the radius of the circle is 𝑟 = 3   cm r=3cm, then its area is: 𝜋 ( 3 2 ) = 9 𝜋   cm 2 π(3 2 )=9πcm 2 The area of the sector is then: 1 2 ⋅ 9 𝜋 = 4.5 𝜋   cm 2 2 1 ​ ⋅9π=4.5πcm 2 The correct answer is C. 4.5 𝜋   cm 2 4.5πcm 2 . Question 5 The prism has a cross-sectional area in the shape of a sector. Calculate: a. the radius 𝑟 r cm, b. the cross-sectional area of the prism, c. the total surface area of the prism, d. the volume of the prism. Answer 1 a. To find 𝑟 r, we use the given arc length of 20 cm and the angle of 5 0 ∘ 50 ∘ that creates the arc. The arc length, 𝑠 s, of a circle is given by the formula: 𝑠 = 𝜃 36 0 ∘ ( 2 𝜋 𝑟 ) s= 360 ∘ θ ​ (2πr) where 𝑠 = 20   cm s=20cm and 𝜃 = 5 0 ∘ θ=50 ∘ . Therefore: 20 = 50 360 ( 2 𝜋 𝑟 ) 20 = 5 𝜋 𝑟 18 𝑟 = 20 ⋅ 18 5 𝜋 = 360 5 𝜋 ≈ 22.9   cm 20= 360 50 ​ (2πr)20= 18 5πr ​ r= 5π 20⋅18 ​ = 5π 360 ​ ≈22.9cm b. The cross-sectional area of a prism is the area of the sector. The area, 𝐴 A, of a sector is given by: 𝐴 = 𝜃 36 0 ∘ ( 𝜋 𝑟 2 ) A= 360 ∘ θ ​ (πr 2 ) Substituting the values: 𝐴 = 50 360 ( 𝜋 ( 22.9 ) 2 ) ≈ 228.8   cm 2 A= 360 50 ​ (π(22.9) 2 )≈228.8cm 2 c. The surface area of a prism is: 𝑆 = 2 𝐵 + 𝑃 ℎ S=2B+Ph where 𝐵 B is the area of the base, 𝑃 P is the perimeter of the base, and ℎ h is the height of the prism. The sector is the base, so 𝐵 = 228.8   cm 2 B=228.8cm 2 . The perimeter is the arc length plus two times the radius: 𝑃 = 20 + 2 ( 22.9 ) = 65.8   cm P=20+2(22.9)=65.8cm The height of the prism is ℎ = 8   cm h=8cm. Therefore, the surface area is: 𝑆 = 2 ( 228.8 ) + 65.8 ( 8 ) = 457.6 + 526.4 = 984   cm 2 S=2(228.8)+65.8(8)=457.6+526.4=984cm 2 d. The volume of a prism is: 𝑉 = 𝐵 ℎ V=Bh 𝑉 = 228.8 ( 8 ) = 1830.4   cm 3 V=228.8(8)=1830.4cm 3