question
eliminte ?, 1) if x=3 sec ? + 4 tan ? and y=4 sec ?- 3 tan ?. 2) the angles of dipression of the top and bottom of 8 meter tall building from the top of multistoried building are 30â° and 45â° respectively.find the height of multistoried building and the distance between two buildings (?3=1.73)
eliminte ?, 1) if x=3 sec ? + 4 tan ? and y=4 sec ?- 3 tan ?. 2) the angles of dipression of the top and bottom of 8 meter tall building from the top of multistoried building are 30â° and 45â° respectively.find the height of multistoried building and the distance between two buildings (?3=1.73)
given that,l ‖ m and n cuts themlet,∠1 = 65o∠2 = x∠3 = 40o∠1 = ∠4 = 65o (alternate angle) (i)∠3 + ∠4 + ∠5 = 180o (angle sum property)40o + 65o + ∠5 = 180o∠5 = 75onow,∠2 + ∠5 = 180o (linear pair)x + 75o = 180ox = 105o let, y = (1 – 2 tan x)(5 + 4 sin x)we have to find dy/dxas we can observe that y is a product of two functions say u and v where,u = (1 – 2tan x) and v = (5 + 4sin x)∴ y = uvas we know that to find the derivative of product of two function we apply product rule of differentiation.by product rule, we have –…equation 1as, u = (1 – 2tan x)∴ ⇒ ⇒ …..equation 2 {∵ }as, v = 5 + 4sin x⇒ ⇒ …equation 3 {∵ }∴ from equation 1, we can find dy/dx∴ using equation 2 & 3, we get –⇒ ⇒ ∵ sin x = tan x cos x , so we get –⇒ hence, slope of the line ⇒ slope of the lineequation of line :⇒ 3(y - 5) = 2(x - 2)⇒ 2x - 3y + 11 = 0 …equation (i)putting x = x + 3 and y = y + 2 in the above equation we get2(x + 3) - 3(y + 2) + 11 = 0⇒ 2x - 3y + 11 = 0since it again gives the same equation as (i) so it is a point on this line.hence proved given: (i) xy = 4.5cm

(ii) yz = 5 cm

(iii) zx = 6 cm

to construct: a xyz

steps for construction

step 1 : first we will draw a line segment xy of length 4.5cm



step 2 : we have to draw an arc , taking x as center and radius 6cm



step 3 : we have to another arc, with radius 5cm and center at y and mark the intersection of arcs as z.



step 4 : we have to join xz and xy



this is the required triangle `sin^2 30° cos^2 45 ° + 4 tan^2 30° + 1/2 sin^2 90° − 2 cos^2 90° + 1/24 cos^2 0°` ....(i) by trigonometric ratios we have `sin 30^@ = 1/2` `cos 45^@ = 1/sqrt2` `tan 30^2 = 1/sqrt3` `sin 90^@ = 1 cos 90^@ = 0 cos 0^@ = 1` by substituting above values in (i), we get `[1/2]^2 . [1/sqrt2]^2 + 4[1/sqrt3]^2 + 1/2[1]^2 - 2[0]^2 + 1/24 [1]^2` `1/4.1/2 + 4/ 1/3 + 1/2 - 0 + 1/24` `1/8 + 4/3 + 1/2 + 1/24 = 48/24 = 2` (a) g(x) is a horizontal translation of 2 units to the left then a reflection over the x-axis of f(x)
  

(b) h(x) is a horizontal translation of 3 units to the right then a vertical translation os 4 units up of graph of f(x). Correct option is A) 9.378 m2


![](https://lh3.googleusercontent.com/-UueuqyijXPQ/YLtK0v0m9CI/AAAAAAAABro/xFcfQOQAOx8jHAtpqiORvW_8DRFX1yK3ACLcBGAsYHQ/image.png)


result must have significant figures equal to least figures being multiplid do area of metal sheet 9.378 m2  (four significant digits) The given case is SAS (two sides and included angle) so use the Law of Cosines:
  

\(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{a}{c}{\cos{{B}}}\)
  

\(\displaystyle{b}^{{2}}={5}^{{2}}+{3}^{{2}}-{2}{\left({5}\right)}{\left({3}\right)}{\cos{{92}}}°\)
  

\(\displaystyle{b}^{{2}}={34}-{30}{\cos{{92}}}°\)
  

\(\displaystyle{b}=\sqrt{{{34}-{30}{\cos{{92}}}°}}\)
  

b ~ 5.92 finding the correct answer here is important, but perhaps equally important is finding it using the method of matrix reduction, since those are the instructions.the augmented matrix is4 -2 -5 -514 -1 -2 -364 -2 4 -24you'll have to forgive the poor formatting. this answer box doesn't have a proper math symbol editor.using elementary row operations, the reduced row echelon form of the augmented matrix is1 0 0 -60 1 0 60 0 1 3from which we conclude that the system has a unique solution (-6,6,3). hello again curtin,you will need to visit the water parks site and click on each attraction (for each of the water parks)that you are interested in. each one will list their height requirements or other special restrictions. an example of this is the downhill double dipper (cool dude!), which is located at blizzard beach and has a height requirement of 48 (122cm) or taller (listed underneath the description of the attraction). one of my favorites at typhoon lagoon is humunga kowabunga with a height restriction of 48 (122cm) or taller as well. enjoy your time at the water park's and remember to use that sunscreen...it can get real hot there. have a magical day!brigitta