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40)Questions realted to Trignometry A2 ..
Q40 the underlined part
Q44 (ii) b
Q45
Q46
Answers :
ZaqZainab
Thought blocker
RoOkaYya G
R = square root ( 1^2 + 1 ^2) = square root 2
alpha :
tan alpha = 1/1
alpha=tan inverse 1
= pi/4
Rcos (theta + alpha) = square root 2 (theta + pi/4) <----------ANSWER
cos theta + 1/ root 2 = sin theta
cos theta- sin theta = - 1 / root 2
Rcos (theta + alpha) = - 1 / root 2
square root 2 (theta + pi/4) = - 1 / root 2
solve for theta :
cos (theta+ pi/4) = - 0.5
(theta+ pi/4) = cos inverse 0.5
(theta+ pi/4) = pi/3
cos is negative so in quadrant S and T
in quadrant S (theta+ pi/4) = (pi -pi/3) = 2pi/3
in quadrant T (theta+ pi/4) = pi + pi/3 = 5 pi/3
theta = (2pi/3 - pi/4) and (5pi/3- pi/4)
=5pi/ 12 and 14pi/12
for acute angle, theta = 5pi/12 <---------------- ANSWER
44) ii) b)
solving part i) ull get it in R cos form as :
25sin(theta +73.7)....using this derive 25 cos (theta -73.7)
25 cos (theta-73.7) = -3
cos (theta-73.7) = -3/25
(theta-73.7) = cos inverse (3/25)
since cos is negative the solutions r found in S andT quadrant
in S, (theta-73.7) = 180 -83.1 = 96.9
in T,(theta-73.7) =180+ 83.1 = 263.1
(theta-73.7) =96.9 and 263.1
theta =(96.9+73.7) and (263.1 +73.7)
theta= 170.6 and 336.8
theta = 170.6 for 0< theta< 180 <-----------ANSWER
45)
equate the give equation of tan theta to 1/5
cross multiply
ull get x^2 +384 =5(8x)
x^2 + 384-40x =0
x^2 -40x +384 = 0
solving the quadratic :
(x-16)(x-24) =0
x= 16 and x=24 <------------ANSWER
ill post 46 in a while dinner