Maths Study Material and Solved Assignments

Maths

TASK 1
1.1 In ∆ABC, A = 530 , B = 610 and length of a = 12.6cm
So, b and c can be calculated by following way -
using law of sine -
a = b = c
Sin A Sin B Sin C
from, given ∆ -
A + B + C = 180
530 + 610 + C = 180
C = 660
therefore,
12.6 = b = c
Sin 530 Sin 610 Sin 660
12.6 = b = c
0.79 0.87 0.91
so, b = 13.87 cm
while, c = 14.51 cm
B
Here, perpendicular = 15 unit
hypotenuse = 17 unit
then, base can be determined by -

b2 = h2 – p2
= 172 – 152
= 289 – 225
= 64
or, b = √64 = 8 unit
Sin Ø = perpendicular / hypotenuse
Sin Ø = 15/17 unit
TASK 2
a) Amplitude and period of waveform
y = 4 cos(2θ + 450)
Amplitude of given equation is 4
and, period of above equation = 2 π / 2θ = π / θ
a) Amplitude and period of waveform
y = 6 sin (t - 300)
Amplitude of given equation is 6
and, period of above equation = 2 π / t
c) To Prove -
sin 2 x (sec x + cosec x) = 1 + tan x
cosx tanx
taking LHS side of given equation-
sin 2 x (sec x + cosec x)
cosx tanx
» sin 2 x (1/cosx + 1/sinx )
cox . sinx/cosx
» sin 2 x (sin x + cos x)/ sinx cosx
cox . sinx/cosx
» sinx + cos x
cox
» tanx + 1 = RHS Hence Proved
2

Task 2
Given Indical equation -
x2 (2x-3)
(32) + (3) = 27
2x2 (2x-3) 3
(3) + (3) = 3
Using exponent rules,
2x2 + 2x – 3 3
3 = 3
3