Simplify cos2x(sec2x+sec2xtan2x)
tan x
tanx secx
sec2x
cosec2x
Correct answer is C
cos2x(sec2x+sec2xtan2x)
= cos2xsec2x+cos2xsec2xtan2x
= 1+tan2x
= 1+sin2xcos2x
= cos2x+sin2xcos2x
= 1cos2x=sec2x
60 (tan 62o - tan 64o)
60 (cot 64o - cot 62o)
60 (cot 62o - cot 64o)
60 (tan 64o - tan 62o)
Correct answer is D
BC60 = tan621
BC = 60 tan 62
AC60 = tan621
AC = 60 tan 64
AB = AC - BC
= 60(tan 64o - tan 62o)
If the exterior angles of a pentagon are x°, (x + 5)°, (x + 10)°, (x + 15)° and (x + 20)°, find x
118o
72o
62o
36o
Correct answer is C
The sum of the exterior angles of a regular polygon = 360°.
\therefore x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360°
5x + 50 = 360° \implies 5x = 360° - 50° = 310°
x = \frac{310°}{5} = 62°
8\sqrt{3}
\frac{16}{\sqrt{3}}
\sqrt{3}
\frac{10}{\sqrt{3}}
Correct answer is A
\frac{x}{4} = \frac{\tan 60}{1}
x = 4 tan60
= 4\sqrt{3}
BD = 2x
= 8\sqrt{3}
The area of a square is 144 sq cm. Find the length of its diagonal
11\sqrt{3cm}
12cm
12\sqrt{2cm}
13cm
Correct answer is C
BD = \sqrt{x^2 + x^2}
= \sqrt{12^2 + 12^2}
= \sqrt{144 + 144}
= 2(144)
= 12\sqrt{2cm}