In the given figure, find the value of cos alpha.
Answers
Answer:
cos: base / hypothesis
EC/DC
Answer:
Step-by-step explanation:
From right angled ΔACD,
By Pythagoras theorem we get
AC
2
=AD
2
+CD
2
AD
2
=AC
2
−CD
2
AD
2
=(13)
2
−(5)
2
AD
2
=169−25
AD
2
=144
AD
2
=12
2
AD=12
From right angled ΔABD,
By Pythagoras angled ΔABD
By Pythagoras theorem we get
AB
2
=AD
2
+BD
2
AB
2
=400
AB
2
=(20)
2
AB=20
(i) tan x = perpendicular/Base (in right angled ΔACD)
=CD/AD
= 5/12
(ii) cos y = Base/Hypotenuse (in right angled ΔABD)
= BD/AB
= (20)/12 – (5/3)
Cot y = Base/Perpendicular (in right ΔABD)
=BD/AB
= 16/20 = 4/5
(iii) ) cos y = Hypotenuse/ perpendicular (in right angled ΔABD)
BD/AB
= 20/12
= 5/3
Cot y = Base/Perpendicular (in right ΔABD)
AB/AD
= 16/12
= 4/3
Cosec
2
y−cot
2
y=(5/3)
2
−(4/3)
2
= (25/9) – (16/9)
= (25-16)/9
= 9/9
= 1
Hence, Cosec
2
y−cot
2
y=1
(iv) sin x = Perpendicular/Hypotenuse (in right angled ΔACD)
= AD/AB
= 12/20
= 3/5
Cot y = Base/Perpendicular (in right angled ΔABD)
= BD/AD
= 16/12
= 4/3
(5/sin x) + (3/sin y) – 3cot y
=5/(5/13)+3/(3/5)–3×4/3
=5×13/5+3×5/3–3×4/3
=1×13/1+1×5/1–1×4/1
= 13 + 5 – 4 = 18 – 4
= 14
Hence 5/sin x + 3/sin y – 3cot y = 14