In figure ∆ ABC is a right angled at B
∆ ℎ ∆BRS
Is right angled at R. AB =18 cm, BC = 7.5 cm RS = 5
cm ∟BSR = x0 and ∟SAB = y0
i) The value of tan x is
a) 5/13 b) 12/13 c) 6/5 d)5/6
1
ii) The value of sin y is
a) 3/13 b)5/13 c)7/13 d) 4/13
1
iii) The value of cos y is
a) 11/13 b) 12/13 c) 5/13 d) 6/5
1
iv) Length AS is
a) 13 b) 12 c) 6 d)none
1
v) ∟SRB=?
a) 90 b) 120 c)150 d)180
Answers
Answer
In the given figure, triangle ABC is right angled at B. D is the foot of the perpendicular from B to AC. Given that BC=3cm and AB=4cm. find:
i) tan∠DBC
ii) sin∠DBA
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ANSWER
Let CD=y
then AD=(5−y)
and BD=x
In △BCD
CD
2
+BD
2
=BC
2
(pythagoras theorem)
⇒y
2
+x
2
=9−(1)
In △ABD
AD
2
+BD
2
=16
⇒(5−y)
2
+x
2
=16−(2)
Subtracting (2) from (1) we get
(5−y)
2
−y
2
=7
⇒25+y
2
−10y−y
2
=7
⇒10y=18
⇒y=1.8
∴CD=1.8,AD=3.2,BD=2.4
Now,
(i)tan∠DBC=
BD
CD
=
2.4
1.8
=
4
3
(ii)sin∠DBA=
AB
AD
=
4
3.2
=
5
4
solution
Answer:
10th
Maths
Introduction to Trigonometry
Trigonometric Ratios
In a right angle triangle A...
MATHS
In a right angle triangle ABC right angle at B , ∠ABC=θ , AB = 2cm and BC = 1cm . Find the value of sin
2
θ+tan
2
θ
MEDIUM
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ANSWER
We have,
AB=2cm and BC=1cm
Using Pythagoras theorem in ΔCAB, we have
AC
2
=AB
2
+BC
2
⇒AC
2
=2
2
+1=5
⇒AC=
5
Thus, Base=BC=1cm, Perpendicular =AB=2cm
and, Hypotenuse =AC=
5
cm
sinθ=
AC
AB
=
5
2
, and tan θ=
BC
AB
=
1
2
=2
Now, sin
2
θ+tan
2
θ=(sinθ)
2
+(tanθ)
2
=(
5
2
)
2
+(2)
2
=
5
4
+4=
5
24
solution