Trigonometrical Ratios

Trigonometric Ratios

.

There are six trigonometric ratios which relate the sides of a right triangle to its angles. Trigonometric ratios are generally used to calculate the unknown lengths and angles in a right triangle. The six trigonometric ratios are tabulated below: -

1) Sinθ \(\frac{perpendicuar}{hypotenuse}\) (\(\frac{p}{h}\))
2) Cosθ \(\frac{base}{hypotenuse}\) (\(\frac{b}{h}\))
3) Tanθ \(\frac{perpendicuar}{base}\) (\(\frac{p}{b}\))
4) Cosecθ \(\frac{hypotenuse}{perpendicuar}\) (\(\frac{h}{p}\))
5) Secθ \(\frac{hypotenuse}{base}\) (\(\frac{h}{b}\))
6) Cotθ \(\frac{base}{perpendicuar}\) (\(\frac{b}{p}\))

NOTE: - Since, Sinθ ≠ Sin×θ, Cosθ ≠ Cos×θ, etc.

Operation of Trigonometric Ratios

As we know that the trigonometric ratios are numbers like fraction, decimal or whole numbers. We can operate the operations of addition, subtraction, multiplication and division on them that we do in algebra.

Operations In algebra In Trigonometry
Addition 2a + 4a = 6a
4x2 + 7x2 = 11x2
2sinθ + 4sinθ = 6sinθ
4cos2θ + 7cos2θ = 11cos2θ
Subtraction 11z − 6z = 5z
8x2− 5x2 = 3x2
11tanθ − 6tanθ = 5tanθ
8sec2θ − 5sec2θ = 3sec2θ
Multiplication 7a × 4a = 28a2
5y2× 3y2= 15y4
7sinθ × 4sinθ = 28sin2θ
5tan2θ × 4tan2θ = 20tan4θ
Division \(\frac{q^{5}}{q}\) = q4
10b ÷ 5b = 2
\(\frac{cosθ^{5}}{cosθ}\) = cosθ4
10secθ ÷ 5secθ = 2


Trigonometric Expressions

.

Trigonometric Expressions are the expressions in which variables are written under the signs of trigonometric functions.

Simplification and factorisation

We can simplify or factorise the trigonometric expressions as in algebra. For examples,

  1. (a+b) (a−b) = a2− b2
    (SinA + CosA) (SinA − CosA) = Sin2A − Cos2A
  2. (a+b)2= a2+ 2ab + b2
    (SinA + CosA)2= Sin2A + 2SinA.CosA + Cos2A
  3. a3+b3= (a+b) (a2−ab+b2)
    Sin3A + Cos3A = (SinA + CosA)(Sin2A − 2SinA.CosA + Cos2A)
  4. a3−b3= (a+b) (a2+ab+b2)
    Sin3A − Cos3A = (SinA − CosA) (Sin2A + SinA.CosA + Cos2A)



Relation between the Trigonometric Ratios of an Angle

Simply there are four fundamental relations of trigonometric ratios. They are,

  1. Reciprocal Relation
  2. Quotient Relation
  3. Pythagoras Relation
  4. Derived Relation

Reciprocal Relation

.

Reciprocal relations of trigonometric ratios are explained to represent the relationship between the three pairs of trigonometric ratios as well as their reciprocals. The reciprocals relations are given below: -

Sinθ = \(\frac{p}{h}\) and Cosecθ = \(\frac{h}{p}\)
Then, sinθ × cosecθ = \(\frac{p}{h}\) × \(\frac{h}{p}\) = 1

∴ Sinθ × cosecθ = 1


sinθ = \(\frac{1}{cosecθ }\), cosecθ = \(\frac{1}{sinθ }\)
Also, cosθ = \(\frac{b}{h}\) and secθ = \(\frac{h}{b}\)
Then, cosθ × secθ = \(\frac{b}{h}\) × \(\frac{h}{b}\) = 1

∴ cosθ × secθ = 1


cosθ = \(\frac{1}{secθ}\), secθ \(\frac{1}{cosθ}\)
And, tanθ = \(\frac{p}{b}\), cotθ = \(\frac{b}{p}\)
Then tanθ × cotθ = 1

∴ tanθ × cotθ = 1


tanθ = \(\frac{1}{cotθ}\), cotθ = \(\frac{1}{tanθ}\)

Quotient Relation
We have, sinθ = \(\frac{p}{h}\), cosθ = \(\frac{b}{h}\)
Then, \(\frac{sinθ}{cosθ}\) = \(\frac{p}{h}\) × \(\frac{h}{b}\) = \(\frac{p}{b}\)
We have tanθ = \(\frac{p}{b}\)
∴ tanθ = \(\frac{sinθ}{cosθ}\)
Similarly,
cotθ = \(\frac{cosθ}{sinθ}\)

Pythagoras Relation
From the right angled triangle ABC,
CA2= AB2 + BC2
or, h2= p2 + b2
Dividing both sides
or, \(\frac{h^{2}}{h^{2}}\) = \(\frac{p^{2}+b^{2}}{h^{2}}\)
or, 1 = \(\frac{p^{2}}{h^{2}}\) + \(\frac{b^{2}}{h^{2}}\)
or, \(\frac{p^{2}}{h^{2}}\) + \(\frac{b^{2}}{h^{2}}\) = 1
or, (\(\frac{p^{2}}{h^{2}}\)) + (\(\frac{b^{2}}{h^{2}}\)) = 1
or, (sinθ)2 + (cosθ)2 = 1
∴ sin2θ = 1 − cos2θ
cos2θ = 1 − sin2θ
Also,
sinθ = \(\sqrt{1-cos^{2}}{θ}\)
or, cosθ = \(\sqrt{1−sin^{2}}{θ}\)

Again,
p2+ b2 = h2
or,h2− p 2= b2
Dividing both sides by b2
or, \(\frac{h^{2}}{b{2}}\) - \(\frac{p^{2}}{b^{2}}\) = \(\frac{b^{2}}{b^{2}}\)
or, (\(\frac{h}{b}\))2 + (\(\frac{p}{b}\))2= 1
or, (secθ)2− (tanθ)2 = 1
or, sec2θ − tan2θ = 1
or, sec2θ = 1 + tan2θ
or, tan2θ = sec2θ − 1
Also,
secθ = \(\sqrt{1+tan^{2}}{θ}\)
or, tanθ = \(\sqrt{sec^{2} θ − 1}\)

Again,
p2+ b 2= h2
or, h2− b 2= p2
Dividing both sides by p2
\(\frac{h^{2}}{p{2}}\) − \(\frac{b^{2}}{p^{2}}\) = \(\frac{p^{2}}{p^{2}}\)
or, (cosecθ)2− (cotθ)2 = 1
or, cosec2θ − cot2θ = 1
or, cosec2θ = 1 + cot2θ
or, cot2θ = cosec2θ − 1
Also, cotθ = \(\sqrt{cosec^{2} θ − 1}\)
or, cosecθ = \(\sqrt{1+cot^{2}}{θ}\)

Methods of proving Trigonometric Identities

To prove trigonometric identities, we may follow any one of the following: -

  • Begin from the left-hand side (L.H.S.) and deduct it to the right-hand side (R.H.S.), if L.H.S. is more complex.
  • Begin from R.H.S. and deduct it to L.H.S., if L.H.S. is complex.
  • Reduce both the L.H.S. and the R.H.S. to the same expression if both the expression are complex.
  • By transposition or cross multiplication, change the identity into the appropriate form. Then show that the new L.H.S. = the new R.H.S.

Conversion of a Trigonometric Ratios

With the use of two methods, we can express the trigonometric ratios in terms of other ratios of the same angle. The two methods are as followings: -

  • Using trigonometric relations i.e Trigonometric formulae.
  • Using Pythagoras theorem.
NOTE: - We can also find remaining ratios if the value of any trigonometric ratio of an angle is given.


Trigonometric Ratios of some angles

As trigonometry have different angles, for those different angles trigonometric ratios also have different values. In the trigonometry the angles 0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, 360° are taken as standard angles.
We can find the trigonometric ratios of 0° and 90°. The following table shows the values of trigonometric ratios of the angles 0°, 30°, 45°, 60° and 90°.

Angles 30° 45° 60° 90°
sinθ 0 \(\frac{1}{2}\) \(\frac{1}{\sqrt2}\) \(\frac{\sqrt{3}}{2}\) 1
cosθ 1 \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt2}\) \(\frac{1}{2}\) 0
tanθ 0 \(\frac{1}{\sqrt3}\) 1 \(\sqrt{3}\)
cosecθ 2 \(\sqrt{2}\) \(\frac{2}{\sqrt3}\) 1
secθ 1 \(\frac{2}{\sqrt3}\) \(\sqrt{2}\) 2
cotθ \(\sqrt{3}\) 1 \(\frac{1}{\sqrt3}\) 0

Procedures to find the trigonometric ratios

The standard angles i.e 0°, 30°, 45°, 60°, 90° in prompt way is given below: -

Process 1: -
Put the number from 0 to 4 as follows: -

30° 45° 60° 90°
0 1 2 3 4

Process 2: -

Divide each number by 4

30° 45° 60° 90°
\(\frac{0}{4}\) \(\frac{1}{4}\) \(\frac{2}{4}\) \(\frac{3}{4}\) \(\frac{4}{4}\)

Process 3: -
Take square root of all the numbers

30° 45° 60° 90°
\(\sqrt{\frac{0}{4}}\) \(\sqrt{\frac{1}{4}}\) \(\sqrt{\frac{2}{4}}\) \(\sqrt{\frac{3}{4}}\) \(\sqrt{\frac{4}{4}}\)

Process 4: -
The values which are obtained at first are the values of sine of the standard angles.

Angles 30° 45° 60° 90°
sin 0 \(\frac{1}{2}\) \(\frac{1}{\sqrt2}\) \(\frac{\sqrt{3}}{2}\) 1

Process 5: -
Then reversing the order, the value of cosine of theb standard angles are obtained.

Angles 30° 45° 60° 90°
cos 1 \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt2}\) \(\frac{1}{2}\) 0

Process: -
After dividing the value of sine by the value of cosine, the values of tangent can be obtained: -

Angle 30° 45° 60° 90°
tan 0 \(\frac{1}{\sqrt3}\) 1 \(\sqrt{3}\)

NOTE: - The reciprocal of sine, cosine and tangent are the values of cosecant, secant and cotangent respectively.


Complementary Angles

Two angles are called complementary when those angles are added up to 90° or the sum of two angles is 90°.
For example,
60° and 30° are complementary angles.
5° and 85° are complementary angles.
30° and 60°, 50° and 40°, (90° - θ) and θ are complementary angles.

Trigonometric Ratios of Complementary Angles

.

Here ABC is a right angle triangle right angled at B.
Let, ∠ABC = θ, then ∠CAB = 90° - θ
Now taking θ as an angle of reference,
AB = Perpendicular (p)
BC = Base (b)
CA = Hyptenuse (h)
Sinθ = \(\frac{p}{h}\) = \(\frac{AB}{CA}\) .............. .(i)
Cosθ = \(\frac{b}{h}\) = \(\frac{BC}{CA}\) ..............(ii)
Tanθ = \(\frac{p}{b}\) = \(\frac{AB}{BC}\) ..............(iii)
Cotθ = \(\frac{b}{p}\) = \(\frac{BC}{AB}\) ...............(iv)
Taking (90° - θ) as an angle of reference,
BC = Perpendicular (p)
AB = Base (b)
CA = Hypotenuse (h)
Now,
Sinθ(90° - θ) = \(\frac{p}{h}\) = \(\frac{BC}{CA}\) = Cosθ
Cosθ(90° - θ) = \(\frac{b}{h}\) = \(\frac{AB}{CA}\) = Sinθ
Tanθ(90° - θ) = \(\frac{p}{b}\) = \(\frac{BC}{AB}\) = Cotθ
On the other hand,
cosec(90° - θ) = secθ
sec(90° - θ) = cosecθ
cot(90° - θ) = tanθ

(sinθ)2 + (cosθ)2 = 1 sec2θ = 1+tan2θ cosec2θ - cot2θ = 1
sin2θ = 1-cos2θ sec2θ - tan2θ =1 cosec2θ = 1 + cot2θ
, sinθ= \(\sqrt{1-cos^{2}}{θ}\) tan2θ= sec2θ - 1 cot2θ = cosec2θ - 1
cosθ = \(\sqrt{1-sin^{2}}{θ}\)

secθ = \(\sqrt{1+tan^{2}}{θ}\)

cotθ =\(\sqrt{cosec^{2} θ-1}\)
sec2θ - tan2θ =1 tanθ = \(\sqrt{sec^{2} θ-1}\) cosecθ = \(\sqrt{1+cot^{2}}{θ}\)

Solution:

From the given figure,

Hypotenuse (h) = 13cm

Perpendicular (p) = 5cm

Base (b) = 12cm

Now,

\(\therefore\) sinα = \(\frac{p}{h}\) = \(\frac{5}{13}\)

\(\therefore\) cosα = \(\frac{b}{h}\) = \(\frac{12}{13}\)

\(\therefore\) tanα = \(\frac{p}{b}\) = \(\frac{5}{12}\)

Solution:

From the given figure,

Hypotenuse (h) = 10cm

Perpendicular (p) = 6cm

Base (b) = 8cm

Now,

\(\therefore\) cosecθ = \(\frac{h}{p}\) = \(\frac{10}{6}\) = \(\frac{5}{3}\)

\(\therefore\) secθ = \(\frac{h}{b}\) = \(\frac{10}{8}\) = \(\frac{5}{4}\)

\(\therefore\) cotθ = \(\frac{b}{p}\) = \(\frac{8}{6}\) = \(\frac{4}{3}\)

Solution:

= 18sinθ − 7sinθ

= (18 − 7)sinθ

= 11sinθ

Solution:

L.H.S. = \(\frac{sinθ.cosecθ}{secθ}\)

= sinθ cosecθ÷ secθ

= sinθ× \(\frac{1}{sinθ}\)÷ \(\frac{1}{cosθ}\)

= sinθ× \(\frac{1}{sinθ}\)× cosθ

= cosθ

= R.H.S. proved.

Solution:

L.H.S. = secθ \(\sqrt{1− cos^2θ}\)

= \(\frac{1}{cosθ}\) . sinθ

= \(\frac{sinθ}{cosθ}\)

= tanθ

= R.H.S. proved.

Solution:

L.H.S. = sin2θ + cos2θ.tan2θ

= sin2θ + cos2θ \(\frac{sin^2θ}{cos^2θ}\)

= sin2θ + sin2θ

= 2sin2θ

= R.H.S. proved

Solution:

L.H.S. = (sinA + cosA)2 + (sinA − cosA)2

= sin2A + 2sinA.cosA + cos2A + sin2A − 2sinA.cosA + cos2A

= (sin2A + cos2A) + (sin2A + cos2A)

= 1 + 1

= 2

= R.H.S. proved

Solution:

L.H.S. = \(\frac{1}{tanA + cotA}\)

= \(\frac {1}{\frac {sinA}{cosA} + \frac {cosA}{sinA}}\)

= \(\frac{1}{\frac{sin^2A + cos^2A}{sinA.cosA}}\)

= \(\frac{1}{\frac{1}{sinA.cosA}}\)

= sinA.cosA

= R.H.S. proved

Solution:

L.H.S. = \(\frac{sin^4θ− cos^4θ}{sinθ + cosθ}\)

= \(\frac{(sin^2θ)^2 − (cos^2θ)^2}{sinθ + cosθ}\)

= \(\frac{(sin^2θ + cos^2θ) (sin^2θ− cos^2θ)}{sinθ + cosθ}\)

= \(\frac{1× (sinθ + cosθ) (sinθ− cosθ)}{sinθ + cosθ}\)

= sinθ− cosθ

= R.H.S. proved

Solution:

L.H.S. = \(\frac{cosA}{1 + sinA} + \frac{cosA}{1− sinA}\)

= \(\frac{cosA (1− sinA) + cosA (1 + sinA)}{(1 + sinA) (1− sinA)}\)

= \(\frac{cosA− sinA.cosA + sinA.cosA}{1− sin^2A}\)

= \(\frac{2cosA}{cos^2A}\)

= \(\frac{2}{cosA}\)

= 2secA

= R.H.S. proved

Solution:

L.H.S.
= \(\frac{1− sin^4}{cos4A}\)

= \(\frac{1− sin^4A}{cos^4A}\) − \(\frac{sin^4A}{cos^4A}\)

= sec4A − tan4A

= (sec2A)2− (tan2A)2

= (sec2A− tan2A) (sec2A + tan2A)

= (1 + tan2A + tan2A)

= 1 + 2tan2A

= R.H.S. proved

Solution:

L.H.S
= (xcosθ + ysinθ)2+ (xsinθ− yccosθ)2

= (xcosθ)2+ 2xysinθ.cosθ + (ysinθ)2+ (xsinθ)2− 2xysinθ.cosθ + (ycosθ)2

= x2cos2θ + y2sin2θ + x2sin2θ+ y2cos2θ

= x2(cos2θ + sin2θ) + y2(sin2θ + cos2θ)

= (x2+ y2) (cos2θ + sin2θ)

= (x2+ y2) \(\times\) 1 [\(\because\) (cos2θ + sin2θ) =1]

= (x2+ y2) R.H.S. proved

Solution:

L.H.S. =\(\sqrt\frac{1 + cosθ}{1 – cosθ}\)

= \(\sqrt{\frac{1 + cosθ}{1− cosθ}× \frac{1 + cosθ}{1− cosθ}}\)

= \(\sqrt\frac{(1 + cosθ)^2}{1− cos^2θ}\)

= \(\sqrt\frac{(1 + cosθ)^2}{sin^2θ}\)

= \(\frac{1 + cosθ}{sinθ}\)

= \(\frac{1}{sinθ} + \frac{cosθ}{sinθ}\)

= cosecθ + cotθ

= R.H.S. proved

Solution:

Let, ABC be a right angled triangle where ∠B = 90° and ∠C = θ = k

Let, sinθ = k

Then,

or, sinθ = \(\frac{k}{1}\) = \(\frac{p}{h}\)

∴ p = k, h = 1

Now, from right angled ΔABC,

or, b = \(\sqrt{h^2 − p^2}\)

or, b = \(\sqrt{1 − k^2}\)

or, b = \(\sqrt{1− sin^2θ}\)

Then,

cosecθ = \(\frac{h}{p}\) = \(\frac{1}{k}\) = \(\frac{1}{sinθ}\)

cosθ = \(\frac{b}{h}\) = \(\sqrt\frac{1− sin^2θ}{1}\) = \(\sqrt{1−sin^2θ}\)

secθ = \(\frac{h}{b}\) = \(\frac{1}{\sqrt{1− k^2}}\)

tanθ = \(\frac{p}{b}\) = \(\frac{k}{\sqrt{1− k^2}}\) = \(\frac{sinθ}{\sqrt{1− sin^2θ}}\)

cotθ = \(\frac{b}{p}\) = \(\frac{\sqrt{1− sin^2θ}}{sinθ}\)

Solutions:

Here,

cosα = \(\frac{5}{13}\) = \(\frac{b}{h}\)

Then,

b = 5, h = 13, p= ?

By using Pythagoras theorem, we have,

or, p2= h2− b2

or, p = \(\sqrt{h^2 − b^2}\)

or, p = \(\sqrt{13^2 − 5^2}\)

or, p = \(\sqrt{169 − 25}\)

or, p = \(\sqrt{144}\)

∴ p = 12

Now,

secα = \(\frac{h}{b}\) = \(\frac{13}{5}\)

sinα = \(\frac{p}{h}\) = \(\frac{12}{13}\)

cosecα = \(\frac{h}{p}\) = \(\frac{13}{12}\)

tanα = \(\frac{p}{b}\) = \(\frac{12}{5}\)

cotα = \(\frac{b}{p}\) = \(\frac{5}{12}\)

Solution:

Here,

or, sinθ − cosθ = 0

or, sinθ = cosθ

or, tanθ = 1

We have,

tanθ = \(\frac{p}{b}\) = \(\frac{1}{1}\)

Then,

p = 1, b = 1

By Pythagoras theorem,

or, h2 = p2 + b2

or, h = \(\sqrt{p^2 + b^2}\)

or, h = \(\sqrt{1^2 +1^2}\)

or, h = \(\sqrt{1 + 1}\)

or, h = \(\sqrt{2}\)

∴ h = \(\sqrt{2}\)

Now,

cosecθ = \(\frac{h}{p}\) = \(\frac{\sqrt{2}}{1}\) = \(\sqrt{2}\)

sinθ = \(\frac{p}{h}\) = \(\frac{1}{\sqrt{2}}\)

Solution:

Here,

or, sinθ = \(\frac{m}{n}\) = \(\frac{p}{h}\)

Then,

p = m, h = n

We have,

or, b2 = h2− p2

or, b2 = n2− m2

or, b = \(\sqrt{n^2 − m^2}\)

Now,

or, tanθ = \(\frac{p}{b}\)

or, tanθ = \(\frac{m}{\sqrt{n^2− m^2}}\)

or, \(\sqrt{n^2− m^2}\) tanθ = m

\(\therefore\) L.H.S = R.H.S proved

Solution:

Here,

or, 3cotθ = 4

We have,

or, cotθ = \(\frac{b}{p}\) = \(\frac{4}{3}\)

Then,

b = 4, p = 3, h = ?

By Pythagoras theorem,

or, h2 = p2 + b2

or, h2 = 32 + 42

or, h2 = 9 + 16

or, h2 = \(\sqrt{25}\)

∴ h = 5

Now,

= \(\frac{3cosθ + 2sinθ}{3cosθ− 2sinθ}\)

= {3 \((\frac{b}{h}) + 2 (\frac{p}{h})\)}÷ {3 \((\frac{b}{h})− 2 (\frac{p}{h})\)}

= (3× \(\frac{4}{5}\) + 2× \(\frac{3}{5}\))÷ (3× \(\frac{4}{5}\)− 2× \(\frac{3}{5}\))

= (\(\frac{12}{5} + \frac{6}{5}\))÷ (\(\frac{12}{5}− \frac{6}{5}\))

= \(\frac{18}{5}\)÷ \(\frac{6}{5}\)

= \(\frac{18}{5}\)× \(\frac{5}{6}\)

= 3

0%
  • sinθ is equal to ______.

    (frac{h}{p})
    (frac{p}{h})
    (frac{p}{b})
    (frac{b}{h})
  • cosθ is equal to ________.

    (frac{p}{b})
    (frac{p}{h})
    (frac{b}{h})
    (frac{b}{p})
  • tanθ is equal to _________.

    (frac{p}{b})
    (frac{b}{h})
    (frac{p}{h})
    (frac{b}{p})
  • cosecθ is equal to ______.

    (frac{b}{p})
    (frac{p}{h})
    (frac{h}{p})
    (frac{b}{h})
  • secθ is equal to _________.

    (frac{h}{b})
    (frac{h}{p})
    (frac{b}{p})
    (frac{b}{h})
  • cotθ is equal to ________.

    (frac{b}{p})
    (frac{p}{b})
    (frac{h}{b})
    (frac{b}{h})
  • Solve for x.
    (frac{15}{18})=(frac{x}{6})

    4
    3
    6
    5
  • Solve for x.
    (frac{12}{z})=(frac{8}{6})

    9
    8
    4
    6
  • Solve for y.
    (frac{1}{3})=(frac{y}{24})

    12
    16
    8
    14
  • Find the value of trigonometrical ratios of sin ( heta), if tan  ( heta) = (frac{3}{4})

    (frac{4}{5})
    (frac{3}{4})
    (frac{3}{5})
    (frac{2}{3})
  • Find the value of trigonometrical ratios of sin ( heta), if cosec  ( heta) = 2

    (frac{1}{4})
    (frac{2}{4})
    (frac{3}{4})
    (frac{1}{2})
  • Find the value of trigonometrical ratios of  cot ( heta), if sec  ( heta) = 2

    2
    (frac{1}{2})
    (frac{2}{3})
    1
  • Solve for u.

    (frac{u}{8})=(frac{27}{12})

    16
    12
    15
    18
  • While researching universities, Jack learns that a typical class at Clarksville Technical University has 15 boys and 9 girls, and a typical class at Lawrence County University has 12boys and 4 girls. Which university has classes with a lower ratio of boy to girl?

    Question incorrect
    Lawrence County University
    Clarksville Technical University
    neither; the ratios are equivalent
  • An auto dealership sells minivans and sedans. In January, they sold 18 minivans and 22 sedans. In February, they sold 9 minivans and 11 sedans. During which month did the auto dealership sell a lower ratio of minivans to sedans?

    January
    February
    neither; the ratios are equivalent
    Question Incorrect
  • You scored /15


    Take test again

Any Questions on Trigonometrical Ratios ?

Please Wait...

Discussions about this note

Forum Time Replies Report
Ankit

Solve please

In the adjoining figure find out perpendicular base hypotenuse from the given right angled triangle with a and n as the reference angles .also give reason

in the adjoinig figure find out perpendicular base hypotenuse from the given right angled triangle with a and b as reference angles. also give reason