CBSE 10th Maths Chapter 8 - Introduction to Trigonometry-Exercise 8.1

Solution:

In a given triangle ABC, right angled at B = $\angle$B = 90$^\circ$

Given: AB = 24 cm and BC = 7 cm

According to the Pythagoras Theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

By applying Pythagoras theorem, we get

AC$^{2}$=$AB^{2}+BC^{2}$

AC$^{2}$ = $(24)^{2}+7^{2}$

AC$^{2}$ = (576+49)

AC$^{2}$ = 625cm$^{2}$

AC = $\sqrt{625}$ = 25

Therefore, AC = 25 cm

(i) To find Sin (A), Cos (A)

We know that sine (or) Sin function is equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes

Sin (A) = $\dfrac{Opposite side }{Hypotenuse}$ = $\dfrac{BC}{AC}$ = $\dfrac{7}{25}$

Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,

Cos (A) = $\dfrac{Adjacent side}{Hypotenuse}$ = $\dfrac{AB}{AC}$ = $\dfrac{24}{25}$

(ii) To find Sin (C), Cos (C)

Sin (C) = $\dfrac{AB}{AC}$ = $\dfrac{24}{25}$

Cos (C) = $\dfrac{BC}{AC}$ = $\dfrac{7}{25}$

Solution:

In the given triangle PQR, the given triangle is right angled at Q and the given measures are:

PR = 13cm,

PQ = 12cm

Since the given triangle is right angled triangle, to find the side QR, apply the Pythagorean theorem

According to Pythagorean theorem,

In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.

PR$^{2}$ = $QR^{2}+ PQ^{2}$

Substitute the values of PR and PQ

13$^{2}$ = $QR^{2}+12^{2}$

169 = QR$^{2}$+144

Therefore, QR$^{2}$ = 169−144

QR$^{2}$= 25

QR = $\sqrt{25}$ = 5

Therefore, the side QR = 5 cm

To find tan P – cot R:

According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes

tan (P) = $\dfrac{Opposite side }{Adjacent side}$ = $\dfrac{QR}{PQ}$ = $\dfrac{5}{12}$

Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,

Cot (R) = $\dfrac{Adjacent side}{Opposite side}$ = $\dfrac{QR}{PQ}$ = $\dfrac{5}{12}$

Therefore,

tan (P) – cot (R) = $\dfrac{5}{12}$ – $\dfrac{5}{12}$ = 0

Therefore, tan(P) – cot(R) = 0

Solution:

Let us assume a right angled triangle ABC, right angled at B

Given: Sin A = $\dfrac{3}{4}$

We know that, Sin function is equal to the ratio of length of the opposite side to the hypotenuse side.

Therefore, Sin A = $\dfrac{Opposite side }{Hypotenuse}$= $\dfrac{3}{4}$

Let BC be 3k and AC will be 4k

where k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

AC$^{2}$=AB$^{2}$ + BC$^{2}$

Substitute the value of AC and BC

(4k)$^{2}$=AB$^{2}$ + (3k)$^{2}$

16k$^{2}$−9k$^{2}$=AB$^{2}$

AB$^{2}$=7k$^{2}$

Therefore, AB = $\sqrt{7}k$

Now, we have to find the value of cos A and tan A

We know that,

Cos (A) = $\dfrac{Adjacent side}{Hypotenuse}$

Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get

$\dfrac{AB}{AC}$ = $\dfrac{\sqrt{7}k}{4k}$ = $\dfrac{\sqrt{7}}{4}$

Therefore, cos (A) = $\dfrac{\sqrt{7}}{4}$

tan(A) = $\dfrac{Opposite side}{Adjacent side}$

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

$\dfrac{BC}{AB}$ = $\dfrac{3k}{\sqrt{7}k}$ = $\dfrac{3}{\sqrt{7}}$

Therefore, tan A = $\dfrac{3}{\sqrt{7}}$

Solution:

Let us assume a right angled triangle ABC, right angled at B

Given: 15 cot A = 8

So, Cot A = $\dfrac{8}{15}$

We know that, cot function is the equal to the ratio of length of the adjacent side to the opposite side.

Therefore, cot A = $\dfrac{Adjacent side}{Opposite side}$ = $\dfrac{AB}{BC}$ = $\dfrac{8}{15}$

Let AB be 8k and BC will be 15k

Where, k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

$AC^{2}$=$AB^{2} + BC^{2}$

Substitute the value of AB and BC

$AC^{2}$= $(8k)^{2}+ (15k)^{2}$

$AC^{2}$= $64k^{2} + 225k^{2}$

$AC^{2}$= $289k^{2}$

Therefore, AC = 17k

Now, we have to find the value of sin A and sec A

We know that,

Sin (A) = $\dfrac{Opposite side }{Hypotenuse}$

Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get

Sin A = $\dfrac{BC}{AC}$ = $\dfrac{15k}{17k}$ = $\dfrac{15}{17}$

Therefore, sin A = $\dfrac{15}{17}$

Since secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side.

Sec (A) = $\dfrac{Hypotenuse}{Adjacent side}$

Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,

$\dfrac{AC}{AB}$ = $\dfrac{17k}{8k}$ = $\dfrac{17}{8}$

Therefore sec (A) = $\dfrac{17}{8}$

Solution:

We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side

Let us assume a right angled triangle ABC, right angled at B

sec θ = $\dfrac{13}{12}$ = $\dfrac{Hypotenuse}{Adjacent side}$ = $\dfrac{AC}{AB}$

Let AC be 13k and AB will be 12k

Where, k is a positive real number.

According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,

$AC^{2}$=$AB^{2} + BC^{2}$

Substitute the value of AB and AC

$(13k)^{2}$= $(12k)^{2} + BC^{2}$

$169k^{2}$= $144k^{2} + BC^{2}$

$169k^{2}$= $144k^{2} + BC^{2}$

$BC^{2}$ = $169k^{2} – 144k^{2}$

$BC^{2}$= $25k^{2}$

Therefore, BC = 5k

Now, substitute the corresponding values in all other trigonometric ratios

So,

Sin θ = $\dfrac{Opposite Side}{Hypotenuse}$ =$\dfrac{ BC}{AC}$ = $\dfrac{5}{13}$

Cos θ = $\dfrac{Adjacent Side}{Hypotenuse}$ = $\dfrac{AB}{AC}$ = $\dfrac{12}{13}$

tan θ = $\dfrac{Opposite Side}{Adjacent Side}$ = $\dfrac{BC}{AB}$ = $\dfrac{5}{12}$

Cosec θ = $\dfrac{Hypotenuse}{Opposite Side}$ = $\dfrac{AC}{BC}$ = $\dfrac{13}{5}$

cot θ = $\dfrac{Adjacent Side}{Opposite Side}$ = $\dfrac{AB}{BC}$ = $\dfrac{12}{5}$

Solution:

Let us assume the triangle ABC in which CD$\perp$AB

Give that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cos ratio is written as

$\dfrac{AD}{AC}$ = $\dfrac{BD}{BC}$

Now, interchange the terms, we get

$\dfrac{AD}{BD}$ = $\dfrac{AC}{BC}$

Let take a constant value

$\dfrac{AD}{BD}$ = $\dfrac{AC}{BC}$ = k

Now consider the equation as

AD = k BD …(1)

AC = k BC …(2)

By applying Pythagoras theorem in △CAD and △CBD we get,

$CD^{2}$ = $BC^{2}– BD^{2}$ … (3)

$CD^{2}$ =$AC^{2} −AD^{2}$ ….(4)

From the equations (3) and (4) we get,

$AC^{2}−AD^{2}$ = $BC^{2}−BD^{2}$

Now substitute the equations (1) and (2) in (3) and (4)

$K^{2}(BC^{2}−BD^{2})$=$(BC^{2}−BD^{2}) k^{2}$=1

Putting this value in equation, we obtain

AC = BC

$\angle$A=$\angle$B (Angles opposite to equal side are equal-isosceles triangle)

Solution:

Let us assume a $\triangle$ABC in which $\angle$B = 90$^\circ$ and $\angle$C = θ

Given:

cot θ = $\dfrac{BC}{AB}$ = $\dfrac{7 }{8}$

Let BC = 7k and AB = 8k, where k is a positive real number

According to Pythagoras theorem in $\triangle$ABC we get.

$AC^{2}$ = $AB^{2}+BC^{2}$

$AC^{2}$ = $(8k)^{2}+(7k)^{2}$

$AC^{2}$ = $64k^{2}+49k^{2}$

$AC^{2}$ = $113k^{2}$

AC = $\sqrt{113}k$

According to the sine and cos function ratios, it is written as

sin θ = $\dfrac{AB}{AC}$ = $\dfrac{Opposite Side}{Hypotenuse}$ = $\dfrac{8k}{\sqrt{113}k}$= $\dfrac{8}{\sqrt{113}}$ and

cos θ = $\dfrac{Adjacent Side}{Hypotenuse}$ = $\dfrac{BC}{AC}$ = $\dfrac{7k}{\sqrt{113}k}$ = $\dfrac{7}{\sqrt{113}}$

Now apply the values of sin function and cos function:

(i) $\dfrac{(1+sin θ)(1-sin θ)}{(1+cos θ)(1-cos θ)}$=$\dfrac{1-sin^{2} θ}{1-cos^{2} θ}$ =$\dfrac{1-(\dfrac{9}{\sqrt{113}})^{2}}{1-(\dfrac{7}{\sqrt{113}})^{2}}$ =$\dfrac{49}{64}$

(ii)$cot^{2}θ$=$(\dfrac{7}{8})^{2}$=$\dfrac{49}{64}$

Solution:

Let $\triangle$ABC in which $\angle$B=90°

We know that, cot function is the reciprocal of tan function and it is written as

cot(A) = $\dfrac{AB}{BC}$ = $\dfrac{4}{3}$

Let AB = 4k an BC =3k, where k is a positive real number.

According to the Pythagorean theorem,

$AC^{2}$=$AB^{2}+BC^{2}$

$AC^{2}$=$(4k)^{2}+(3k)^{2}$

$AC^{2}$=$16k^{2}+9k^{2}$

$AC^{2}$=$25k^{2}$

AC=5k

Now, apply the values corresponding to the ratios

tan(A) = $\dfrac{BC}{AB}$ = $\dfrac{3}{4}$

sin (A) = $\dfrac{BC}{AC}$ = $\dfrac{3}{5}$

cos (A) = $\dfrac{AB}{AC}$ = $\dfrac{4}{5}$

Now compare the left hand side(LHS) with right hand side(RHS)

L.H.S=$\dfrac{1-tan^{2}A}{1+tan^{2}A}$=$\dfrac{1-(\dfrac{3}{4})^{2}}{1+(\dfrac{3}{4})^{2}}$=$\dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}$=$\dfrac{7}{25}$

R.H.S=$cos^{2}A$=$sin^{2}A$=$(\dfrac{4}{5})^{2})-(\dfrac{3}{5})^{2})$=$\dfrac{16}{25}-\dfrac{9}{25}$=$\dfrac{7}{25}$

Since, both the LHS and RHS = $\dfrac{7}{25}$

R.H.S. =L.H.S.

Hence, $\dfrac{1-tan^{2}A}{1+tan^{2}A}$= $cos^{2}A$=$sin^{2}A$ is proved

Solution:

Let $\triangle$ABC in which $\angle$B=90$^\circ$

tan A = $\dfrac{BC}{AB}$ = $\dfrac{1}{\sqrt{3}}$

Let BC = 1k and AB = $\sqrt{3} k$,

Where k is the positive real number of the problem

By Pythagoras theorem in $\triangle$ ABC we get:

AC$^{2}$=$AB^{2}+BC^{2}$

AC$^{2}$=$(\sqrt{3} k)^{2}+(k)^{2}$

AC$^{2}$=$3k^{2}+k^{2}$

AC$^{2}$=$4k^{2}$

AC = 2k

Now find the values of cos A, Sin A

Sin A = $\dfrac{BC}{AC}$ = $\dfrac{1}{2}$

Cos A = $\dfrac{AB}{AC}$ =$\dfrac{ \sqrt{3}}{2}$

Then find the values of cos C and sin C

Sin C = $\dfrac{AB}{AC}$ = $\dfrac{\sqrt{3}}{2}$

Cos C =$\dfrac{ BC}{AC}$ = $\dfrac{1}{2}$

Now, substitute the values in the given problem

(i) sin A cos C + cos A sin C =$ (\dfrac{1}{2}) ×(\dfrac{1}{2})+ \dfrac{ \sqrt{3}}{2}×\dfrac{ \sqrt{3}}{2}$ =$\dfrac{1}{4} + \dfrac{3}{4}$ = 1

(ii) cos A cos C – sin A sin C = $(\dfrac{\sqrt{3}}{2} )(\dfrac{1}{2}) – (\dfrac{1}{2}) (\dfrac{\sqrt{3}}{2})$ = 0

Solution:

In a given triangle PQR, right angled at Q, the following measures are

PQ = 5 cm

PR + QR = 25 cm

Now let us assume, QR = x

PR = 25-QR

PR = 25- x

According to the Pythagorean Theorem,

$PR^{2}$ = $PQ^{2} + QR^{2}$

Substitute the value of PR as x

$(25- x) ^{2}$ = $5^{2} + x^{2}$

$252 + x^{2} – 50x $= $25 + x^{2}$

$625 + x^{2} -50x -25 – x^{2}$ = 0

-50x = -600

x= $\dfrac{-600}{-50}$

x = 12 = QR

Now, find the value of PR

PR = 25- QR

Substitute the value of QR

PR = 25-12

PR = 13

Now, substitute the value to the given problem

(1) sin p = $\dfrac{Opposite Side}{Hypotenuse}$ = $\dfrac{QR}{PR}$ = $\dfrac{12}{13}$

(2) Cos p = $\dfrac{Adjacent Side}{Hypotenuse}$ = $\dfrac{PQ}{PR}$ = $\dfrac{5}{13}$

(3) tan p =$\dfrac{Opposite Side}{Adjacent side}$ = $\dfrac{QR}{PQ}$ = $\dfrac{12}{5}$