Problem Set 8 : Standard IX : Maths II : Trigonometry

Choose the correct alternative answer for following multiple choice questions:

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(A) sin θ = cos (90 − θ)

D = 1

2tan 45° + cos 45° − sin 45°

= 2(1) + \(\displaystyle \frac{\sqrt{2}}{2}\) − \(\displaystyle \frac{\sqrt{2}}{2}\)

= 2 + 0

= 2

∴ The correct option is: (C) 2

\(\displaystyle \frac{\text {cos }28^{\circ}}{\text {sin } 62^{\circ}}\)

= \(\displaystyle \frac{\text {sin } (90^{\circ} - 28^{\circ})}{\text {sin }62^{\circ}}\) ... [ ∵ \(\displaystyle \text {cos }\theta = {\text {sin } (90 - \theta)}\) ]

= \(\displaystyle \frac{\text {cos }28^{\circ}}{\text {cos } 28^{\circ}}\)

= \(\displaystyle 1\)

∴ The correct option is: (D) 1

In right angled \(\triangle\)TSU,
TS = 5, \(\angle\)S = 90°, SU = 12
then find sin T, cos T, tan T.
Similarly, find sin U, cos U, tan U.

Solution:

In right angled \( \triangle \text {TSU} \),

TU² = TS² + SU² ... (Pythagoras’ Theorem)

∴ TU² = 5² + 12²

∴ TU² = 25 + 144

∴ TU² = 169

∴ TU = \(\displaystyle \sqrt {169}\)

∴ TU = 13 ... (i)

Now,

\(\displaystyle \sin \text{T} = \frac {\text{SU}}{\text {TU}} = \frac {12}{13}\)

\(\displaystyle \cos \text{T} = \frac {\text{ST}}{\text {TU}} = \frac {5}{13}\)

\(\displaystyle \tan \text{T} = \frac {\text{SU}}{\text {ST}} = \frac {12}{5}\)

Also,

\(\displaystyle \sin \text{U} = \frac {\text{ST}}{\text {TU}} = \frac {5}{13}\)

\(\displaystyle \cos \text{U} = \frac {\text{SU}}{\text {TU}} = \frac {12}{13}\)

\(\displaystyle \tan \text{U} = \frac {\text{ST}}{\text {SU}} = \frac {5}{12}\)

In right angled \(\triangle\)XYZ,
\(\angle\)X = 90°, XZ = 8 cm, YZ = 17 cm.
Find sin Y, cos Y, tan Y,
sin Z, cos Z, tan Z.

Solution:

In right angled \( \triangle \text {ZXY} \),

XZ² + XY² = YZ² ... (Pythagoras’ Theorem)

∴ 8² + XY² = 17²

∴ 64 + XY² = 289

∴ XY² = 289 − 64

∴ XY² = 225

∴ XY = \(\displaystyle \sqrt {225}\)

∴ XY = 15 cm ... (i)

Now,

\(\displaystyle \sin \text{Y} = \frac {\text{XZ}}{\text {YZ}} = \frac {8}{17}\)

\(\displaystyle \cos \text{Y} = \frac {\text{XY}}{\text {YZ}} = \frac {15}{17}\)

\(\displaystyle \tan \text{Y} = \frac {\text{XZ}}{\text {XY}} = \frac {8}{15}\)

Also,

\(\displaystyle \sin \text{Z} = \frac {\text{XY}}{\text {YZ}} = \frac {15}{17}\)

\(\displaystyle \cos \text{Z} = \frac {\text{XZ}}{\text {YZ}} = \frac {8}{17}\)

\(\displaystyle \tan \text{Z} = \frac {\text{XY}}{\text {XZ}} = \frac {15}{8}\)

In right angled \(\triangle\)LMN,
if \(\angle\)N = θ,
\(\angle\)M = 90°, \(\displaystyle \text{cos }\theta = \frac{24}{25}\), find sin θ and tan θ.
Similarly, find (sin² θ ) and (cos² θ ).

Solution:

cos θ = \(\displaystyle \frac {\text{MN}}{\text {LN}}\)

But, cos θ = \(\displaystyle \frac {24}{25}\) ... (Given)

∴ \(\displaystyle \frac {\text{MN}}{\text {LN}}\) = \(\displaystyle \frac {24}{25}\)

Let, MN = 24k and LN = 25k.

Now, using Pythagoras’ Theorem, we can find the length of LM.

MN² + LM² = LN²

∴ (24k)² + LM² = (25k)²

∴ 576k² + LM² = 625k²

∴ LM² = 625k² − 576k²

∴ LM² = 49k²

∴ LM = \(\displaystyle \sqrt {49k^2}\)

∴ LM = 7k ... (i)

Now, sin θ = \(\displaystyle \frac {\text{LM}}{\text {LN}}\)

∴ sin θ = \(\displaystyle \frac {7k}{25k}\)

∴ sin θ = \(\displaystyle \frac {7}{25}\) ... (ii)

∴ sin² θ = \(\displaystyle \left(\frac {7}{25}\right)^2\)

∴ sin² θ = \(\displaystyle \frac {49}{625}\) ... (iii)

And tan θ = \(\displaystyle \frac {\text{LM}}{\text {MN}}\)

∴ tan θ = \(\displaystyle \frac {7k}{24k}\)

∴ tan θ = \(\displaystyle \frac {7}{24}\) ... (iv)

Also, cos θ = \(\displaystyle \frac {24}{25}\) ... (Given)

∴ cos² θ = \(\displaystyle \left(\frac {24}{25}\right)^2\)

∴ cos² θ = \(\displaystyle \frac {576}{625}\) ... (v)

Fill in the blanks:

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sin 20° = cos 70°

tan 30° × tan 60° = 1

cos 40° = sin 50°

Choose the correct alternative answer for following multiple choice questions: Click on the question to view the answer

Which of the following statements is true? sin θ = cos (90 − θ) cos θ = tan (90 − θ) sin θ = tan (90 − θ) tan θ = tan (90 − θ)

Which of the following is the value of sin 90°? \(\displaystyle \frac{\sqrt{3}}{2}\) 0 \(\displaystyle \frac{1}{2}\) 1

2tan 45° + cos 45° − sin 45° = ? 0 1 2 3

\(\displaystyle \frac{\text {cos }28^{\circ}}{\text {sin } 62^{\circ}}\) = ? 2 − 1 0 1

In right angled \(\triangle\)TSU,TS = 5, \(\angle\)S = 90°, SU = 12then find sin T, cos T, tan T. Similarly, find sin U, cos U, tan U.

Solution:

In right angled \(\triangle\)XYZ,\(\angle\)X = 90°, XZ = 8 cm, YZ = 17 cm.Find sin Y, cos Y, tan Y, sin Z, cos Z, tan Z.

Solution:

In right angled \(\triangle\)LMN,if \(\angle\)N = θ,\(\angle\)M = 90°, \(\displaystyle \text{cos }\theta = \frac{24}{25}\), find sin θ and tan θ.Similarly, find (sin² θ ) and (cos² θ ).

Solution:

Fill in the blanks: Click on the question to view the answer

sin 20° = °

tan 30° × ° = 1

cos 40° = °

Choose the correct alternative answer for following multiple choice questions:

Click on the question to view the answer

In right angled \(\triangle\)TSU,
TS = 5, \(\angle\)S = 90°, SU = 12
then find sin T, cos T, tan T.
Similarly, find sin U, cos U, tan U.

In right angled \(\triangle\)XYZ,
\(\angle\)X = 90°, XZ = 8 cm, YZ = 17 cm.
Find sin Y, cos Y, tan Y,
sin Z, cos Z, tan Z.

In right angled \(\triangle\)LMN,
if \(\angle\)N = θ,
\(\angle\)M = 90°, \(\displaystyle \text{cos }\theta = \frac{24}{25}\), find sin θ and tan θ.
Similarly, find (sin² θ ) and (cos² θ ).

Fill in the blanks:

Click on the question to view the answer