10th Standard Syllabus & Materials

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TN 10th Tamil உயிரின்ஓசை - துணைப்பாடம் -பிருமம் Sample Question Papers Study Material - QB365 Set A

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TN 10th Tamil அறம், தத்துவம், சிந்தனை - கவிதை பேழை (செய்யுள்) -தேம்பாவணி * Sample Question Papers Study Material - QB365 Set A

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10th Maths - Trigonometry Five Marks Questions

Published on: 09/10/2019

Trigonometry

Download Tamil Nadu 10th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.

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Std: 10th Standard - Maths

chapter: - Model Question paper

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Questions:

Generic 5

10 x 5 = 50

The shadow of a tower, when the angle of elevation of the sum is 45^o is found to be 10 metres, longer than when it is 60^o. find the height of the tower

The angle of elevation of a tower at a point is 45^o, After going 20 meters towards the foot of the tower the angle of elevation of the tower becomes 60^o calculate the height of the tower.

If tanθ+sinθ=P; tanθ-sinθ=q P.T P²-q²=4\(\sqrt{pq}\)

\(P.T\left( \frac { 1+{ tan }^{ 2 }A }{ 1+{ cot }^{ 2 }A } \right) ={ \left( \frac { 1-tan\quad A }{ 1-cot\quad A } \right) }^{ 2 }={ tan }^{ 2 }A\)

If 15tan²θ+4 sec²θ=23 then find the value of (secθ+cosecθ)²-sin² θ

P.T (1+tan∝tan∝tanβ)²+(tan∝-tanβ)²=sec²∝sec²β.

If ATB=90^o then prove that
\(\sqrt { \frac { tanA\quad tanB+tanA\quad cotB }{ sinA\quad secB } } -\frac { { Sin }^{ 2 }A }{ { Cos }^{ 2 }A } =tanA\)

If sin 3A = cos (A - 26°), where 3A is an acute angle, find the value at A.

Express the ratios cos A, tan A and sec A in terms of sin A.

10.

If sin (A - B) = \(\frac12\), cos (A + B) = \(\frac12\), 0^o < A + ≤ 90°, A > B, find A and B.

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Answers Key:

Generic 5

10 x 5 = 50

We are given that sin 3A = cos (A - 26°) ...(1)
Since sin 3A = cos(90° - 3A) we can write (1) as
cos (90° - 3A) = cos (A - 26°)
Since 90° - 3A and A - 26° are both acute angles
90° - 3A = A - 26°
which gives A = 29°

Since
cos²A + sin²A = 1 therefore
cos²A = 1 - sin²A
i.e., cos A = 土 \(\sqrt { 1-{ sin }^{ 2 }A } \)
This gives cos A = \(\sqrt { 1-{ sin }^{ 2 }A } \)
Hence, \(tanA=\frac { sinA }{ cosA } =\frac { sinA }{ \sqrt { { 1-sin }^{ 2 }A } } \)
and \(secA=\frac { 1 }{ cosA } =\frac { 1 }{ \sqrt { 1-{ sin }^{ 2 }A } } \)

10.

Since, sin(A - B) = \(\frac12\), ∴ A-B = 30° ..... (1)
Also, since cos (A + B) = \(\frac12\)
∴ A + B = 60° ...(2)
Solving (1) and (2)
A - B + A + B = 30^o + 60^o
2A = 90^o
A = 45^o
We get,
A = 45° and B = 15°