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Introduction to Trigonometry
CH08 Question Bank
CBSE_CH08_QUESTION_BANK
Grade 10
Question:
[Case Study]
A triangular sail for a small sailing boat is shaped like a right triangle, with sides $5$ m, $12$ m, and $13$ m as shown, and the angle at one corner marked $\phi$.
(a) Find $\sin\phi$ and $\cos\phi$. [1 Mark]
(b) Find $\sec\phi$ and $\text{cosec}\,\phi$. [1 Mark]
(c) Verify that $1+\tan^2\phi=\sec^2\phi$ for this sail. [1 Mark]
(d) Find the value of $\tan\phi \times \cot\phi$. [1 Mark]
A triangular sail for a small sailing boat is shaped like a right triangle, with sides $5$ m, $12$ m, and $13$ m as shown, and the angle at one corner marked $\phi$.
(a) Find $\sin\phi$ and $\cos\phi$. [1 Mark]
(b) Find $\sec\phi$ and $\text{cosec}\,\phi$. [1 Mark]
(c) Verify that $1+\tan^2\phi=\sec^2\phi$ for this sail. [1 Mark]
(d) Find the value of $\tan\phi \times \cot\phi$. [1 Mark]
Step-by-Step Solution
Key Concept: Case study on trigonometry ratios and identities.
(a) Find $\sin\phi$ and $\cos\phi$. [1 Mark]
$\sin\phi=\dfrac{5}{13}$; $\cos\phi=\dfrac{12}{13}$. [1.0 Mark]
(b) Find $\sec\phi$ and $\text{cosec}\,\phi$. [1 Mark]
$\sec\phi=\dfrac{13}{12}$; $\text{cosec}\,\phi=\dfrac{13}{5}$. [1.0 Mark]
(c) Verify that $1+\tan^2\phi=\sec^2\phi$ for this sail. [1 Mark]
$\tan\phi=\dfrac{5}{12}$; $1+\tan^2\phi=1+\dfrac{25}{144}=\dfrac{169}{144}$; and $\sec^2\phi=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}$. Both sides match, verified. [1.0 Mark]
(d) Find the value of $\tan\phi \times \cot\phi$. [1 Mark]
$\tan\phi\times\cot\phi=1$ (since $\cot\phi$ is the reciprocal of $\tan\phi$, for any value of $\phi$ where both are defined). [1.0 Mark]
$\sin\phi=\dfrac{5}{13}$; $\cos\phi=\dfrac{12}{13}$. [1.0 Mark]
(b) Find $\sec\phi$ and $\text{cosec}\,\phi$. [1 Mark]
$\sec\phi=\dfrac{13}{12}$; $\text{cosec}\,\phi=\dfrac{13}{5}$. [1.0 Mark]
(c) Verify that $1+\tan^2\phi=\sec^2\phi$ for this sail. [1 Mark]
$\tan\phi=\dfrac{5}{12}$; $1+\tan^2\phi=1+\dfrac{25}{144}=\dfrac{169}{144}$; and $\sec^2\phi=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}$. Both sides match, verified. [1.0 Mark]
(d) Find the value of $\tan\phi \times \cot\phi$. [1 Mark]
$\tan\phi\times\cot\phi=1$ (since $\cot\phi$ is the reciprocal of $\tan\phi$, for any value of $\phi$ where both are defined). [1.0 Mark]
Correct Answer: $\sin\phi=\dfrac{5}{13}$; $\cos\phi=\dfrac{12}{13}$. [1.0 Mark] | $\sec\phi=\dfrac{13}{12}$; $\text{cosec}\,\phi=\dfrac{13}{5}$. [1.0 Mark] | $\tan\phi=\dfrac{5}{12}$; $1+\tan^2\phi=1+\dfrac{25}{144}=\dfrac{169}{144}$; and $\sec^2\phi=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}$. Both sides match, verified. [1.0 Mark] | $\tan\phi\times\cot\phi=1$ (since $\cot\phi$ is the reciprocal of $\tan\phi$, for any value of $\phi$ where both are defined). [1.0 Mark]
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