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Q11(ii):
State whether the following are true or false. Justify your answer. (ii) $\sec A = \frac{12}{5}$ for some value of angle $A$.
Solution :
Given: A trigonometric ratio $\sec A = \frac{12}{5}$ for an angle $A$.
To Find: Determine whether the statement "$\sec A = \frac{12}{5}$ for some value of angle $A$" is true or false, and provide a justification.
Step 1: Definition of the Secant Ratio
In a right-angled triangle, for an acute angle $A$, the trigonometric ratio $\sec A$ is defined as the ratio of the length of the hypotenuse to the length of the side adjacent to angle $A$.
$\sec A = \frac{\text{Hypotenuse}}{\text{Adjacent side}}$
Step 2: Analyzing the given value
Given $\sec A = \frac{12}{5}$.
Comparing this to the definition, we can assume:
Hypotenuse $= 12k$
Adjacent side $= 5k$
where $k$ is a positive constant.
Step 3: Applying the Pythagorean Theorem
In any right-angled triangle, the hypotenuse is the longest side. Let the third side (opposite to angle $A$) be $BC$. According to the Pythagorean theorem:
$(\text{Hypotenuse})^2 = (\text{Adjacent side})^2 + (\text{Opposite side})^2$
$(12k)^2 = (5k)^2 + (BC)^2$
$144k^2 = 25k^2 + (BC)^2$
$(BC)^2 = 144k^2 - 25k^2$
$(BC)^2 = 119k^2$
$BC = \sqrt{119}k \approx 10.9k$
Step 4: Logical Justification
Since $12k > 5k$ and $12k > 10.9k$, the hypotenuse is indeed the longest side of the triangle. In a right-angled triangle, the ratio $\frac{\text{Hypotenuse}}{\text{Adjacent}}$ must always be greater than or equal to $1$ because the hypotenuse is always greater than or equal to any other side. Since $\frac{12}{5} = 2.4$, which is greater than $1$, this value is mathematically possible for an angle $A$.
Final Answer: The statement is True. Since the hypotenuse is the longest side in a right-angled triangle, the ratio $\sec A = \frac{\text{Hypotenuse}}{\text{Adjacent}}$ can take any value greater than or equal to $1$. As $\frac{12}{5} = 2.4 > 1$, it is a valid value for $\sec A$.
More Questions from Class 10 Mathematics Introduction to Trigonometry EXERCISE 8.1
- Q1(i): In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine : (i) $\sin A, \cos A$
- Q1(ii): In $\triangle ABC$, right-angled at $B$, $AB = 24$ cm, $BC = 7$ cm. Determine : (ii) $\sin C, \cos C$
- Q2: In Fig. 8.13, find $\tan P – \cot R$.
- Q3: If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.
- Q4: Given $15 \cot A = 8$, find $\sin A$ and $\sec A$.
- Q5: Given $\sec \theta = \frac{13}{12}$, calculate all other trigonometric ratios.
- Q6: If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.
- Q7(i): If $\cot \theta = \frac{7}{8}$, evaluate : (i) $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$
- Q7(ii): If $\cot \theta = \frac{7}{8}$, evaluate : (ii) $\cot^2 \theta$
- Q8: If $3 \cot A = 4$, check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A – \sin^2 A$ or not.
- Q9(i): In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of: (i) $\sin A \cos C + \cos A \sin C$
- Q9(ii): In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of: (ii) $\cos A \cos C – \sin A \sin C$
- Q10: In $\triangle PQR$, right-angled at $Q$, $PR + QR = 25$ cm and $PQ = 5$ cm. Determine the values of $\sin P, \cos P$ and $\tan P$.
- Q11(i): State whether the following are true or false. Justify your answer. (i) The value of $\tan A$ is always less than 1.
- Q11(iii): State whether the following are true or false. Justify your answer. (iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$.
- Q11(iv): State whether the following are true or false. Justify your answer. (iv) $\cot A$ is the product of cot and $A$.
- Q11(v): State whether the following are true or false. Justify your answer. (v) $\sin \theta = \frac{4}{3}$ for some angle $\theta$.
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
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