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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 $\tan P$.
Solution :
Given:
In $\triangle PQR$, $\angle Q = 90^\circ$.
The length of side $PQ = 5$ cm.
The sum of the lengths of the hypotenuse and the other side is $PR + QR = 25$ cm.
To Find:
The values of $\sin P$, $\cos P$, and $\tan P$.
Step 1: Expressing sides in terms of a single variable
Let $QR = x$ cm. Since $PR + QR = 25$ cm, we can express $PR$ as:
$PR = 25 - x$
Step 2: Applying the Pythagorean Theorem
In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides [Pythagorean Theorem: $Hypotenuse^2 = Base^2 + Perpendicular^2$].
$PR^2 = PQ^2 + QR^2$
Substitute the known values and expressions:
$(25 - x)^2 = 5^2 + x^2$
Step 3: Solving for $x$
Expand the left side using the identity $(a - b)^2 = a^2 - 2ab + b^2$:
$625 - 50x + x^2 = 25 + x^2$
Subtract $x^2$ from both sides:
$625 - 50x = 25$
Rearrange to solve for $x$:
$625 - 25 = 50x$
$600 = 50x$
$x = \frac{600}{50} = 12$
Thus, $QR = 12$ cm.
Now, find $PR$: $PR = 25 - 12 = 13$ cm.
Step 4: Calculating Trigonometric Ratios
For $\angle P$, the side opposite is $QR = 12$ cm, the side adjacent is $PQ = 5$ cm, and the hypotenuse is $PR = 13$ cm.
Using the definitions of trigonometric ratios:
$\sin P = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{12}{13}$
$\cos P = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{5}{13}$
$\tan P = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{QR}{PQ} = \frac{12}{5}$
Final Answer:
$\sin P = \frac{12}{13}, \cos P = \frac{5}{13}, \tan P = \frac{12}{5}$
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) $\sin C, \cos C$
- 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(ii): State whether the following are true or false. Justify your answer. (ii) $\sec A = \frac{12}{5}$ for some value of angle $A$.
- 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 $\theta$.
- 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) $\cos A \cos C – \sin A \sin C$
CBSE Solutions for Class 10 Mathematics Introduction to Trigonometry
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