Class 10th Mathematics Chapter 8 Introduction To Trigonometry

Key Concepts: Application in Real Life

c) $36.87^\circ$
[Solution Description]
To find the angle, we can use the cosine function. The cosine of the $\angle$ $\theta$ between the ladder and the ground is given by the adjacent side (height) divided by the hypotenuse (length of the ladder):

$$\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{8}{10} = 0.8$$

Then, we need to find the angle whose cosine is 0.8. Using inverse cosine ($\cos^{-1}$), we get:

$$\theta = cos^{-1}(0.8) \approx 36.87^\circ$$

Hence, the angle is approximately $36.87^\circ$.

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Key Concepts: Basic Ratios

b) $\frac{\sqrt{2}}{2}$
[Solution Description]

The sine of an angle of $45^\circ$ in a right triangle is known as:
\[
\sin 45^\circ = \frac{\sqrt{2}}{2}
\]

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Key Concepts: Identity Transformation

a) $cos^2 A$
[Solution Description]
We know from the Pythagorean identity that $cos^2 A + sin^2 A = 1$. Rearranging this gives us:

\[
\cos^2 A = 1 – \sin^2 A
\]

Therefore, $1 – sin^2 A$ is equal to $cos^2 A$.

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Key Concepts: Angle Definitions

d) Undefined
[Solution Description] The value of $tan 90^\circ$ is undefined because it involves dividing by zero $(\cot 90^\circ = \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0})$.

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Key Concepts: Angle Identification

d) $90^\circ$
[Solution Description] Cosine of angle A equals 0 at 90°. Hence, if cos A = 0, then angle A is $90^\circ$.

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Key Concepts: Trigonometric Ratios

b) sin
[Solution Description]
Trigonometric ratios include sine, cosine, tangent, cosecant, secant, and cotangent.

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Key Concepts: Relationship between Trigonometric Ratios of Complementary Angles

b) $\sin A = \cos(90^\circ – A)$
[Solution Description] For complementary angles A and B such that $A + B = 90^\circ$: sin A = cos B, cos A = sin B, tan A = cot B, etc. Therefore, the statement “$\sin A = \cos(90^\circ – A)$” is correct.

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Key Concepts: Identity Application

b) $\frac{25}{16}$
[Solution Description]

Using the identity $1 + tan^2 A = sec^2 A$, we have:
$$
tan^2 A = ( \frac{3}{4})^2 = \frac{9}{16}
$$
Now substitute into the identity:
$$
1 + \frac{9}{16} = sec^2 A
$$
To solve for $sec^2 A$,
$$
sec^2 A = 1 + \frac{9}{16} = \frac{16}{16} + \frac{9}{16} = \frac{25}{16}
$$

Key Concepts: Hypotenuse Relation

b) 1
[Solution Description] Cosecant is the reciprocal of sine. Since $sin 90^\circ = 1$, $csc 90^\circ = \frac{1}{1} = 1$.

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Key Concepts: Pythagorean Identity Application

a) $25/24$
[Solution Description] Using the identity $1 + tan^2 A = sec^2 A$, we substitute $tan A = 7/24$
\[
\sec^2 A = 1 + \left( \frac{7}{24} \right)^2 = 1 + \frac{49}{576} = \frac{625}{576}
\]
Taking the square root, $sec A = \frac{25}{24}$.

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Key Concepts: Complex Expression

b) $sec A + tan A$
[Solution Description]

We begin with the expression:
$$
\frac{sec A}{sec A – tan A}
$$
Multiply numerator and denominator by the conjugate of the denominator $(sec A + tan A)$:
$$
\frac{sec A(sec A + tan A)}{(sec A – tan A)(sec A + tan A)} = \frac{sec^2 A + sec A tan A}{sec^2 A – tan^2 A}
$$
Since $sec^2 A – tan^2 A = (sec^2 A – 1) = 1$ by the identity $1 + tan^2 A = sec^2 A$,
$$
= sec A + tan A
$$
The expression simplifies to $sec A + tan A$.

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