Exercise 8.1: Introduction to Trigonometry — NCERT Solutions

In right triangle $ABC$, right-angled at $B$, the hypotenuse is $AC$.

Using Pythagoras theorem: $AC^2 = AB^2 + BC^2 = 24^2 + 7^2 = 576 + 49 = 625$ So $AC = 25$ cm.

For angle $A$:

• Side opposite to $A$ = $BC = 7$
• Side adjacent to $A$ = $AB = 24$
• Hypotenuse = $AC = 25$

Therefore: $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{7}{25}$ $\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{24}{25}$

Given $\tan A = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{3}{4}$.

Let the opposite side $= 3k$ and adjacent side $= 4k$ for some positive $k$.

By Pythagoras theorem, hypotenuse: $= \sqrt{(3k)^2 + (4k)^2} = \sqrt{9k^2 + 16k^2} = \sqrt{25k^2} = 5k$

Therefore: $\sin A = \frac{3k}{5k} = \frac{3}{5}$ $\cos A = \frac{4k}{5k} = \frac{4}{5}$

Given $15\cot A = 8$, so $\cot A = \dfrac{8}{15} = \dfrac{\text{adjacent}}{\text{opposite}}$.

Let adjacent $= 8k$ and opposite $= 15k$.

Hypotenuse $= \sqrt{(8k)^2 + (15k)^2} = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k$.

Therefore: $\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{15k}{17k} = \frac{15}{17}$ $\sec A = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{17k}{8k} = \frac{17}{8}$

Given $\sin \theta = \dfrac{5}{13} = \dfrac{\text{opposite}}{\text{hypotenuse}}$.

Let opposite $= 5k$, hypotenuse $= 13k$.

Adjacent $= \sqrt{(13k)^2 - (5k)^2} = \sqrt{169k^2 - 25k^2} = \sqrt{144k^2} = 12k$.

So $\cos\theta = \dfrac{12}{13}$.

Now substitute: $\frac{\cos\theta - \sin\theta}{\cos\theta + \sin\theta} = \frac{\frac{12}{13} - \frac{5}{13}}{\frac{12}{13} + \frac{5}{13}} = \frac{\frac{7}{13}}{\frac{17}{13}} = \frac{7}{17}$

Let $QR = x$ cm. Then $PR = (25 - x)$ cm and $PQ = 5$ cm.

By Pythagoras theorem ($PR$ is the hypotenuse): $PR^2 = PQ^2 + QR^2$ $(25 - x)^2 = 5^2 + x^2$ $625 - 50x + x^2 = 25 + x^2$ $625 - 50x = 25$ $50x = 600 \implies x = 12$

So $QR = 12$ cm and $PR = 25 - 12 = 13$ cm.

For angle $P$: opposite $= QR = 12$, adjacent $= PQ = 5$, hypotenuse $= PR = 13$. $\sin P = \frac{12}{13}, \quad \cos P = \frac{5}{13}, \quad \tan P = \frac{12}{5}$

Consider a right triangle $ABC$ right-angled at $C$, with $\angle A$ and $\angle B$ acute.

By definition: $\cos A = \frac{AC}{AB}, \qquad \cos B = \frac{BC}{AB}$

Given $\cos A = \cos B$: $\frac{AC}{AB} = \frac{BC}{AB}$

Since the denominators are equal: $AC = BC$

In a triangle, angles opposite equal sides are equal. Here $AC = BC$, so the angles opposite them are equal: $\angle B = \angle A$

Hence $\angle A = \angle B$.