Method I:

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

But, cos θ = \(\displaystyle \frac {35}{37}\) ... (Given)

∴ \(\displaystyle \frac {\text{AB}}{\text {AC}}\) = \(\displaystyle \frac {35}{37}\)

Let, AB = 35k and AC = 37k.

Now, using Pythagoras’ Theorem, we can find the length of BC.

  AB² + BC² = AC²

∴ (35k)² + BC² = (37k)²

∴ 1225k² + BC² = 1369k²

∴ BC² = 1369k² − 1225k²

∴ BC² = 144k²

∴ BC = \(\displaystyle \sqrt {144k^2}\)

∴ BC = 12k ... (i)

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

∴ sin θ = \(\displaystyle \frac {12k}{37k}\)

∴ sin θ = \(\displaystyle \frac {12}{37}\)

And tan θ = \(\displaystyle \frac {\text{BC}}{\text {AB}}\)

∴ tan θ = \(\displaystyle \frac {12k}{35k}\)

∴ tan θ = \(\displaystyle \frac {12}{35}\)

---

Method II:

Here, we will use a trigonometric identity.

We know that, sin² θ + cos² θ = 1

∴ sin² θ + \(\displaystyle \left(\frac {35}{37}\right)^2\) = 1

∴ sin² θ + \(\displaystyle \frac {1225}{1369}\) = 1

∴ sin² θ = 1 − \(\displaystyle \frac {1225}{1369}\)

∴ sin² θ = \(\displaystyle \frac {1369}{1369}\) − \(\displaystyle \frac {1225}{1369}\)

∴ sin² θ = \(\displaystyle \frac {144}{1369}\)

∴ sin θ = \(\displaystyle \sqrt {\frac {144}{1369}}\)

∴ sin θ = \(\displaystyle \frac {12}{37}\)

Now, tan θ = \(\displaystyle \frac {\text{sin } \theta}{\text{cos } \theta}\)

∴ tan θ = \(\displaystyle \frac {\frac {12}{37}}{\frac {35}{37}}\)

∴ tan θ = \(\displaystyle \frac {12}{37} \times \frac {37}{35}\)

∴ tan θ = \(\displaystyle \frac {12}{35}\)

Method I:

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

But, sin θ = \(\displaystyle \frac {11}{61}\) ... (Given)

∴ \(\displaystyle \frac {\text{BC}}{\text {AC}}\) = \(\displaystyle \frac {11}{61}\)

Let, BC = 11k and AC = 61k.

Now, using Pythagoras’ Theorem, we can find the length of AB.

  AB² + BC² = AC²

∴ AB² + (11k)² = (61k)²

∴ AB² + 121k² = 3721k²

∴ AB² = 3721k² − 121k²

∴ AB² = 3600k²

∴ AB = \(\displaystyle \sqrt {3600k^2}\)

∴ AB = 60k ... (i)

Now, cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

∴ cos θ = \(\displaystyle \frac {60k}{61k}\)

∴ cos θ = \(\displaystyle \frac {60}{61}\)

And tan θ = \(\displaystyle \frac {\text{BC}}{\text {AB}}\)

∴ tan θ = \(\displaystyle \frac {11k}{60k}\)

∴ tan θ = \(\displaystyle \frac {11}{60}\)

---

Method II:

Here, we will use a trigonometric identity.

We know that, sin² θ + cos² θ = 1

∴ \(\displaystyle \left(\frac {11}{61}\right)^2\) + cos² θ = 1

∴ \(\displaystyle \frac {121}{3721}\) + cos² θ = 1

∴ cos² θ = 1 − \(\displaystyle \frac {121}{3721}\)

∴ cos² θ = \(\displaystyle \frac {3721}{3721}\) − \(\displaystyle \frac {121}{3721}\)

∴ cos² θ = \(\displaystyle \frac {3600}{3721}\)

∴ cos θ = \(\displaystyle \sqrt {\frac {3600}{3721}}\)

∴ cos θ = \(\displaystyle \frac {60}{61}\)

Now, tan θ = \(\displaystyle \frac {\text{sin } \theta}{\text{cos } \theta}\)

∴ tan θ = \(\displaystyle \frac {\frac {11}{61}}{\frac {60}{61}}\)

∴ tan θ = \(\displaystyle \frac {11}{61} \times \frac {61}{60}\)

∴ tan θ = \(\displaystyle \frac {11}{60}\)

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, tan θ = \(\displaystyle \frac {\text{BC}}{\text {AB}}\)

But, tan θ = \(\displaystyle \frac {1}{1}\) ... (Given)

∴ \(\displaystyle \frac {\text{BC}}{\text {AB}}\) = \(\displaystyle \frac {1}{1}\)

∴ BC = AB = 1

Let, BC = 1k and AB = 1k.

Now, using Pythagoras’ Theorem, we can find the length of AC.

  AC² = AB² + BC²

∴ AC² = (1k)² + (1k)²

∴ AC² = 1k² + 1k²

∴ AC² = 2k²

∴ AC = \(\displaystyle \sqrt {2k^2}\)

∴ AC = \(\displaystyle \sqrt {2}k\) ... (i)

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

∴ sin θ = \(\displaystyle \frac {1k}{\sqrt {2}k}\)

∴ sin θ = \(\displaystyle \frac {1}{\sqrt {2}}\)

And cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

∴ cos θ = \(\displaystyle \frac {1k}{\sqrt {2}k}\)

∴ cos θ = \(\displaystyle \frac {1}{\sqrt {2}}\)

Method I:

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

But, sin θ = \(\displaystyle \frac {1}{2}\) ... (Given)

∴ \(\displaystyle \frac {\text{BC}}{\text {AC}}\) = \(\displaystyle \frac {1}{2}\)

Let, BC = 1k and AC = 2k.

Now, using Pythagoras’ Theorem, we can find the length of AB.

  AB² + BC² = AC²

∴ AB² + (1k)² = (2k)²

∴ AB² + 1k² = 4k²

∴ AB² = 4k² − 1k²

∴ AB² = 3k²

∴ AB = \(\displaystyle \sqrt {3k^2}\)

∴ AB = \(\sqrt {3}k\) ... (i)

Now, cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

∴ cos θ = \(\displaystyle \frac {\sqrt {3}k}{2k}\)

∴ cos θ = \(\displaystyle \frac {\sqrt {3}}{2}\)

And tan θ = \(\displaystyle \frac {\text{BC}}{\text {AB}}\)

∴ tan θ = \(\displaystyle \frac {1k}{\sqrt {3}k}\)

∴ tan θ = \(\displaystyle \frac {1}{\sqrt {3}}\)

---

Method II:

Here, we will use a trigonometric identity.

We know that, sin² θ + cos² θ = 1

∴ \(\displaystyle \left(\frac {1}{2}\right)^2\) + cos² θ = 1

∴ \(\displaystyle \frac {1}{4}\) + cos² θ = 1

∴ cos² θ = 1 − \(\displaystyle \frac {1}{4}\)

∴ cos² θ = \(\displaystyle \frac {4}{4}\) − \(\displaystyle \frac {1}{4}\)

∴ cos² θ = \(\displaystyle \frac {3}{4}\)

∴ cos θ = \(\displaystyle \sqrt {\frac {3}{4}}\)

∴ cos θ = \(\displaystyle \frac {\sqrt {3}}{2}\)

Now, tan θ = \(\displaystyle \frac {\text{sin } \theta}{\text{cos } \theta}\)

∴ tan θ = \(\displaystyle \frac {\frac {1}{2}}{\frac {\sqrt {3}}{2}}\)

∴ tan θ = \(\displaystyle \frac {1}{2} \times \frac {2}{\sqrt {3}}\)

∴ tan θ = \(\displaystyle \frac {1}{\sqrt {3}}\)

Method I:

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

But, cos θ = \(\displaystyle \frac {1}{\sqrt {3}}\) ... (Given)

∴ \(\displaystyle \frac {\text{AB}}{\text {AC}}\) = \(\displaystyle \frac {1}{\sqrt {3}}\)

Let, AB = 1k and AC = \(\sqrt {3}\)k.

Now, using Pythagoras’ Theorem, we can find the length of BC.

  AB² + BC² = AC²

∴ (1k)² + BC² = (\(\sqrt {3}\)k)²

∴ 1k² + BC² = 3k²

∴ BC² = 3k² − 1k²

∴ BC² = 2k²

∴ BC = \(\displaystyle \sqrt {2k^2}\)

∴ BC = \(\sqrt {2}\)k ... (i)

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

∴ sin θ = \(\displaystyle \frac {\sqrt {2}k}{\sqrt {3}k}\)

∴ sin θ = \(\displaystyle \frac {\sqrt {2}}{\sqrt {3}}\)

And tan θ = \(\displaystyle \frac {\text{BC}}{\text {AB}}\)

∴ tan θ = \(\displaystyle \frac {\sqrt {2}k}{1k}\)

∴ tan θ = \(\sqrt {2}\)

---

Method II:

Here, we will use a trigonometric identity.

We know that, sin² θ + cos² θ = 1

∴ sin² θ + \(\displaystyle \left(\frac {1}{\sqrt {3}}\right)^2\) = 1

∴ sin² θ + \(\displaystyle \frac {1}{3}\) = 1

∴ sin² θ = 1 − \(\displaystyle \frac {1}{3}\)

∴ sin² θ = \(\displaystyle \frac {3}{3}\) − \(\displaystyle \frac {1}{3}\)

∴ sin² θ = \(\displaystyle \frac {2}{3}\)

∴ sin θ = \(\displaystyle {\frac {\sqrt 2}{\sqrt 3}}\)

Now, tan θ = \(\displaystyle \frac {\text{sin } \theta}{\text{cos } \theta}\)

∴ tan θ = \(\displaystyle \frac {\frac {\sqrt {2}}{\sqrt {3}}}{\frac {1}{\sqrt {3}}}\)

∴ tan θ = \(\displaystyle \frac {\sqrt {2}}{\sqrt {3}} \times \frac {\sqrt {3}}{1}\)

∴ tan θ = \(\displaystyle \sqrt {2}\)

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, tan θ = \(\displaystyle \frac {\text{BC}}{\text {AB}}\)

But, tan θ = \(\displaystyle \frac {21}{20}\) ... (Given)

∴ \(\displaystyle \frac {\text{BC}}{\text {AB}}\) = \(\displaystyle \frac {21}{20}\)

Let, BC = 21k and AB = 20k.

Now, using Pythagoras’ Theorem, we can find the length of AC.

  AC² = AB² + BC²

∴ AC² = (20k)² + (21k)²

∴ AC² = 400k² + 441k²

∴ AC² = 841k²

∴ AC = \(\displaystyle \sqrt {841k^2}\)

∴ AC = \(\displaystyle 29k\) ... (i)

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

∴ sin θ = \(\displaystyle \frac {21k}{29k}\)

∴ sin θ = \(\displaystyle \frac {21}{29}\)

And cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

∴ cos θ = \(\displaystyle \frac {20k}{29k}\)

∴ cos θ = \(\displaystyle \frac {20}{29}\)

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, tan θ = \(\displaystyle \frac {\text{BC}}{\text {AB}}\)

But, tan θ = \(\displaystyle \frac {8}{15}\) ... (Given)

∴ \(\displaystyle \frac {\text{BC}}{\text {AB}}\) = \(\displaystyle \frac {8}{15}\)

Let, BC = 8k and AB = 15k.

Now, using Pythagoras’ Theorem, we can find the length of AC.

  AC² = AB² + BC²

∴ AC² = (15k)² + (8k)²

∴ AC² = 225k² + 64k²

∴ AC² = 289k²

∴ AC = \(\displaystyle \sqrt {289k^2}\)

∴ AC = \(\displaystyle 17k\) ... (i)

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

∴ sin θ = \(\displaystyle \frac {8k}{17k}\)

∴ sin θ = \(\displaystyle \frac {8}{17}\)

And cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

∴ cos θ = \(\displaystyle \frac {15k}{17k}\)

∴ cos θ = \(\displaystyle \frac {15}{17}\)

Method I:

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

But, sin θ = \(\displaystyle \frac {3}{5}\) ... (Given)

∴ \(\displaystyle \frac {\text{BC}}{\text {AC}}\) = \(\displaystyle \frac {3}{5}\)

Let, BC = 3k and AC = 5k.

Now, using Pythagoras’ Theorem, we can find the length of AB.

  AB² + BC² = AC²

∴ AB² + (3k)² = (5k)²

∴ AB² + 9k² = 25k²

∴ AB² = 25k² − 9k²

∴ AB² = 16k²

∴ AB = \(\displaystyle \sqrt {16k^2}\)

∴ AB = 4k ... (i)

Now, cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

∴ cos θ = \(\displaystyle \frac {4k}{5k}\)

∴ cos θ = \(\displaystyle \frac {4}{5}\)

And tan θ = \(\displaystyle \frac {\text{BC}}{\text {AB}}\)

∴ tan θ = \(\displaystyle \frac {3k}{4k}\)

∴ tan θ = \(\displaystyle \frac {3}{4}\)

---

Method II:

Here, we will use a trigonometric identity.

We know that, sin² θ + cos² θ = 1

∴ \(\displaystyle \left(\frac {3}{5}\right)^2\) + cos² θ = 1

∴ \(\displaystyle \frac {9}{25}\) + cos² θ = 1

∴ cos² θ = 1 − \(\displaystyle \frac {9}{25}\)

∴ cos² θ = \(\displaystyle \frac {25}{25}\) − \(\displaystyle \frac {9}{25}\)

∴ cos² θ = \(\displaystyle \frac {16}{25}\)

∴ cos θ = \(\displaystyle \sqrt {\frac {16}{25}}\)

∴ cos θ = \(\displaystyle \frac {4}{5}\)

Now, tan θ = \(\displaystyle \frac {\text{sin } \theta}{\text{cos } \theta}\)

∴ tan θ = \(\displaystyle \frac {\frac {3}{5}}{\frac {4}{5}}\)

∴ tan θ = \(\displaystyle \frac {3}{5} \times \frac {5}{4}\)

∴ tan θ = \(\displaystyle \frac {3}{4}\)

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, tan θ = \(\displaystyle \frac {\text{BC}}{\text {AB}}\)

But, tan θ = \(\displaystyle \frac {1}{2\sqrt 2}\) ... (Given)

∴ \(\displaystyle \frac {\text{BC}}{\text {AB}}\) = \(\displaystyle \frac {1}{2\sqrt 2}\)

Let, BC = 1k and AB = 2\(\sqrt 2\)k.

Now, using Pythagoras’ Theorem, we can find the length of AC.

  AC² = AB² + BC²

∴ AC² = (1k)² + (2\(\sqrt 2\)k)²

∴ AC² = 1k² + 8k²

∴ AC² = 9k²

∴ AC = \(\displaystyle \sqrt {9k^2}\)

∴ AC = \(\displaystyle 3k\) ... (i)

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

∴ sin θ = \(\displaystyle \frac {1k}{3k}\)

∴ sin θ = \(\displaystyle \frac {1}{3}\)

And cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

∴ cos θ = \(\displaystyle \frac {2\sqrt 2k}{3k}\)

∴ cos θ = \(\displaystyle \frac {2\sqrt 2}{3}\)

\(\displaystyle 5 \text {sin} 30 ^{\circ} + 3 \text {tan} 45 ^{\circ}\)

\(\displaystyle = 5 \times \frac {1}{2} + 3 \times 1\)

\(\displaystyle = \frac {5}{2} + 3\)

\(\displaystyle = \frac {5}{2}\) + \(\displaystyle \frac {3}{1}\)

\(\displaystyle = \frac {5}{2}\) + \(\displaystyle \frac {6}{2}\)

\(\displaystyle = \frac {11}{2}\)

∴ 5sin 30° + 3tan 45° = \(\displaystyle \frac {11}{2}\)

\(\displaystyle \frac {4}{5} \text{tan}^2 60^{\circ} + 3\text{ sin}^2 60^{\circ} \)

\(\displaystyle = \frac {4}{5} \times (\sqrt 3)^2 + 3 \times \left(\frac {\sqrt 3}{2}\right)^2\)

\(\displaystyle = \frac {4}{5} \times 3 + 3 \times \frac {3}{4}\)

\(\displaystyle = \frac {12}{5} + \frac {9}{4}\)

\(\displaystyle = \frac {48}{20} + \frac {45}{20}\)

\(\displaystyle = \frac {93}{20}\)

∴ \(\displaystyle \frac {4}{5} \text{tan}^2 60^{\circ} + 3\text{sin}^2 60^{\circ} \) = \(\displaystyle \frac {93}{20}\)

\(\displaystyle 2 \text{sin } 30 ^{\circ} + \text{cos } 0^{\circ} + 3\text{sin } 90^{\circ} \)

\(\displaystyle = 2 \times \frac {1}{2} + 1 + 3 \times 1\)

\(\displaystyle = 1 + 1 + 3\)

\(\displaystyle = 5\)

∴ \(\displaystyle 2 \text{sin } 30 ^{\circ} + \text{cos } 0^{\circ} + 3\text{sin } 90^{\circ} \) = \(\displaystyle 5\)

\(\displaystyle \frac {\text{tan } 60^{\circ}}{\text{sin } 60^{\circ} + \text{cos } 60^{\circ}}\)

\(\displaystyle = \frac {\sqrt 3}{\frac {\sqrt 3}{2} + \frac {1}{2}}\)

\(\displaystyle = \frac {\sqrt 3}{\frac {\sqrt 3 + 1}{2}}\)

\(\displaystyle = {\sqrt 3} \times {\frac {2}{\sqrt 3 + 1}}\)

\(\displaystyle = \frac {2\sqrt 3}{\sqrt 3 + 1} \)

∴ \(\displaystyle \frac {\text{tan } 60^{\circ}}{\text{sin } 60^{\circ} + \text{cos } 60^{\circ}}\) = \(\displaystyle \frac {2\sqrt 3}{\sqrt 3 + 1}\)

\(\displaystyle \text{cos}^2 45^{\circ} + \text{ sin}^2 30^{\circ} \)

\(\displaystyle = \left(\frac {1}{\sqrt 2}\right)^2 + \left(\frac {1}{2}\right)^2\)

\(\displaystyle = \frac {1}{2} + \frac {1}{4}\)

\(\displaystyle = \frac {2}{4} + \frac {1}{4}\)

\(\displaystyle = \frac {3}{4}\)

∴ \(\displaystyle \text{cos}^2 45^{\circ} + \text{ sin}^2 30^{\circ} \) = \(\displaystyle \frac {3}{4}\)

\(\displaystyle \text{cos } 60^{\circ} \times \text{cos } 30^{\circ} + \text{sin } 60^{\circ} \times \text{sin } 30^{\circ}\)

\(\displaystyle = \frac {1}{2} \times \frac {\sqrt 3}{2} + \frac {\sqrt 3}{2} \times \frac {1}{2}\)

\(\displaystyle = \frac {\sqrt 3}{4} + \frac {\sqrt 3}{4}\)

\(\displaystyle = \frac {2\sqrt 3}{4}\)

\(\displaystyle = \frac {\sqrt 3}{2}\)

∴ \(\displaystyle \text{cos } 60^{\circ} \times \text{cos } 30^{\circ} + \text{sin } 60^{\circ} \times \text{sin } 30^{\circ}\) = \(\displaystyle \frac {\sqrt 3}{2}\)

Method I:

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

But, sin θ = \(\displaystyle \frac {4}{5}\) ... (Given)

∴ \(\displaystyle \frac {\text{BC}}{\text {AC}}\) = \(\displaystyle \frac {4}{5}\)

Let, BC = 4k and AC = 5k.

Now, using Pythagoras’ Theorem, we can find the length of AB.

  AB² + BC² = AC²

∴ AB² + (4k)² = (5k)²

∴ AB² + 16k² = 25k²

∴ AB² = 25k² − 16k²

∴ AB² = 9k²

∴ AB = \(\displaystyle \sqrt {9k^2}\)

∴ AB = 3k ... (i)

Now, cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

∴ cos θ = \(\displaystyle \frac {3k}{5k}\)

∴ cos θ = \(\displaystyle \frac {3}{5}\)

---

Method II:

Here, we will use a trigonometric identity.

We know that, sin² θ + cos² θ = 1

∴ \(\displaystyle \left(\frac {4}{5}\right)^2\) + cos² θ = 1

∴ \(\displaystyle \frac {16}{25}\) + cos² θ = 1

∴ cos² θ = 1 − \(\displaystyle \frac {16}{25}\)

∴ cos² θ = \(\displaystyle \frac {25}{25}\) − \(\displaystyle \frac {16}{25}\)

∴ cos² θ = \(\displaystyle \frac {9}{25}\)

∴ cos θ = \(\displaystyle \sqrt {\frac {9}{25}}\)

∴ cos θ = \(\displaystyle \frac {3}{5}\)

Method I:

Here, we will use Pythagoras’ Theorem

In right angled \( \triangle \text {ABC} \),
\( \angle \text {B} \) = 90°, \( \angle \text {A} = \theta\).

Now, cos θ = \(\displaystyle \frac {\text{AB}}{\text {AC}}\)

But, cos θ = \(\displaystyle \frac {15}{17}\) ... (Given)

∴ \(\displaystyle \frac {\text{AB}}{\text {AC}}\) = \(\displaystyle \frac {15}{17}\)

Let, AB = 15k and AC = 17k.

Now, using Pythagoras’ Theorem, we can find the length of BC.

  AB² + BC² = AC²

∴ (15k)² + BC² = (17k)²

∴ 225k² + BC² = 289k²

∴ BC² = 289k² − 225k²

∴ BC² = 64k²

∴ BC = \(\displaystyle \sqrt {64k^2}\)

∴ BC = 8k ... (i)

Now, sin θ = \(\displaystyle \frac {\text{BC}}{\text {AC}}\)

∴ sin θ = \(\displaystyle \frac {8k}{17k}\)

∴ sin θ = \(\displaystyle \frac {8}{17}\)

---

Method II:

Here, we will use a trigonometric identity.

We know that, sin² θ + cos² θ = 1

∴ sin² θ + \(\displaystyle \left(\frac {15}{17}\right)^2\) = 1

∴ sin² θ + \(\displaystyle \frac {225}{289}\) = 1

∴ sin² θ = 1 − \(\displaystyle \frac {225}{289}\)

∴ sin² θ = \(\displaystyle \frac {289}{289}\) − \(\displaystyle \frac {225}{289}\)

∴ sin² θ = \(\displaystyle \frac {64}{289}\)

∴ sin θ = \(\displaystyle \sqrt {\frac {64}{289}}\)

∴ sin θ = \(\displaystyle \frac {8}{17}\)