1. In the following table, a ratio is given in each column. Find the remaining two ratios in the column and complete the table:
[These are 9 separate problems. We will solve them one by one.]
| (i) | sin θ |
|---|---|
| cos θ | \(\displaystyle \frac {35}{37}\) |
| tan θ |
Solution:
We can solve this using two different methods. In the first example, I will show you both the methods.
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 = \(\displaystyle \frac {12}{35}\)
| (ii) | sin θ | \(\displaystyle \frac {11}{61}\) |
|---|---|---|
| cos θ | ||
| tan θ |
Solution:
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 = \(\displaystyle \frac {11}{60}\)
| (iii) | sin θ |
|---|---|
| cos θ | |
| tan θ | 1 |
Solution:
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 = \(\displaystyle \frac {1}{\sqrt {2}}\)
| (iv) | sin θ | \(\displaystyle \frac {1}{2}\) |
|---|---|---|
| cos θ | ||
| tan θ |
Solution:
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 = \(\displaystyle \frac {1}{\sqrt {3}}\)
| (v) | sin θ |
|---|---|
| cos θ | \(\displaystyle \frac {1}{\sqrt {3}}\) |
| tan θ |
Solution:
We can solve this using two different methods. In the first example, I will show you both the methods.
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 = \(\displaystyle \sqrt {2}\)
| (vi) | sin θ |
|---|---|
| cos θ | |
| tan θ | \(\displaystyle \frac {21}{20}\) |
Solution:
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 = \(\displaystyle \frac {20}{29}\)
| (vii) | sin θ |
|---|---|
| cos θ | |
| tan θ | \(\displaystyle \frac {8}{15}\) |
Solution:
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 = \(\displaystyle \frac {15}{17}\)
| (viii) | sin θ | \(\displaystyle \frac {3}{5}\) |
|---|---|---|
| cos θ | ||
| tan θ |
Solution:
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 = \(\displaystyle \frac {3}{4}\)
| (ix) | sin θ |
|---|---|
| cos θ | |
| tan θ | \(\displaystyle \frac {1}{2\sqrt 2}\) |
Solution:
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 = \(\displaystyle \frac {2\sqrt 2}{3}\)
2. Find the values of:
(i) \(\displaystyle 5 \text {sin} 30 ^{\circ} + 3 \text {tan} 45 ^{\circ}\)
Solution:
\(\displaystyle 5 \text {sin} 30 ^{\circ} + 3 \text {tan} 45 ^{\circ}\)
\(\displaystyle = 5 \times \frac {1}{2} + 3 \times 1\)
\(\displaystyle = \frac {5}{2} + 3\)
\(\displaystyle = \frac {5}{2}\) + \(\displaystyle \frac {3}{1}\)
\(\displaystyle = \frac {5}{2}\) + \(\displaystyle \frac {6}{2}\)
\(\displaystyle = \frac {11}{2}\)
∴ 5sin 30° + 3tan 45° = \(\displaystyle \frac {11}{2}\)
2. Find the values of:
(ii) \(\displaystyle \frac {4}{5} \text{tan}^2 60^{\circ} + 3\text{ sin}^2 60^{\circ} \)
Solution:
\(\displaystyle \frac {4}{5} \text{tan}^2 60^{\circ} + 3\text{ sin}^2 60^{\circ} \)
\(\displaystyle = \frac {4}{5} \times (\sqrt 3)^2 + 3 \times \left(\frac {\sqrt 3}{2}\right)^2\)
\(\displaystyle = \frac {4}{5} \times 3 + 3 \times \frac {3}{4}\)
\(\displaystyle = \frac {12}{5} + \frac {9}{4}\)
\(\displaystyle = \frac {48}{20} + \frac {45}{20}\)
\(\displaystyle = \frac {93}{20}\)
∴ \(\displaystyle \frac {4}{5} \text{tan}^2 60^{\circ} + 3\text{sin}^2 60^{\circ} \) = \(\displaystyle \frac {93}{20}\)
2. Find the values of:
(iii) \(\displaystyle 2 \text{ sin } 30 ^{\circ} + \text{ cos } 0^{\circ} + 3\text{ sin } 90^{\circ} \)
Solution:
\(\displaystyle 2 \text{sin } 30 ^{\circ} + \text{cos } 0^{\circ} + 3\text{sin } 90^{\circ} \)
\(\displaystyle = 2 \times \frac {1}{2} + 1 + 3 \times 1\)
\(\displaystyle = 1 + 1 + 3\)
\(\displaystyle = 5\)
∴ \(\displaystyle 2 \text{sin } 30 ^{\circ} + \text{cos } 0^{\circ} + 3\text{sin } 90^{\circ} \) = \(\displaystyle 5\)
2. Find the values of:
(iv) \(\displaystyle \frac {\text{tan } 60^{\circ}}{\text{sin } 60^{\circ} + \text{cos } 60^{\circ}} \)
Solution:
\(\displaystyle \frac {\text{tan } 60^{\circ}}{\text{sin } 60^{\circ} + \text{cos } 60^{\circ}}\)
\(\displaystyle = \frac {\sqrt 3}{\frac {\sqrt 3}{2} + \frac {1}{2}}\)
\(\displaystyle = \frac {\sqrt 3}{\frac {\sqrt 3 + 1}{2}}\)
\(\displaystyle = {\sqrt 3} \times {\frac {2}{\sqrt 3 + 1}}\)
\(\displaystyle = \frac {2\sqrt 3}{\sqrt 3 + 1} \)
∴ \(\displaystyle \frac {\text{tan } 60^{\circ}}{\text{sin } 60^{\circ} + \text{cos } 60^{\circ}}\) = \(\displaystyle \frac {2\sqrt 3}{\sqrt 3 + 1}\)
2. Find the values of:
(v) \(\displaystyle \text{cos}^2 45^{\circ} + \text{ sin}^2 30^{\circ} \)
Solution:
\(\displaystyle \text{cos}^2 45^{\circ} + \text{ sin}^2 30^{\circ} \)
\(\displaystyle = \left(\frac {1}{\sqrt 2}\right)^2 + \left(\frac {1}{2}\right)^2\)
\(\displaystyle = \frac {1}{2} + \frac {1}{4}\)
\(\displaystyle = \frac {2}{4} + \frac {1}{4}\)
\(\displaystyle = \frac {3}{4}\)
∴ \(\displaystyle \text{cos}^2 45^{\circ} + \text{ sin}^2 30^{\circ} \) = \(\displaystyle \frac {3}{4}\)
2. Find the values of:
(vi) \(\displaystyle \text{cos } 60^{\circ} \times \text{cos } 30^{\circ} + \text{sin } 60^{\circ} \times \text{sin } 30^{\circ}\)
Solution:
\(\displaystyle \text{cos } 60^{\circ} \times \text{cos } 30^{\circ} + \text{sin } 60^{\circ} \times \text{sin } 30^{\circ}\)
\(\displaystyle = \frac {1}{2} \times \frac {\sqrt 3}{2} + \frac {\sqrt 3}{2} \times \frac {1}{2}\)
\(\displaystyle = \frac {\sqrt 3}{4} + \frac {\sqrt 3}{4}\)
\(\displaystyle = \frac {2\sqrt 3}{4}\)
\(\displaystyle = \frac {\sqrt 3}{2}\)
∴ \(\displaystyle \text{cos } 60^{\circ} \times \text{cos } 30^{\circ} + \text{sin } 60^{\circ} \times \text{sin } 30^{\circ}\) = \(\displaystyle \frac {\sqrt 3}{2}\)
3. If \(\displaystyle \text{sin } \theta = \frac {4}{5} \text{, then find } \text{cos } \theta\)
Solution:
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 = \(\displaystyle \frac {3}{5}\)
4. If \(\displaystyle \text{cos } \theta = \frac {15}{17} \text{, then find } \text{sin } \theta\)
Solution:
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 = \(\displaystyle \frac {8}{17}\)
This page was last modified on
28 February 2026 at 19:44