Question

10th std. in triangle ABC,right-angled at b,if tan a=1/3 find the value of: (i) sin A cos C + cos A sin C (ii) cos A Cos - sin A sin C. ​

Answer 1

$\huge\boxed{\mathtt\red{Solution :-}}$

Assume the attachment :-

Given :-

• tan A = 1/3

To Find :-

• sin A cos C + cos A sin C = ?
• cos A Cos - sin A sin C = ?

Assumption :-

• Let the perpendicular be 1 unit.
• Then, base be 3 units

Solving :-

Now, by applying Pythagoras theorem :-

$\large\boxed{\mathtt\blue{Hypotenuse² = Base² + Perpendicular²}}$

✒ H² = (3)²+(1)²

✒ H² = 10

∴ H = √10 units.

Now, Finding all values :-

• sin A = $\large\mathtt{\frac{1}{√10}}$
• cos A = $\large\mathtt{\frac{3}{√10}}$
• sin C = $\large\mathtt{\frac{3}{√10}}$
• cos C = $\large\mathtt{\frac{1}{√10}}$

Now, putting the values in the Questions :-

(i). sin A cos C + cos A sin C

✒ $\large\mathtt{\frac{1}{√10}×\frac{1}{√10} + \frac{3}{√10}×\frac{3}{√10}}$

✒ $\large\mathtt{\frac{1}{10}+\frac{9}{10}}$

✒ 1

$\large\boxed{\mathtt\red{∴sin\: A \:cos \:C + cos\: A \:sin \:C = 1}}$

(ii). cos A cos C - sin A sin C

✒ $\large\mathtt{\frac{3}{√10}×\frac{1}{√10} - \frac{1}{√10}×\frac{3}{√10}}$

✒ $\large\mathtt{\frac{3}{10}-\frac{3}{10}}$

✒ 0

$\large\boxed{\mathtt\red{∴cos \:A \:cos C\: - sin\: A \:sin\: C=0}}$