Question

if a and b are acute angles, such that sin a = 3/5 and tan b = 5/12. find cos (a+b)

Answer 1

Step-by-step explanation:

5 The simple interest on a sum of money for 2 years at 12% per annum is 1380. Fi

(i) the sum of money.

(ii) the compound interest on this sum for one year payable half-yearly at the same

rate.

Answer 2

Solution:-

Given

$\rm \: \sin A = \frac{3}{5}$

$\rm \: \tan \: B = \frac{5}{12}$

Now, take

$\rm \: \sin A = \frac{3}{5} = \frac{p}{h}$

$\rm \: p = 3 \: \: b \: = x \: \: and \: h = 5$

Using phythogoeros theorem

$\rm {h}^{2} = {p}^{2} + b {}^{2}$

$\rm \: (5) {}^{2} = (3) {}^{2} + x {}^{2}$

$\rm \: 25 = 9 + {x}^{2}$

$\rm \: 25 - 9 = x {}^{2}$

$\rm \: {x}^{2} = 16$

$\rm \: x = 4$

$\rm \: b = 4$

Now , we get

$\rm \: \cos(A) = \frac{b}{h} = \frac{4}{5}$

$\rm \: \sin A = \frac{3}{5} = \frac{p}{h}$

Now , take

$\rm \: \tan \: B = \frac{5}{12} = \frac{p}{b}$

$\rm \: p = 5 \: \: \: b \: = \: 12 \: \: and \: \: h = x$

Using phythogoeros theorem

$\rm {h}^{2} = {p}^{2} + b {}^{2}$

$\rm \: {x}^{2} = (5) {}^{2} + (12) {}^{2}$

$\rm \: {x}^{2} = 25 + 144$

$\rm \: {x}^{2} = 169$

$\rm \: x = 13$

$\rm \: h = 13$

So , we get

$\rm \sin B = \frac{p}{h} = \frac{5}{13}$

$\rm \: \cos B = \frac{b}{h } = \frac{12}{13}$

Now take

$\rm \: \cos(a + b)$

$\rm \: \cos( A+ B) = \cos(A) \cos(B) - \sin(A) \sin(B)$

So put the value we get

$\rm \: \frac{4}{5} \times \frac{12}{13} - \frac{3}{5} \times \frac{5}{13}$

$\rm \: \frac{48}{65} - \frac{15}{65}$

$\rm \: \frac{33}{65}$

Answer

$\to\rm \: \frac{33}{65}$