Question

A &B are both obtuse angles and sinA=5/13 and cosB= -4/5 then find sin(A +B)
1 point
-56/65
56/65
65/56
0​

Answer 1

Answer:

galat questions hai kuoki right angle triangle me ek angle 90° ho gaya aur baki angle 90° ka sum hoga to

Answer 2

Answer:

The value of  $sin(A+B)$  is  $\frac{-56}{65}$ .

Explanation:

• The sine of an angle is equal to the perpendicular divided by the hypotenuse of the right-angled triangle.
• The cosine of an angle is equal to the base divided by the hypotenuse of the right-angled triangle.
• The Sine addition formula is used to calculate the sine of the sum of the two angles.

The formula used to solve the problem is:

$sin(A+B)=sinA*cosB + cosA*sinB$

where, $A$ and $B$ are the angle values.

Step 1:

Given the value of  $sinA=\frac{5}{13}$.

We know that $cosA=\sqrt{1-sin^{2}A }$

So, $cosA=\sqrt{1-(\frac{5}{13})^{2} }$

$cosA=\sqrt{\frac{144}{169} }$

$cosA=\frac{12}{13}$

Given the value of  $cosB=\frac{-4}{5}$.

We know that $sinB=\sqrt{1-cos^{2}B }$

So, $sinB=-\sqrt{1-(\frac{-4}{5})^{2} }$

$sinB=-\sqrt{\frac{9}{25} }$

$sinB=-\frac{3}{5}$

Step 2:

Applying the above formula:

$sin(A+B)=(\frac{5}{13}) *(\frac{-4}{5}) + (\frac{12}{13}) *(\frac{-3}{5})$

$sin(A+B)=(\frac{-20}{65} )+(\frac{-36}{65} )$

$sin(A+B)=\frac{-56}{65}$

So, the value of $sin(A+B)$  is  $\frac{-56}{65}$ .