Advertisements
Advertisements
Question
Use tables to find cosine of 9° 23’ + 15° 54’
Advertisements
Solution
cos (9° 23’ + 15° 54’) = cos 24° 77’
= cos 25° 17’
= cos (25° 12’ + 5’)
= 0.9048 − 0.0006
= 0.9042
APPEARS IN
RELATED QUESTIONS
Evaluate cosec 31° − sec 59°
if `tan theta = 1/sqrt2` find the value of `(cosec^2 theta - sec^2 theta)/(cosec^2 theta + cot^2 theta)`
For triangle ABC, show that: `sin (A + B)/2 = cos C/2`
Use tables to find sine of 62° 57'
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
If A + B = 90°, then \[\frac{\tan A \tan B + \tan A \cot B}{\sin A \sec B} - \frac{\sin^2 B}{\cos^2 A}\]
In the following figure the value of cos ϕ is

If ∆ABC is right angled at C, then the value of cos (A + B) is ______.
Prove that:
\[\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2\]
Express the following in term of angles between 0° and 45° :
sin 59° + tan 63°
