Advertisements
Advertisements
Question
If sin (A – B) = `1/2` and cos (A + B) = `1/2, 0^0 ≤ (A + B) ≤ 90^0` and A > B, then find A and B.
Solution
Here, sin (A – B) = `1/2`
`⇒ sin (A – B) = 30^0 [∵ sin 30^0 = 1/2]`
`⇒ (A – B) = 30^0` …….(i)
`Also, cos (A + B) = 1/2`
`⇒ cos (A + B) = cos 60^0 [∵ cos 60^0 = 1/2]`
`⇒ A + B = 60^0` ….(ii)
Solving (i) and (ii), we get:
`A = 45^0 and B = 15^0`
APPEARS IN
RELATED QUESTIONS
In right angled triangle ΔABC at B, ∠A = ∠C. Find the values of Sin A cos C + Cos A Sin C
If sec θ = `5/4 ` show that `((sin θ - 2 cos θ))/(( tan θ - cot θ)) = 12/7`
If tan `theta = a/b`, show that `((a sin theta - b cos theta))/((a sin theta + bcos theta))= ((a^2-b^2))/(a^2+b^2)`
If a right ΔABC , right-angled at B, if tan A=1 then verify that 2sin A . cos A = 1
If A = 300 , verify that:
(ii) cos 2A = `(1- tan^2A)/(1+tan^2A)`
If 5 cot θ = 12, find the value of : Cosec θ+ sec θ
In the diagram, given below, triangle ABC is right-angled at B and BD is perpendicular to AC.
Find:
(i) cos ∠DBC
(ii) cot ∠DBA
In triangle ABC, AB = AC = 15 cm and BC = 18 cm. Find:
- cos B
- sin C
- tan2 B - sec2 B + 2
In ΔABC, ∠B = 90°. If AB = 12units and BC = 5units, find: sinA
In the given figure, PQR is a triangle, in which QS ⊥ PR, QS = 3 cm, PS = 4 cm and QR = 12 cm, find the value of: cot2P - cosec2P