Advertisements
Advertisements
Question
In triangle ABC, AB = AC = 15 cm and BC = 18 cm, find cos ∠ABC.
Solution
Consider the figure below :
In the isosceles ΔABC, AB = AC = 15cm and BC =18cm the perpendicular drawn from angle A to the side BC divides the side BC into two equal parts BD = DC = 9 cm
cos ∠ABC = `"base"/"hypotenuse" = "BD"/"AB" = (9)/(15) = (3)/(5)`
APPEARS IN
RELATED QUESTIONS
If `sin A = 9/41` compute cos 𝐴 𝑎𝑛𝑑 tan 𝐴
If θ = 30° verify `tan 2 theta = (2 tan theta)/(1 - tan^2 theta)`
If sin θ = `3/4` show that `sqrt((cosec^2theta - cot^2theta)/(sec^2theta-1)) =sqrt(7)/3`
In a ΔABC , ∠B = 90° , AB = 12 cm and BC = 5 cm Find
(i) cos A (ii) cosec A (iii) cos C (iv) cosec C
Show that:
(i)` (1-sin 60^0)/(cos 60^0)=(tan60^0-1)/(tan60^0+1)`
Show that:
(ii) `(cos30^0+sin 60^0)/(1+sin30^0+cos60^0)=cos 30^0`
Verify each of the following:
(iii) `2 sin 30^0 cos 30^0`
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.
cos 40° = sin ______°
If sin A = `(7)/(25)`, find the value of : `"cos A" + (1)/"cot A"`