Advertisements
Advertisements
प्रश्न
Use the given figure to find :
(i) sin xo
(ii) cos yo
(iii) 3 tan xo - 2 sin yo + 4 cos yo.
उत्तर
Consider the given figure : 
Since the triangle is a right-angled triangle, so using Pythagorean Theorem
AD2 = 82 + 62
AD2 = 64 + 36 = 100
AD = 10
Also
BC2 = AC2 – AB2
BC2 = 172 – 82 = 225
BC = 15
(i) sin x° = `"perpendicular"/"hypotenuse" = (8)/(17)`
(ii) cos y° = `"base"/"hypotenuse" = (6)/(10) =(3)/(5)`
(iii) sin y° = `"perpendicular"/"hypotenuse" = "AB"/"AD" = (8)/(10) = (4)/(5)`
cos y° = `"base"/"hypotenuse" = (6)/(10) = (3)/(5)`
tan x° = `"perpendicular"/"base" = "AB"/"BC" = (8)/(15)`
Therefore
3tan x° – 2sin y° + 4 cos y°
= `3 (8/15) – 2 (4/5) + 4 (3/5)`
= `(8)/(5) – (8)/(5) + (12)/(5)`
= `2(2)/(5)`
APPEARS IN
संबंधित प्रश्न
In ΔPQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R.
If ∠A and ∠P are acute angles such that tan A = tan P, then show that ∠A = ∠P.
If Sin (A + B) = 1 and cos (A – B) = 1, 0° < A + B ≤ 90° A ≥ B. Find A & B
Show that:
(ii) `(cos30^0+sin 60^0)/(1+sin30^0+cos60^0)=cos 30^0`
If A = 300 , verify that:
(ii) cos 2A = `(1- tan^2A)/(1+tan^2A)`
From the following figure, find:
(i) y
(ii) sin x°
(iii) (sec x° - tan x°) (sec x° + tan x°)
In the given figure, triangle ABC is right-angled at B. D is the foot of the perpendicular from B to AC. Given that BC = 3 cm and AB = 4 cm.
find :
- tan ∠DBC
- sin ∠DBA
If sin A = cos A, find the value of 2 tan2A - 2 sec2 A + 5.
If cosec θ = `(29)/(20)`, find the value of: `("sec" θ)/("tan" θ - "cosec" θ)`
From the given figure, find the values of cos C
