Advertisements
Advertisements
प्रश्न
In a right-angled triangle, it is given that A is an acute angle and tan A = `(5) /(12)`.
find the value of :
(i) cos A
(ii) sin A
(iii) ` (cosA+sinA)/(cosA– sin A)`
उत्तर
Consider the diagram below :
tan A = `(5)/(12)`
i.e.`"perpendicular"/"base" = (5)/(12) ⇒ "BC"/"AB" = (5)/(12)`
Therefore if length of AB = 12x, length of BC = 5x
Since
AB2 + BC2 = AC2 ...[ Using Pythagoras Theorem ]
(12x)2 + (5x)2 = AC2
AC2 = 144x2 + 25x2
AC2 = 169x2
∴ AC = 13x ...( hypotenuse)
(i) cos A = `"base"/"hypotenuse" = "AB"/"AC" = (12x)/(13x) = 12/13`
(ii) sin A = `"perpendicular"/"hypotenuse" = (5x)/(13x) = 5/13`
(iii) `(cos "A" + sin "A")/(cos "A" – sin "A")`
= `(12/13+5/13)/(12/13 – 5/13)`
= `(17/13)/(17/7)`
= `17/7`
= `2(3)/(7)`
APPEARS IN
संबंधित प्रश्न
If cot θ = 2 find all the values of all T-ratios of θ .
If cosec θ= 2 show that `(cot θ +sin θ /(1+cos θ )) =2`
If a right ΔABC , right-angled at B, if tan A=1 then verify that 2sin A . cos A = 1
Evaluate:
`(5 cos^2 60^circ + 4 sec^2 30^circ - tan^2 45^circ)/(sin^2 30^circ + cos^2 30^circ)`
If A = 450, verify that :
(i) sin 2A = 2 sin A cos A
Using the formula, tan 2A =`(2 tan A )/(1- tan^2 A)` find the value of tan 600, it being given that tan 300 = `1/sqrt(3)`.
In triangle ABC, AB = AC = 15 cm and BC = 18 cm. Find:
- cos B
- sin C
- tan2 B - sec2 B + 2
If 5 cos θ = 3, evaluate : `(co secθ – cot θ)/(co secθ + cot θ)`
In ΔABC, ∠B = 90°. If AB = 12units and BC = 5units, find: tan A
In ΔABC, ∠B = 90°. If AB = 12units and BC = 5units, find: cot C