Advertisements
Advertisements
प्रश्न
In the given figure, AD is the median on BC from A. If AD = 8 cm and BC = 12 cm, find the value of `(1)/("sin"^2 x) - (1)/("tan"^2 x)`
उत्तर
Since AD is median on BC, we have
BD = DC = `(1)/(2) xx "BC" = (1)/(2) xx 12` = 6cm
ΔADB is a right-angled triangle.
∴ AB2
= AD2 + BD2
= 82 + 62
= 64 + 36
= 100
⇒ AB = 10cm
ΔADC is a right-angled triangle.
∴ AC2
= AD2 + DC2
= 82 + 62
= 64 + 36
= 100
⇒ AC = 10cm
`(1)/("sin"^2 x) - (1)/("tan"^2 x)`
= `(1)/(4/5)^2 - (1)/(4/3)^2`
= `(25)/(16) - (9)/(16)`
= `(16)/(16)`
= 1.
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.
In a ΔABC, right angled at A, if tan C = `sqrt3` , find the value of sin B cos C + cos B sin C.
If A = B = 60°, verify that sin (A − B) = sin A cos B − cos A sin B
If A, B, C are the interior angles of a ΔABC, show that `cos[(B+C)/2] = sin A/2`
If cosec θ= 2 show that `(cot θ +sin θ /(1+cos θ )) =2`
If sin ∝ = `1/2` prove that (3cos∝ - `4cos^2` ∝)=0
From the following figure, find the values of:
- sin A
- cos A
- cot A
- sec C
- cosec C
- tan C
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)`
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
From the given figure, find the values of tan C