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 sin 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
sin x
= `"AD"/"AB"`
= `(8)/(10)`
= `(4)/(5)`.
APPEARS IN
संबंधित प्रश्न
If 3 cot A = 4, Check whether `((1-tan^2 A)/(1+tan^2 A)) = cos^2 "A" - sin^2 "A"` or not.
If cos θ=0.6 show that (5sin θ -3tan θ) = 0
If 3 cot `theta = 2, `show that `((4 sin theta - 4 cos theta))/((2 sin theta + 6 cos theta ))=1/3`
Use the given figure to find :
(i) sin xo
(ii) cos yo
(iii) 3 tan xo - 2 sin yo + 4 cos yo.
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 ΔABC, ∠A = 90°. If AB = 5 units and AC = 12 units, find: tan B.
If sin θ = `(8)/(17)`, find the other five trigonometric ratios.
If tan = 0.75, find the other trigonometric ratios for A.
From the given figure, find all the trigonometric ratios of angle B
A boy standing at a point O finds his kite flying at a point P with distance OP = 25 m. It is at a height of 5 m from the ground. When the thread is extended by 10 m from P, it reaches a point Q. What will be the height QN of the kite from the ground? (use trigonometric ratios)
