Advertisements
Advertisements
प्रश्न
In the given figure, AD is perpendicular to BC. Find: 15 tan y
उत्तर
ΔADB is a right-angled triangle.
∴ AB2
= AB2 + BD2
= 122 + 162
= 144 + 256
= 400
⇒ AB = 20cm
ΔADC is a right-angled triangle.
∴ AC2
= AD2 + DC2
= 122 + 92
= 144 + 81
= 225
⇒ AC = 15cm
15 tan y
= `15 xx "CD"/"AD"`
= `15 xx (9)/(12)`
= `(45)/(4)`.
APPEARS IN
संबंधित प्रश्न
In the given figure, ∠Q = 90°, PS is a median om QR from P, and RT divides PQ in the ratio 1 : 2. Find: `("tan" ∠"TSQ")/("tan"∠"PRQ")`
In a right-angled triangle ABC, ∠B = 90°, BD = 3, DC = 4, and AC = 13. A point D is inside the triangle such as ∠BDC = 90°.
Find the values of 2 tan ∠BAC - sin ∠BCD
If 24cosθ = 7 sinθ, find sinθ + cosθ.
If 8tanA = 15, find sinA - cosA.
If 5cosθ = 3, find the value of `(4cosθ - sinθ)/(2cosθ + sinθ)`
If cotθ = `(1)/sqrt(3)`, show that `(1 - cos^2θ)/(2 - sin^2θ) = (3)/(5)`
If cosecθ = `1(9)/(20)`, show that `(1 - sinθ + cosθ)/(1 + sinθ + cosθ) = (3)/(7)`
If b tanθ = a, find the values of `(cosθ + sinθ)/(cosθ - sinθ)`.
If 3 tanθ = 4, prove that `sqrt(secθ - "cosec"θ)/(sqrt(secθ - "cosec"θ)) = (1)/sqrt(7)`.
If tan θ = `"m"/"n"`, show that `"m sin θ - n cos θ"/"m sinθ + n cos θ" = ("m"^2 - "n"^2)/("m"^2 + "n"^2)`
