What is the value of h in a right angled triangle with 31° and 15cm?

1 Answer

Answer :

Following the symbols in the image:Assuming 15cm corresponds to a (the line adjacent to the angle), then you need to use the cosine formulacos(ø) = a/hcos(31º) = 15cm/hh*cos(31º) = (15cm/h) * hh*cos(31º) = 15cm * 1h*cos(31º)/cos(31º) = 15cm/cos(31º)h*1 = 15cm/cos(31º)h = 16.398

Related questions

Description : If A(5,2), B(2,-2) and C(-2,t) are the vertices of a right angled triangle with ∠B = 900 , then the value of t is: (a) -1 (b) 1 (c) -2 (d) 2

Last Answer : (b) 1

Description : In right triangle ABC, right angled at C, if tan A = 1, then the value of 2 sin A cos A is (a) 0 (b) 1 (c) – 1 (d) 2

Last Answer : (b) 1

Description : Is it possible for a right angled triangle with sides 3 and 4 units long to have a hypotenuse 6 units in length?

Last Answer : answer:I'm not quite getting you. It isn't actually a triangle when the hypotenuse has these indentations, right? The hypotenuse isn't a straight line as you describe it. If the other sides are 3 ... and 5.00001, you don't have a straight line. Unless I'm misunderstanding what you're suggesting.

Description : The centre of gravity of a plane lamina will not be at its geometrical centre if it is a a.Square b.Equilateral triangle c.Circle d.Rectangle e.Right angled triangle

Last Answer : e. Right angled triangle

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

Description : ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC ad D. -Maths 9th

Last Answer : Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 ... congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Last Answer : NEED ANSWER

Description : triangle ABC is right angled at A. AL is drawn perpendicular to BC. Prove that /_ BAL = /_ ACB -Maths 9th

Last Answer : This answer was deleted by our moderators...

Description : ABC is a triangle right-angled at C. A line through the mid-point of hypotenuse AB and parallel to BC intersects AC at D. Show that -Maths 9th

Last Answer : Solution :-

Description : The base of a right-angled triangle measures 4 cm and its hypotenuse measures 5 cm. Find the area of the triangle. -Maths 9th

Last Answer : In right-angled triangle ABC AB2 + BC2 = AC2 (By Pythagoras Theorem) ⇒ AB2 + 42 = 52 ⇒ AB2 = 25 – 16 = 9 5 cm ⇒ AB = 3 cm ∴ Area of △ABC = 1/2 BC x AB = 1/2 x 4 x 3 = 6cm2

Description : If the length of hypotenuse of a right angled triangle is 5 cm and its area is 6 sq cm, then what are the lengths of the remaining sides? -Maths 9th

Last Answer : Let one of the remaining sides be x cm.Then, other side = \(\sqrt{5^2-x^2}\) cm∴ Area = \(rac{1}{2} imes{x} imes\sqrt{25-x^2}\) = 6⇒ \(x\sqrt{25-x^2}\) = 12 ⇒ x2(25 - x2) = 144⇒ 25x2 - x4 = 144 ⇒ x4 - 25x2 ... (x2 - 16) (x2 - 9) = 0 ⇒ x2 = 16 or x2 = 9 ⇒ x = 4 or 3∴ The two sides are 4 cm and 3 cm.

Description : In a right-angled triangle ABC, D is the foot of the perpendicular from B on the hypotenuse AC -Maths 9th

Last Answer : Area of ΔABC = \(rac{1}{2}\) x 3 x 4 cm2 = 6 cm2. Also, AC = \(\sqrt{3^2+4^2}\) = 5 cm.∴ Area of ΔABC = \(rac{1}{2}\) x BD x AC ⇒ 6 = \(rac{1}{2}\) BD x 5 ⇒ BD = \(rac{12}{5}\) cm.Now in ΔABD, AD = \(\ ... \(rac{1}{2}\)x AD x BD = \(rac{1}{2}\) x \(rac{9}{5}\) x \(rac{12}{5}\) = \(rac{54}{25}\) cm2.

Description : The hypotenuse of an isosceles right-angled triangle is q. If we describe equilateral triangles (outwards) on all its three sides, -Maths 9th

Last Answer : (b) \(rac{q^2}{4}\) (2√3 + 1).AC = q, ∠ABC = 90º ⇒ q = \(\sqrt{AB^2+BC^2}\)⇒ q = \(\sqrt{2x^2}\)⇒ q2 = 2x2 ⇒ \(x\) = \(rac{q}{\sqrt2}\)∴ Area of the re-entrant hexagon = Sum of areas of (ΔABC + ΔADC ... (rac{\sqrt3}{4}\)q2 + \(rac{\sqrt3}{8}\)q2 + \(rac{\sqrt3q^2}{8}\) = \(rac{q^2}{4}\) (2√3 + 1).

Description : A piece of paper is in the shape of a right-angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. -Maths 9th

Last Answer : (d) 14.365Given, ST || RQ∴ \(rac{ ext{Area of ΔSPT}}{ ext{Area of ΔRPQ}}\) = \(rac{ST^2}{RQ^2}\)Also, given ST = \(\bigg(1-rac{35}{100}\bigg)RQ\) = (0.65) RQ∴ \(rac{ST}{RQ}\) = 0.65 ⇒ \(\bigg(rac ... ΔRPQ}}\) = 0.4225 ⇒ \(rac{ ext{Area of ΔSPT}}{{34}}\) = 0.4225⇒ Area of ΔSPT = 0.4225 x 34 = 14.365

Description : Let a, b, c be the lengths of the sides of a right angled triangle, the hypotenuse having the length c, then a + b is -Maths 9th

Last Answer : answer:

Description : If A is the area of the right angled triangle and b is one of the sides containing the right angle, then what is the length of the -Maths 9th

Last Answer : answer:

Description : A city has a park shaped as a right angled triangle. The length of the longest side of this park is 80 m. The Mayor of the city -Maths 9th

Last Answer : answer:

Description : Let ABC be a right angled triangle with AC as its hypotenuse. Then, -Maths 9th

Last Answer : answer:

Description : Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. Also find the length of the hypotenuse -Maths 9th

Last Answer : Let the co-ordinates of any point on the x-axis be (x, 0). Then distance between (x, 0) and (– 4, 8) is 10 units.⇒ \(\sqrt{(x+4)^2+(0-8)^2}\) = 10 ⇒ x2 + 8x + 16 + 64 = 100 ⇒ x2 + 8x – 20 = 0 ⇒ (x + 10) (x – 2) = 0 ⇒ x = –10 or 2 ∴ The required points are (– 10, 0) and (2, 0).

Description : What is a right-angled triangle ?

Last Answer : : A triangle which has three right angles is called a right angled triangle. The sum of any two angles of a right-angled triangle is always greater than 900. The arms of a right-angled ... the center , and the perpendicular center of a right -angled triangle are all located inside the triangle.

Last Answer : : A triangle whose three angles are smaller than one right angle ( 90 0) is called a right-angled triangle.

Description : Is right angled triangle a right angle?

Last Answer : Yes and a right angle is 90 degrees

Description : Name the type of triangle formed. (a) Right angled (b) Equilateral (c) Isosceles (d) Scalene

Last Answer : (d) Scalene

Description : Find the area of a right angled triangle with sides of 90 degree unit and the functions described by L = cos y and M = sin x. a) 0 b) 45 c) 90 d) 180

Last Answer : d) 180

Description : The C.G. of a plane lamina will not be at its geometrical centre in the case of a  (A) Right angled triangle  (B) Equilateral triangle  (C) Square  (D) Circle

Last Answer : (A) Right angled triangle 

Description : The sides of a triangle are 15cm, 8cm and 17cm. Find its area -Maths 9th

Last Answer : Answer:- As we know herons formula = √[s(s-a)(s-b)(s-c)] Now here a=15cm, b=8cm, c=17cm And, s=20cm Then, area is=√[20*5*12*3] =√3600 =60cm² Therefore the Area is 60cm². Regards, Rishikesh

Description : What is the perimeter and height of a triangle that has a base length of 15cm with base angles of 28.67 degrees and 15.5 degrees?

Last Answer : Using the sine formulae of a/A=b/B=c/C and A/a=B/b=C/c intrigonometry the perimeter of the triangle is 31.08 cm with aheight of 2.76 cm both rounded to two decimal places.

Description : Areas of two similar triangles are 36 cm 2 and 100 cm 2 . If the length of a side of thelarger triangle is 20 cm, then the length of the corresponding side of the smaller triangle is: (a) 12cm (b) 13cm (c) 14cm (d) 15cm

Last Answer : (a) 12cm

Description : If the coefficient of discharge is 0.6, then the discharge over a right angled notch is (A) 0.417 H 5/2 (B) 1.417 H 5/2 (C) 4.171 H 5/2 (D) 7.141 H 5/2

Last Answer : Answer: Option B

Description : The discharge over a right angled notch is (where H = Height of liquid above the apex of notch) (A) (8/15) Cd. 2g. H (B) (8/15) Cd. 2g. H3/2 (C) (8/15) Cd. 2g. H² (D) (8/15) Cd. 2g. H5/2

Last Answer : Answer: Option D

Description : ABC is an acute angled triangle. CD is the altitude through C. If AB = 8 units, CD = 6 units, find the distance -Maths 9th

Last Answer : answer:

Description : In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The value of tan C is: a 12/7 b 24/7 c 20/7 d 7/24

Last Answer : b 24/7

Description : A right angled A ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. -Maths 9th

Last Answer : When rt. ∠ed △ABC is revolved about AB = 4 cm, it forms a right circular cone of radius 3 cm and height 4 cm . Slant height of the cone is 5 cm. Volume of cone = 1 / 3 πr2h 1 / 3 22 / 7 3 3 4 = ... Total surface area of the solid = πr2 + πrl = πr (r + l ) = 22 / 7 3 8 = 75.43 cm2

Description : On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Last Answer : According to question ∠BAC = ∠BDC.

Description : A right angled A ABC with sides 3 cm, 4 cm and 5 cm is revolved about the fixed side of 4 cm. -Maths 9th

Last Answer : When rt. ∠ed △ABC is revolved about AB = 4 cm, it forms a right circular cone of radius 3 cm and height 4 cm . Slant height of the cone is 5 cm. Volume of cone = 1 / 3 πr2h 1 / 3 22 / 7 3 3 4 = ... Total surface area of the solid = πr2 + πrl = πr (r + l ) = 22 / 7 3 8 = 75.43 cm2

Description : On a common hypotenuse AB, two right angled triangles, ACB and ADB are situated on opposite sides. -Maths 9th

Last Answer : According to question ∠BAC = ∠BDC.

Description : A ray incident normally on one face of a right-angled isosceles glass prism in air is deviated through `90^(@)`. What can you say about the refractive

Last Answer : A ray incident normally on one face of a right-angled isosceles glass prism in air is deviated ... you say about the refractive index of glass ?

Description : What is the small index of refractive of the material of a right angled prism with equal sides for which a ray of light entering one of the sides norm

Last Answer : What is the small index of refractive of the material of a right angled prism with equal sides ... of the sides normally will be totally reflected ?

Description : How many right-angled triangles are there in the given image? A. 20 B. 24 C. 28 D. 32

Last Answer : B

Description : A convex mirror and a concave mirror of radius 10cm each are placed 15cm apart facing each other. An object is placed midway between them. If the refl

Last Answer : A convex mirror and a concave mirror of radius 10cm each are placed 15cm apart facing each other ... convex mirror, the position of the final image is

Description : How many liters to fill a raised garden bed 1.2m x 1.2m x 15cm?

Last Answer : 1 m = 100 cm1000 cu cm = 1 litre→ volume = 1.2 m × 1.2 m × 15 cm= (1.2 × 100) cm × (1.2 × 100) cm × 15 cm= 216000 cu cm= 216000 ÷ 1000 litres= 216 litres

Description : Find the turns in an solenoid of inductance 23.4mH , current 2A and area 15cm. a) 900 b) 400 c) 498 d) 658

Last Answer : c) 498

Description : Find the magnetic flux density when a flux of 28 units is enclosed in an area of 15cm. a) 178.33 b) 186.67 c) 192.67 d) 124.33

Last Answer : b) 186.67

Description : Puddling is done in standing water of depth. b. 10-15cm c. 15-20cm d. None of the above

Last Answer : Ans - a. 5-10cm

Description : The diameter of fine tubes in Cochran boiler is of order of _______ a. 2cm b. 6cm c. 8cm d. 15cm

Last Answer : b. 6cm

Description : A cylindrical drum of 15cm radius and 70cm height is full of water. The water is emptied into a rectangular tank 225cm long and 70cm wide. Find the height of water level in the tank. a) 3.5cm b) 2.25cm c) 3.142cm d) 1.57cm

Last Answer : c) 3.142cm

Description : A threshing cylinder of 15cm radius is rotating at 700rpm if peripheral velocity (m/sec) will be a. 11 b.105 c. 210 d. 660

Last Answer : a. 11

Description : Generally weeds seeds are abundant at a depth of a). 1.0-1.5cm b). 5 cm c).10cm d).15cm

Last Answer : a). 1.0-1.5cm

Description : Surface germinating weed seeds are at depth a). 1cm b). 2.5cm c). 7.5cm d).15cm

Last Answer : a). 1cm