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NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Ex 9.1 aplustopper.com/ncert-solution… #NCERTSolutionsforClass10MathsChapter9 #ApplicationOfTrigonometryClass10 #Class10MathsChapter9 #ApplicationsOfTrigonometryClass10Pdf
100/3. Using similar triangles. Detailed solution later as running late now... a/5=(10-a)/10. a=10/3.
あれだね。 90度の角をそれぞれ60度と30度にわけるその線をxとしたら 正三角形と30.30.120の二等辺三角形にする。そしたらなんとxが同じ長さだから10のラインがピッタリ半分になるね。んで、求めたい辺は正三角形の1辺だから10÷2の5だよね はいおわり
同じやつを下にも書き 1辺10cmの二等辺三角形をつくる。 角度に着目すると、30×2の60°になり、正三角形になっていることがわかる。 よって◻︎は10cmの半分の5
算数って事は√が使えない こういう時はもう一つ同じ三角形をくっつけるのが定石 正三角形を作ると2倍の□が10となるので□=5 高校生風に解くなら、□=10sin30°=5
Cara praktis mengingat bidang study Trigonometri ( Sinus s/d Cosecant ) pada sudut istimewa mulai dari 0⁰ s/d 360⁰ tanpa harus di hapal
Calculus 1 drill. 1. 3x²-10x+7 2.3cos(3x) 3.1 4.2x/(x²+1) 5. y =4x -4 6. x² +5x +c 7.1/3 8.2e^(2x) 9. 3 10.3x²-3 =0 x =±1
Class 10th tak sabse jayda intresting chapter trigonometry, algebra or mensuration and geometry hi tha mera sabse tough mana jata tha but agr derivation achhe se practice kar liya or applie karna sikh gya to sabse mzaa inhi chapter ko banane me aata hai mera math fav tha btw
We offer assistance for trigonometry assignments, exams, and homework help with Derivatives of Sine and Cosine, Law of Tangents, Napoleon’s Theorem, Trigonometric Identities, Analytic Trigonometry, Differential Equations, Geometry, Vector Calculus WhatsApp +1 (985) 251-1522
since a triangles angles has to be 180 degrees and the triangle on the right has a 60 degree and 90 degree angle then the 3rd angle in that one is 30 degrees and then you add the 30 degree angle to the 10 degree angle and that makes 40 and then 40 + 90 is 130 so y has to be 50
We offer assistance for trigonometry assignments, exams, and homework help with Derivatives of Sine and Cosine, Law of Tangents, Napoleon’s Theorem, Trigonometric Identities, Inversions, Analytic Trigonometry, Differential Equations, Geometry, Vector Calculus, and Euler’s Formula
We offer assistance for trigonometry assignments, exams, and homework help with Derivatives of Sine and Cosine, Law of Tangents, Napoleon’s Theorem, Trigonometric Identities, Inversions, Analytic Trigonometry, Differential Equations, Geometry, Vector Calculus, and Euler’s Formula
We offer assistance for trigonometry assignments, exams, and homework help with Derivatives of Sine and Cosine, Law of Tangents, Napoleon’s Theorem, Trigonometric Identities, Inversions, Analytic Trigonometry, Differential Equations, Geometry, Vector Calculus, and Euler’s Formula
Let x= interior triangle missing angle. 60deg + x + 90deg = 180deg x + 150deg = 180 deg x =180deg - 150deg = 30deg Outer triangle angles: 10deg+x = 10deg + 30deg = 40deg, 90 deg, y 40 deg + 90 deg + y = 180deg 130deg + y = 180deg y=180deg - 130deg = 50deg
After introducing parts of a circle and some low-stakes quizzing, Year 10 got hands-on today, measuring circumference with string and diameter with a ruler. The final step? Dividing the two and discovering… π! 📏🧵Hands-on measurements, critical thinking and a clear conclusion✅
sin(120° - x) = √2 sin(x - 45°) cos(30° - x) = sin(x) - cos(x) √3/2 cos(x) + 1/2 sin(x) = sin(x) - cos(x) 2(√3/2 + 1) cos(x) = sin(x) 2 + √3 = tan(x) x = atan(2 + √3) = 75°
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