WebAnswer (1 of 5): Applying cosine rule on \angle BDA, we get: c^2 = q^2 + p^2 - 2pq\cos{45^o} Applying sine rule on \triangle BDA, we get: \dfrac{c}{\sin{45^o ...
Given that sin(B)=1/5 and AD=BD , find sin(∠ADC).
WebApr 1, 2024 · TRY THIS Page No : 283 Find sin 3 0 ∘, cos 3 0 ∘, tan 3 0 ∘, cosec 3 0 ∘, sec 3 0 ∘ and cot 3 0 ∘ by using the ratio concepts. (AS 1 ) Sol. Consider an equilateral triangle ABC. Since each angle is 6 0 ∘ in an equilateral triangle, we have ∠ A = ∠ B = ∠ C = 6 0 ∘ and the sides of equilateral triangle is A B = BC = C A = 2 ... WebSo, from BDA and ADC. ∠BDA=∠ADC … [both the angles are equal to 90 ∘] ∠BAD=∠DCA … [from equation (iii)] Therefore, ∠BDA∼∠ADC. BD/AD=AD/DC=AB/AC. Because, corresponding sides of similar triangles are proportional. BD/AD=AD/DC. By cross multiplication we get, AD 2=BD×CD. pulmonary artery normal size
In triangle ABC, the median AD on BC is drawn. If angle ADB=45
Web(b) In the figure (2) given below, ∆ABC is right-angled at B. Given that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the value of sin 2 θ + tan 2 θ. (c) In the figure (3) given below, AD is perpendicular to BC, BD = 15 cm, sin B = 4/5 and tan C = 1. (i) Calculate the lengths of AD, AB, DC and AC. (ii) Show that tan 2 B – 1 ... WebFeb 17, 2024 · P Answered by Master. Yes this is the soln sorry for taking long time. Mathematics. Given: ad = bc and bcd = adc prove: de = ce. Step-by-step answer. P Answered by PhD 44. According to AD=BC So AC=BD. Step-by-step explanation: AD = BC and angle ADC = angle BCD. Web(ii) 9 sin 25 sin 35 sin 35 (9) sin 25 sin 35 9 sin 25 sin 25 (sin 35 sin 25) 9 sin 25 1M 9 sin 25 sin 35 sin 25 3.82 1A x x x x x x x x (b) In ABD, ∠ BDE = 25°+60 ° = 85 ° In BDE, ∠ BED = 180 ° − 50 ° − 85 ° = 45 ° Consider ACD, by the sine formula, 1M sin sin 3.8181 sin 85 cm sin 45 =5.3791 cm 5.38 cm x BE BED BDE BE Consider BEC. pulmonary artery model labeled