Если угол при вершине на 61, 5° больше угла при основании, то в равнобедренном треугольнике угол при основании равен ?

39,5°  

Объяснение:  

Пусть ∠1 = ∠3 - углы при основании, ∠2 - угол при вершине.  

Пусть  ∠1 = ∠3 = х°, тогда ∠2=х+61,5°.  

х+х+х+61,5=180;  3х=118,5;  х=39,5.  

∠1 = ∠3=39,5°

I love autumn holidays. They give a breather in their studies. I just wish they were so short. Autumn holidays last only a week, sometimes ten days. You do not have time to relax properly, as it is time to go to school. But quickly you can not get up early in the morning. And then you barely get out of bed in the morning. Usually I spend autumn vacation at home. The weather is already cold and rainy, so you do not go out on the street especially. At home, I read books, or I dig on the Internet. If the sun looks out, I go out into the yard. In the evenings, we play with checkers, backgammon and other board games. Sometimes, when the weather permits, parents go to the forest for mushrooms. And I, of course, are also taken with me. Therefore, with the word "autumn vacation", I first of all recall the wet forest and mushrooms. Even on the autumn holidays, we go to my grandmother. I stay with her for a few days. By my arrival, my grandmother usually takes tickets to the circus or to a children's play, and we go with her to watch the performance. So that we are not bored on vacation, our class teacher also organizes a cultural campaign, for example, in a museum or an exhibition. During the last autumn holidays we visited the dolphinarium. There were trained sea fur seals, white whales and dolphins. I still have a lot of impressions! So those autumn holidays I spent very well.

Сумма двух векторов: начало второго вектора совмещается с концом первого, сумма же векторов есть вектор, с началом, совпадающим с началом первого, и концом, совпадающим с концом 2-го. Тогда:а). СА=СВ+ВА=а+b.б). СО=СВ+ВО. Но ВО=(1/2)*BD, так как диагонали параллелограмма точкой пересечения делятся пополам. BD=BC+CD, BC=-CB как вектора, равные по модулю, но направленные в противоположные стороны.Значит BD=-a+b=b-a. (1/2)*BD=(b-a)/2.Тогда СО=a+(b-a)/2=(a+b)/2.в). BD=BC+CD=-a+b=b-a.г). СМ=CD+DM=CD+CB/2 (так как точка М - середина вектора DA, а DA=CB как противоположные стороны параллелограмма).СМ=b+а/2.